text:
To install click the Add extension button. That's it.

The source code for the WIKI 2 extension is being checked by specialists of the Mozilla Foundation, Google, and Apple. You could also do it yourself at any point in time.

4,5
Kelly Slayton
Congratulations on this excellent venture… what a great idea!
Alexander Grigorievskiy
I use WIKI 2 every day and almost forgot how the original Wikipedia looks like.
Live Statistics
English Articles
Improved in 24 Hours
What we do. Every page goes through several hundred of perfecting techniques; in live mode. Quite the same Wikipedia. Just better.
.
Leo
Newton
Brights
Milds

# Coordination complex

Cisplatin, PtCl2(NH3)2, is a coordination complex of platinum(II) with two chloride and two ammonia ligands. It is one of the most successful anticancer drugs.

A coordination complex consists of a central atom or ion, which is usually metallic and is called the coordination centre, and a surrounding array of bound molecules or ions, that are in turn known as ligands or complexing agents.[1][2][3] Many metal-containing compounds, especially those of transition metals (d block elements), are coordination complexes.[4]

## Nomenclature and terminology

Coordination complexes are so pervasive that their structures and reactions are described in many ways, sometimes confusingly. The atom within a ligand that is bonded to the central metal atom or ion is called the donor atom. In a typical complex, a metal ion is bonded to several donor atoms, which can be the same or different. A polydentate (multiple bonded) ligand is a molecule or ion that bonds to the central atom through several of the ligand's atoms; ligands with 2, 3, 4 or even 6 bonds to the central atom are common. These complexes are called chelate complexes; the formation of such complexes is called chelation, complexation, and coordination.

The central atom or ion, together with all ligands, comprise the coordination sphere.[5][6] The central atoms or ion and the donor atoms comprise the first coordination sphere.

Coordination refers to the "coordinate covalent bonds" (dipolar bonds) between the ligands and the central atom. Originally, a complex implied a reversible association of molecules, atoms, or ions through such weak chemical bonds. As applied to coordination chemistry, this meaning has evolved. Some metal complexes are formed virtually irreversibly and many are bound together by bonds that are quite strong.[7][8]

The number of donor atoms attached to the central atom or ion is called the coordination number. The most common coordination numbers are 2, 4, and especially 6. A hydrated ion is one kind of a complex ion (or simply a complex), a species formed between a central metal ion and one or more surrounding ligands, molecules or ions that contain at least one lone pair of electrons.

If all the ligands are monodentate, then the number of donor atoms equals the number of ligands. For example, the cobalt(II) hexahydrate ion or the hexaaquacobalt(II) ion [Co(H2O)6]2+ is a hydrated-complex ion that consists of six water molecules attached to a metal ion Co. The oxidation state and the coordination number reflect the number of bonds formed between the metal ion and the ligands in the complex ion. However, the coordination number of Pt(en)2+
2
is 4 (rather than 2) since it has two bidentate ligands, which contain four donor atoms in total.

Any donor atom will give a pair of electrons. There are some donor atoms or groups which can offer more than one pair of electrons. Such are called bidentate (offers two pairs of electrons) or polydentate (offers more than two pairs of electrons). In some cases an atom or a group offers a pair of electrons to two similar or different central metal atoms or acceptors—by division of the electron pair—into a three-center two-electron bond. These are called bridging ligands.

## History

Coordination complexes have been known since the beginning of modern chemistry. Early well-known coordination complexes include dyes such as Prussian blue. Their properties were first well understood in the late 1800s, following the 1869 work of Christian Wilhelm Blomstrand. Blomstrand developed what has come to be known as the complex ion chain theory. The theory claimed that the reason coordination complexes form is because in solution, ions would be bound via ammonia chains.[clarification needed] He compared this effect to the way that various carbohydrate chains form.

Following this theory, Danish scientist Sophus Mads Jørgensen made improvements to it. In his version of the theory, Jørgensen claimed that when a molecule dissociates in a solution there were two possible outcomes: the ions would bind via the ammonia chains Blomstrand had described or the ions would bind directly to the metal.

It was not until 1893 that the most widely accepted version of the theory today was published by Alfred Werner. Werner's work included two important changes to the Blomstrand theory. The first was that Werner described the two possibilities in terms of location in the coordination sphere. He claimed that if the ions were to form a chain, this would occur outside of the coordination sphere while the ions that bound directly to the metal would do so within the coordination sphere.[9] In one of his most important discoveries however Werner disproved the majority of the chain theory. Werner discovered the spatial arrangements of the ligands that were involved in the formation of the complex hexacoordinate cobalt. His theory allows one to understand the difference between a coordinated ligand and a charge balancing ion in a compound, for example the chloride ion in the cobaltammine chlorides and to explain many of the previously inexplicable isomers.

In 1911, Werner first resolved the coordination complex hexol into optical isomers, overthrowing the theory that only carbon compounds could possess chirality.[10]

## Structures

Structure of hexol

The ions or molecules surrounding the central atom are called ligands. Ligands are classified as L or X (or a combination thereof), depending on how many electrons they provide for the bond between ligand and central atom. L ligands provide two electrons from a lone electron pair, resulting in a coordinate covalent bond. X ligands provide one electron, with the central atom providing the other electron, thus forming a regular covalent bond. The ligands are said to be coordinated to the atom. For alkenes, the pi bonds can coordinate to metal atoms. An example is ethylene in the complex [PtCl3(C2H4)].

### Geometry

In coordination chemistry, a structure is first described by its coordination number, the number of ligands attached to the metal (more specifically, the number of donor atoms). Usually one can count the ligands attached, but sometimes even the counting can become ambiguous. Coordination numbers are normally between two and nine, but large numbers of ligands are not uncommon for the lanthanides and actinides. The number of bonds depends on the size, charge, and electron configuration of the metal ion and the ligands. Metal ions may have more than one coordination number.

Typically the chemistry of transition metal complexes is dominated by interactions between s and p molecular orbitals of the donor-atoms in the ligands and the d orbitals of the metal ions. The s, p, and d orbitals of the metal can accommodate 18 electrons (see 18-Electron rule). The maximum coordination number for a certain metal is thus related to the electronic configuration of the metal ion (to be more specific, the number of empty orbitals) and to the ratio of the size of the ligands and the metal ion. Large metals and small ligands lead to high coordination numbers, e.g. [Mo(CN)8]4−. Small metals with large ligands lead to low coordination numbers, e.g. Pt[P(CMe3)]2. Due to their large size, lanthanides, actinides, and early transition metals tend to have high coordination numbers.

Most structures follow the points-on-a-sphere pattern (or, as if the central atom were in the middle of a polyhedron where the corners of that shape are the locations of the ligands), where orbital overlap (between ligand and metal orbitals) and ligand-ligand repulsions tend to lead to certain regular geometries. The most observed geometries are listed below, but there are many cases that deviate from a regular geometry, e.g. due to the use of ligands of diverse types (which results in irregular bond lengths; the coordination atoms do not follow a points-on-a-sphere pattern), due to the size of ligands, or due to electronic effects (see, e.g., Jahn–Teller distortion):

The idealized descriptions of 5-, 7-, 8-, and 9- coordination are often indistinct geometrically from alternative structures with slightly differing L-M-L (ligand-metal-ligand) angles, e.g. the difference between square pyramidal and trigonal bipyramidal structures.[11]

In systems with low d electron count, due to special electronic effects such as (second-order) Jahn–Teller stabilization,[15] certain geometries (in which the coordination atoms do not follow a points-on-a-sphere pattern) are stabilized relative to the other possibilities, e.g. for some compounds the trigonal prismatic geometry is stabilized relative to octahedral structures for six-coordination.

### Isomerism

The arrangement of the ligands is fixed for a given complex, but in some cases it is mutable by a reaction that forms another stable isomer.

There exist many kinds of isomerism in coordination complexes, just as in many other compounds.

#### Stereoisomerism

Stereoisomerism occurs with the same bonds in distinct orientations. Stereoisomerism can be further classified into:[16]

##### Cis–trans isomerism and facial–meridional isomerism

Cis–trans isomerism occurs in octahedral and square planar complexes (but not tetrahedral). When two ligands are adjacent they are said to be cis, when opposite each other, trans. When three identical ligands occupy one face of an octahedron, the isomer is said to be facial, or fac. In a fac isomer, any two identical ligands are adjacent or cis to each other. If these three ligands and the metal ion are in one plane, the isomer is said to be meridional, or mer. A mer isomer can be considered as a combination of a trans and a cis, since it contains both trans and cis pairs of identical ligands.

##### Optical isomerism

Optical isomerism occurs when a complex is not superimposable with its mirror image. It is so called because the two isomers are each optically active, that is, they rotate the plane of polarized light in opposite directions. In the first molecule shown, the symbol Λ (lambda) is used as a prefix to describe the left-handed propeller twist formed by three bidentate ligands. The second molecule is the mirror image of the first, with the symbol Δ (delta) as a prefix for the right-handed propeller twist. The third and fourth molecules are a similar pair of Λ and Δ isomers, in this case with two bidentate ligands and two identical monodentate ligands.[17]

#### Structural isomerism

Structural isomerism occurs when the bonds are themselves different. Four types of structural isomerism are recognized: ionisation isomerism, solvate or hydrate isomerism, linkage isomerism and coordination isomerism.

1. Ionisation isomerism – the isomers give different ions in solution although they have the same composition. This type of isomerism occurs when the counter ion of the complex is also a potential ligand. For example, pentaamminebromocobalt(III) sulphate [Co(NH3)5Br]SO4 is red violet and in solution gives a precipitate with barium chloride, confirming the presence of sulphate ion, while pentaamminesulphatecobalt(III) bromide [Co(NH3)5SO4]Br is red and tests negative for sulphate ion in solution, but instead gives a precipitate of AgBr with silver nitrate.[18]
2. Solvate or hydrate isomerism – the isomers have the same composition but differ with respect to the number of molecules of solvent that serve as ligand vs simply occupying sites in the crystal. Examples: [Cr(H2O)6]Cl3 is violet colored, [CrCl(H2O)5]Cl2·H2O is blue-green, and [CrCl2(H2O)4]Cl·2H2O is dark green. See water of crystallization.[18]
3. Linkage isomerism occurs with ligands with more than one type[clarification needed] of donor atom, known as ambidentate ligands.[19] For example, nitrite can coordinate through O or N.[20] One pair of nitrite linkage isomers have structures (NH3)5CoNO22+ (nitro isomer) and (NH3)5CoONO2+ (nitrito isomer).[19]
4. Coordination isomerism – this occurs when both positive and negative ions of a salt are complex ions and the two isomers differ in the distribution of ligands between the cation and the anion. For example, [Co(NH3)6][Cr(CN)6] and [Cr(NH3)6][Co(CN)6].[18]

## Electronic properties

Many of the properties of transition metal complexes are dictated by their electronic structures. The electronic structure can be described by a relatively ionic model that ascribes formal charges to the metals and ligands. This approach is the essence of crystal field theory (CFT). Crystal field theory, introduced by Hans Bethe in 1929, gives a quantum mechanically based attempt at understanding complexes. But crystal field theory treats all interactions in a complex as ionic and assumes that the ligands can be approximated by negative point charges.

More sophisticated models embrace covalency, and this approach is described by ligand field theory (LFT) and Molecular orbital theory (MO). Ligand field theory, introduced in 1935 and built from molecular orbital theory, can handle a broader range of complexes and can explain complexes in which the interactions are covalent. The chemical applications of group theory can aid in the understanding of crystal or ligand field theory, by allowing simple, symmetry based solutions to the formal equations.

Chemists tend to employ the simplest model required to predict the properties of interest; for this reason, CFT has been a favorite for the discussions when possible. MO and LF theories are more complicated, but provide a more realistic perspective.

The electronic configuration of the complexes gives them some important properties:

Synthesis of copper(II)-tetraphenylporphyrin, a metal complex, from tetraphenylporphyrin and copper(II) acetate monohydrate.

### Color of transition metal complexes

Transition metal complexes often have spectacular colors caused by electronic transitions by the absorption of light. For this reason they are often applied as pigments. Most transitions that are related to colored metal complexes are either d–d transitions or charge transfer bands. In a d–d transition, an electron in a d orbital on the metal is excited by a photon to another d orbital of higher energy, therefore d–d transitions occur only for partially-filled d-orbital complexes (d1–9). For complexes having d0 or d10 configuration, charge transfer is still possible even though d–d transitions are not. A charge transfer band entails promotion of an electron from a metal-based orbital into an empty ligand-based orbital (metal-to-ligand charge transfer or MLCT). The converse also occurs: excitation of an electron in a ligand-based orbital into an empty metal-based orbital (ligand-to-metal charge transfer or LMCT). These phenomena can be observed with the aid of electronic spectroscopy; also known as UV-Vis.[21] For simple compounds with high symmetry, the d–d transitions can be assigned using Tanabe–Sugano diagrams. These assignments are gaining increased support with computational chemistry.

Colours of Various Example Coordination Complexes
Fe2+ Fe3+ Co2+ Cu2+ Al3+ Cr3+
Hydrated Ion [Fe(H2O)6]2+
Pale green
Solution
[Fe(H2O)6]3+
Yellow/brown
Solution
[Co(H2O)6]2+
Pink
Solution
[Cu(H2O)6]2+
Blue
Solution
[Al(H2O)6]3+
Colourless
Solution
[Cr(H2O)6]3+
Green
Solution
(OH), dilute [Fe(H2O)4(OH)2]
Dark green
Precipitate
[Fe(H2O)3(OH)3]
Brown
Precipitate
[Co(H2O)4(OH)2]
Blue/green
Precipitate
[Cu(H2O)4(OH)2]
Blue
Precipitate
[Al(H2O)3(OH)3]
White
Precipitate
[Cr(H2O)3(OH)3]
Green
Precipitate
(OH), concentrated [Fe(H2O)4(OH)2]
Dark green
Precipitate
[Fe(H2O)3(OH)3]
Brown
Precipitate
[Co(H2O)4(OH)2]
Blue/green
Precipitate
[Cu(H2O)4(OH)2]
Blue
Precipitate
[Al(OH)4]
Colourless
Solution
[Cr(OH)6]3−
Green
Solution
NH3, dilute [Fe(NH3)6]2+
Dark green
Precipitate
[Fe(NH3)6]3+
Brown
Precipitate
[Co(NH3)6]2+
Straw coloured
Solution
[Cu(NH3)4(H2O)2]2+
Deep blue
Solution
[Al(NH3)3]3+
White
Precipitate
[Cr(NH3)6]3+
Purple
Solution
NH3, concentrated [Fe(NH3)6]2+
Dark green
Precipitate
[Fe(NH3)6]3+
Brown
Precipitate
[Co(NH3)6]2+
Straw coloured
Solution
[Cu(NH3)4(H2O)2]2+
Deep blue
Solution
[Al(NH3)3]3+
White
Precipitate
[Cr(NH3)6]3+
Purple
Solution
(CO3)2- FeCO3
Dark green
Precipitate
Fe2(CO3)3
Brown
Precipitate+bubbles
CoCO3
Pink
Precipitate
CuCO3
Blue/green
Precipitate

### Colors of lanthanide complexes

Superficially lanthanide complexes are similar to those of the transition metals in that some are colored. However, for the common Ln3+ ions (Ln = lanthanide) the colors are all pale, and hardly influenced by the nature of the ligand. The colors are due to 4f electron transitions. As the 4f orbitals in lanthanides are "buried" in the xenon core and shielded from the ligand by the 5s and 5p orbitals they are therefore not influenced by the ligands to any great extent leading to a much smaller crystal field splitting than in the transition metals. The absorption spectra of an Ln3+ ion approximates to that of the free ion where the electronic states are described by spin-orbit coupling. This contrasts to the transition metals where the ground state is split by the crystal field. Absorptions for Ln3+ are weak as electric dipole transitions are parity forbidden (Laporte forbidden) but can gain intensity due to the effect of a low-symmetry ligand field or mixing with higher electronic states (e.g. d orbitals). f-f absorption bands are extremely sharp which contrasts with those observed for transition metals which generally have broad bands.[22][23] This can lead to extremely unusual effects, such as significant color changes under different forms of lighting.

### Magnetism

Metal complexes that have unpaired electrons are magnetic. Considering only monometallic complexes, unpaired electrons arise because the complex has an odd number of electrons or because electron pairing is destabilized. Thus, monomeric Ti(III) species have one "d-electron" and must be (para)magnetic, regardless of the geometry or the nature of the ligands. Ti(II), with two d-electrons, forms some complexes that have two unpaired electrons and others with none. This effect is illustrated by the compounds TiX2[(CH3)2PCH2CH2P(CH3)2]2: when X = Cl, the complex is paramagnetic (high-spin configuration), whereas when X = CH3, it is diamagnetic (low-spin configuration). It is important to realize that ligands provide an important means of adjusting the ground state properties.

In bi- and polymetallic complexes, in which the individual centres have an odd number of electrons or that are high-spin, the situation is more complicated. If there is interaction (either direct or through ligand) between the two (or more) metal centres, the electrons may couple (antiferromagnetic coupling, resulting in a diamagnetic compound), or they may enhance each other (ferromagnetic coupling). When there is no interaction, the two (or more) individual metal centers behave as if in two separate molecules.

### Reactivity

Complexes show a variety of possible reactivities:[24]

• Electron transfers
Electron transfer (ET) between metal ions can occur via two distinct mechanisms, inner and outer sphere electron transfers. In an inner sphere reaction, a bridging ligand serves as a conduit for ET.
• (Degenerate) ligand exchange
One important indicator of reactivity is the rate of degenerate exchange of ligands. For example, the rate of interchange of coordinate water in [M(H2O)6]n+ complexes varies over 20 orders of magnitude. Complexes where the ligands are released and rebound rapidly are classified as labile. Such labile complexes can be quite stable thermodynamically. Typical labile metal complexes either have low-charge (Na+), electrons in d-orbitals that are antibonding with respect to the ligands (Zn2+), or lack covalency (Ln3+, where Ln is any lanthanide). The lability of a metal complex also depends on the high-spin vs. low-spin configurations when such is possible. Thus, high-spin Fe(II) and Co(III) form labile complexes, whereas low-spin analogues are inert. Cr(III) can exist only in the low-spin state (quartet), which is inert because of its high formal oxidation state, absence of electrons in orbitals that are M–L antibonding, plus some "ligand field stabilization" associated with the d3 configuration.
• Associative processes
Complexes that have unfilled or half-filled orbitals often show the capability to react with substrates. Most substrates have a singlet ground-state; that is, they have lone electron pairs (e.g., water, amines, ethers), so these substrates need an empty orbital to be able to react with a metal centre. Some substrates (e.g., molecular oxygen) have a triplet ground state, which results that metals with half-filled orbitals have a tendency to react with such substrates (it must be said that the dioxygen molecule also has lone pairs, so it is also capable to react as a 'normal' Lewis base).

If the ligands around the metal are carefully chosen, the metal can aid in (stoichiometric or catalytic) transformations of molecules or be used as a sensor.

## Classification

Metal complexes, also known as coordination compounds, include virtually all metal compounds.[25] The study of "coordination chemistry" is the study of "inorganic chemistry" of all alkali and alkaline earth metals, transition metals, lanthanides, actinides, and metalloids. Thus, coordination chemistry is the chemistry of the majority of the periodic table. Metals and metal ions exist, in the condensed phases at least, only surrounded by ligands.

The areas of coordination chemistry can be classified according to the nature of the ligands, in broad terms:

Examples: [Co(EDTA)], [Co(NH3)6]3+, [Fe(C2O4)3]
• Organometallic Chemistry: Ligands are organic (alkenes, alkynes, alkyls) as well as "organic-like" ligands such as phosphines, hydride, and CO.
Example: (C5H5)Fe(CO)2CH3
• Bioinorganic Chemistry: Ligands are those provided by nature, especially including the side chains of amino acids, and many cofactors such as porphyrins.
Example: hemoglobin contains heme, a porphyrin complex of iron
Example: chlorophyll contains a porphyrin complex of magnesium
Many natural ligands are "classical" especially including water.
• Cluster Chemistry: Ligands include all of the above as well as other metal ions or atoms as well.
Example Ru3(CO)12
• In some cases there are combinations of different fields:
Example: [Fe4S4(Scysteinyl)4]2−, in which a cluster is embedded in a biologically active species.

Mineralogy, materials science, and solid state chemistry – as they apply to metal ions – are subsets of coordination chemistry in the sense that the metals are surrounded by ligands. In many cases these ligands are oxides or sulfides, but the metals are coordinated nonetheless, and the principles and guidelines discussed below apply. In hydrates, at least some of the ligands are water molecules. It is true that the focus of mineralogy, materials science, and solid state chemistry differs from the usual focus of coordination or inorganic chemistry. The former are concerned primarily with polymeric structures, properties arising from a collective effects of many highly interconnected metals. In contrast, coordination chemistry focuses on reactivity and properties of complexes containing individual metal atoms or small ensembles of metal atoms.

## Nomenclature of coordination complexes

The basic procedure for naming a complex is:

1. When naming a complex ion, the ligands are named before the metal ion.
2. The ligands' names are given in alphabetical order. Numerical prefixes do not affect the order.
• Multiple occurring monodentate ligands receive a prefix according to the number of occurrences: di-, tri-, tetra-, penta-, or hexa-.
• Multiple occurring polydentate ligands (e.g., ethylenediamine, oxalate) receive bis-, tris-, tetrakis-, etc.
• Anions end in o. This replaces the final 'e' when the anion ends with '-ide', '-ate' or '-ite', e.g. chloride becomes chlorido and sulfate becomes sulfato. Formerly, '-ide' was changed to '-o' (e.g. chloro and cyano), but this rule has been modified in the 2005 IUPAC recommendations and the correct forms for these ligands are now chlorido and cyanido.[26]
• Neutral ligands are given their usual name, with some exceptions: NH3 becomes ammine; H2O becomes aqua or aquo; CO becomes carbonyl; NO becomes nitrosyl.
3. Write the name of the central atom/ion. If the complex is an anion, the central atom's name will end in -ate, and its Latin name will be used if available (except for mercury).
4. The oxidation state of the central atom is to be specified (when it is one of several possible, or zero), and should be written as a Roman numeral (or 0) enclosed in parentheses.
5. Name of the cation should be preceded by the name of anion. (if applicable, as in last example)

Examples:

metal changed to
cobalt cobaltate
aluminium aluminate
chromium chromate
copper cuprate
iron ferrate
[CoCl(NH3)5]SO4 → pentaamminechloridocobalt(III) sulfate
[Cu(H2O)6] 2+ → hexaaquacopper(II) ion
[CuCl5NH3]3− → amminepentachloridocuprate(II) ion
K4[Fe(CN)6] → potassium hexacyanidoferrate(II)
[NiCl4]2− → tetrachloridonickelate(II) ion (The use of chloro- was removed from IUPAC naming convention)[27]

The coordination number of ligands attached to more than one metal (bridging ligands) is indicated by a subscript to the Greek symbol μ placed before the ligand name. Thus the dimer of aluminium trichloride is described by Al2Cl42-Cl)2.

Any anionic group can be electronically stabilized by any cation. An anionic complex can be stabilised by a hydrogen cation, becoming an acidic complex which can dissociate to release the cationic hydrogen. This kind of complex compound has a name with "ic" added after the central metal. For example, H2[Pt(CN)4] has the name tetracyanoplatinic (II) acid.

## Stability constant

The affinity of metal ions for ligands is described by a stability constant, also called the formation constant, and is represented by the symbol Kf. It is the equilibrium constant for its assembly from the constituent metal and ligands, and can be calculated accordingly, as in the following example for a simple case:

(X)Metal(aq) + (Y)Lewis Base(aq) ⇌ (Z)Complex Ion(aq)
${\displaystyle K_{f}={\frac {[{\text{Complex ion}}]^{Z}}{[{\text{Metal ion}}]^{X}[{\text{Lewis base}}]^{Y}}}}$

where X, Y, and Z are the stoichiometric coefficients of each species. Formation constants vary widely. Large values indicate that the metal has high affinity for the ligand, provided the system is at equilibrium.[28]

Sometimes the stability constant will be in a different form known as the constant of destability. This constant is expressed as the inverse of the constant of formation and is denoted as Kd = 1/Kf .[29] This constant represents the reverse reaction for the decomposition of a complex ion into its individual metal and ligand components. When comparing the values for Kd, the larger the value, the more unstable the complex ion is.

As a result of these complex ions forming in solutions they also can play a key role in solubility of other compounds. When a complex ion is formed it can alter the concentrations of its components in the solution. For example:

Ag+
(aq)
+ 2NH4OH(aq) ⇌ Ag(NH3)+
2
+ H2O
AgCl(s) + H2O(l) ⇌ Ag+
(aq)
+ Cl
(aq)

If these reactions both occurred in the same reaction vessel, the solubility of the silver chloride would be increased by the presence of NH4OH because formation of the Diammine argentum(I) complex consumes a significant portion of the free silver ions from the solution. By Le Chatelier's principle, this causes the equilibrium reaction for the dissolving of the silver chloride, which has silver ion as a product, to shift to the right.

This new solubility can be calculated given the values of Kf and Ksp for the original reactions. The solubility is found essentially by combining the two separate equilibria into one combined equilibrium reaction and this combined reaction is the one that determines the new solubility. So Kc, the new solubility constant, is denoted by:

${\displaystyle K_{c}=K_{sp}*K_{f}}$

## Application of coordination compounds

Metals only exist in solution as coordination complexes, it follows then that this class of compounds is useful in a wide variety of ways.

### Bioinorganic chemistry

In bioinorganic chemistry and bioorganometallic chemistry, coordination complexes serve either structural or catalytic functions. An estimated 30% of proteins contain metal ions. Examples include the intensely colored vitamin B12, the heme group in hemoglobin, the cytochromes, the chlorin group in chlorophyll, and carboxypeptidase, a hydrolytic enzyme important in digestion. Another complex ion enzyme is catalase, which decomposes the cell's waste hydrogen peroxide. Synthetic coordination compounds are also used to bind to proteins and especially nucleic acids (e.g. anticancer drug cisplatin).

### Industry

Homogeneous catalysis is a major application of coordination compounds for the production of organic substances. Processes include hydrogenation, hydroformylation, oxidation. In one example, a combination of titanium trichloride and triethylaluminium gives rise to Ziegler–Natta catalysts, used for the polymerization of ethylene and propylene to give polymers of great commercial importance as fibers, films, and plastics.

Nickel, cobalt, and copper can be extracted using hydrometallurgical processes involving complex ions. They are extracted from their ores as ammine complexes. Metals can also be separated using the selective precipitation and solubility of complex ions. Cyanide is used chiefly for extraction of gold and silver from their ores.

Phthalocyanine complexes are an important class of pigments.

### Analysis

At one time, coordination compounds were used to identify the presence of metals in a sample. Qualitative inorganic analysis has largely been superseded by instrumental methods of analysis such as atomic absorption spectroscopy (AAS), inductively coupled plasma atomic emission spectroscopy (ICP-AES) and inductively coupled plasma mass spectrometry (ICP-MS).

## References

1. ^ Lawrance, Geoffrey A. (2010). Introduction to Coordination Chemistry. Wiley. doi:10.1002/9780470687123. ISBN 9780470687123.
2. ^ IUPAC, Compendium of Chemical Terminology, 2nd ed. (the "Gold Book") (1997). Online corrected version:  (2006–) "complex". doi:10.1351/goldbook.C01203
3. ^ IUPAC, Compendium of Chemical Terminology, 2nd ed. (the "Gold Book") (1997). Online corrected version:  (2006–) "coordination entity". doi:10.1351/goldbook.C01330
4. ^ Greenwood, Norman  N.; Earnshaw, Alan (1997). Chemistry of the Elements (2nd ed.). Butterworth-Heinemann. ISBN 978-0-08-037941-8.
5. ^ "Definition of coordination sphere". chemistry-dictionary.com.
6. ^ "What Is A Coordination Compound?". Purdue University Department of Chemistry.
7. ^ Cotton, Frank Albert; Geoffrey Wilkinson; Carlos A. Murillo (1999). Advanced Inorganic Chemistry. p. 1355. ISBN 978-0-471-19957-1.
8. ^ Miessler, Gary L.; Donald Arthur Tarr (1999). Inorganic Chemistry. p. 642. ISBN 978-0-13-841891-5.
9. ^
10. ^ Werner, A. (May 1911). "Zur Kenntnis des asymmetrischen Kobaltatoms. I". Berichte der deutschen chemischen Gesellschaft (in German). 44 (2): 1887–1898. doi:10.1002/cber.19110440297.
11. ^ Wells A.F. (1984) Structural Inorganic Chemistry 5th edition Oxford Science Publications ISBN 0-19-855370-6
12. ^ Angelo R. Rossi; Roald. Hoffmann (1975). "Transition metal pentacoordination". Inorganic Chemistry. 14 (2): 365–374. doi:10.1021/ic50144a032.
13. ^ Roald. Hoffmann; Barbara F. Beier; Earl L. Muetterties; Angelo R. Rossi (1977). "Seven-coordination. A molecular orbital exploration of structure, stereochemistry, and reaction dynamics". Inorganic Chemistry. 16 (3): 511–522. doi:10.1021/ic50169a002.
14. ^ Jeremy K. Burdett; Roald Hoffmann; Robert C. Fay (1978). "Eight-Coordination". Inorganic Chemistry. 17 (9): 2553–2568. doi:10.1021/ic50187a041.
15. ^ Kaupp, Martin (2001). ""Non-VSEPR" Structures and Bonding in d0 Systems". Angew. Chem. Int. Ed. Engl. 40 (1): 3534–3565. doi:10.1002/1521-3773(20011001)40:19<3534::AID-ANIE3534>3.0.CO;2-#.
16. ^ von Zelewsky, A. "Stereochemistry of Coordination Compounds" John Wiley: Chichester, 1995. ISBN 0-471-95599-X.
17. ^ Miessler, Gary L.; Donald Arthur Tarr (1999). "9". Inorganic Chemistry. pp. 315, 316. ISBN 978-0-13-841891-5.
18. ^ a b c Huheey, James E., Inorganic Chemistry (3rd ed., Harper & Row 1983), p.524–5 ISBN 0-06-042987-9
19. ^ a b Jolly, William L. (1984). Modern Inorganic Chemistry. McGraw-Hill. pp. 357–9. ISBN 0-07-032760-2.
20. ^ Huheey, James E., Inorganic Chemistry (3rd ed., Harper & Row 1983), p.513–24 ISBN 0-06-042987-9
21. ^ Harris, D.; Bertolucci, M. (1989). Symmetry and Spectroscopy. Dover Publications. ISBN 9780486661445.
22. ^ Cotton, F. Albert; Wilkinson, Geoffrey; Murillo, Carlos A.; Bochmann, Manfred (1999), Advanced Inorganic Chemistry (6th ed.), New York: Wiley-Interscience, ISBN 0-471-19957-5
23. ^ Cotton, Simon (2006). Lanthanide and Actinide Chemistry. John Wiley & Sons Ltd.
24. ^ R. G. Wilkins Kinetics and Mechanism of Reactions of Transition Metal Complexes, 2nd Edition, VCH, Weinheim, 1991. ISBN 1-56081-125-0
25. ^ Exception: metal vapors, plasmas, and alloys.
26. ^ "Nomenclature of Inorganic Chemistry IUPAC Recommendations 2005" (PDF). IUPAC. section 1.6.4 (p. 10-11). Archived from the original (PDF) on 2014-12-22. Retrieved 2016-03-06.
27. ^ "Nomenclature of Inorganic Chemistry IUPAC Recommendations 2005" (PDF). IUPAC. section 1.6.4 (p. 10-11). Archived from the original (PDF) on 2014-12-22. Retrieved 2016-03-06.
28. ^
29. ^ Stretton, Tom. "Solubility and Complex-ion Equilibria" (PDF).


text:
## Same-fringe using zippers

Recall back in Assignment 4, we asked you to enumerate the "fringe" of a leaf-labeled tree. Both of these trees (here I am drawing the labels in the diagram):

    .                .
/ \              / \
.   3            1   .
/ \                  / \
1   2                2   3

have the same fringe: [1; 2; 3]. We also asked you to write a function that determined when two trees have the same fringe. The way you approached that back then was to enumerate each tree's fringe, and then compare the two lists for equality. Today, and then again in a later class, we'll encounter new ways to approach the problem of determining when two trees have the same fringe.

Supposing you did work out an implementation of the tree zipper, then one way to determine whether two trees have the same fringe would be: go downwards (and leftwards) in each tree as far as possible. Compare the focused leaves. If they're different, stop because the trees have different fringes. If they're the same, then for each tree, move rightward if possible; if it's not (because you're at the rightmost leaf in a subtree), move upwards then try again to move rightwards. Repeat until you are able to move rightwards. Once you do move rightwards, go downwards (and leftwards) as far as possible. Then you'll be focused on the next leaf in the tree's fringe. The operations it takes to get to "the next leaf" may be different for the two trees. For example, in these trees:

    .                .
/ \              / \
.   3            1   .
/ \                  / \
1   2                2   3

you won't move upwards at the same steps. Keep comparing "the next leaves" until they are different, or you exhaust the leaves of only one of the trees (then again the trees have different fringes), or you exhaust the leaves of both trees at the same time, without having found leaves with different labels. In this last case, the trees have the same fringe.

If your trees are very big---say, millions of leaves---you can imagine how this would be quicker and more memory-efficient than traversing each tree to construct a list of its fringe, and then comparing the two lists so built to see if they're equal. For one thing, the zipper method can abort early if the fringes diverge early, without needing to traverse or build a list containing the rest of each tree's fringe.

Let's sketch the implementation of this. We won't provide all the details for an implementation of the tree zipper (you'll need to fill those in), but we will sketch an interface for it.

In these exercises, we'll help ourselves to OCaml's record types. These are nothing more than tuples with a pretty interface. Instead of saying:

# type blah = Blah of int * int * (char -> bool);;

and then having to remember which element in the triple was which:

# let b1 = Blah (1, (fun c -> c = 'M'), 2);;
Error: This expression has type int * (char -> bool) * int
but an expression was expected of type int * int * (char -> bool)
# (* damnit *)
# let b1 = Blah (1, 2, (fun c -> c = 'M'));;
val b1 : blah = Blah (1, 2, <fun>)

records let you attach descriptive labels to the components of the tuple:

# type blah_record = { height : int; weight : int; char_tester : char -> bool };;
# let b2 = { height = 1; weight = 2; char_tester = (fun c -> c = 'M') };;
val b2 : blah_record = {height = 1; weight = 2; char_tester = <fun>}
# let b3 = { height = 1; char_tester = (fun c -> c = 'K'); weight = 3 };; (* also works *)
val b3 : blah_record = {height = 1; weight = 3; char_tester = <fun>}

These were the strategies to extract the components of an unlabeled tuple:

let h = fst some_pair (* accessor functions fst and snd are only predefined for pairs *)

let (h, w, test) = b1 (* works for arbitrary tuples *)

match b1 with
| (h, w, test) -> ... (* same as preceding *)

Here is how you can extract the components of a labeled record:

let h = b2.height (* handy! *)

let {height = h; weight = w; char_tester = test} = b2 in
(* go on to use h, w, and test ... *)

match test with
| {height = h; weight = w; char_tester = test} ->
(* same as preceding *)

Anyway, using record types, we might define the tree zipper interface like so. First, we define a type for leaf-labeled, binary trees:

type 'a tree = Leaf of 'a | Node of 'a tree * 'a tree

Next, the types for our tree zippers:

type 'a zipper = { in_focus: 'a tree; context : 'a context }
and 'a context = Root | Nonroot of 'a nonroot_context
and 'a nonroot_context = { up : 'a context; left: 'a tree option; right: 'a tree option }

Unlike in seminar, here we represent the siblings as 'a tree options rather than 'a tree lists. Since we're dealing with binary trees, each context will have exactly one sibling, either to the right or to the left.

The following function takes an 'a tree and returns an 'a zipper focused on its root:

let new_zipper (t : 'a tree) : 'a zipper =
{in_focus = t; context = Root}

Here are the beginnings of functions to move from one focused tree to another:

let rec move_botleft (z : 'a zipper) : 'a zipper =
(* returns z if the focused node in z has no children *)
(* else returns move_botleft (zipper which results from moving down from z's focused node to its leftmost child) *)
_____ (* YOU SUPPLY THE DEFINITION *)

let rec move_right_or_up (z : 'a zipper) : 'a zipper option =
(* if it's possible to move right in z, returns Some (the result of doing so) *)
(* else if it's not possible to move any further up in z, returns None *)
(* else returns move_right_or_up (result of moving up in z) *)
_____ (* YOU SUPPLY THE DEFINITION *)

1. Your first assignment is to complete the definitions of move_botleft and move_right_or_up. (Really it should be move_right_or_up_..._and_right.)

Having completed that, we can use define a function that enumerates a tree's fringe, step by step, until it's exhausted:

let make_fringe_enumerator (t: 'a tree) =
(* create a zipper focusing the botleft of t *)
let zbotleft = move_botleft (new_zipper t) in
(* create initial state, pointing to zbotleft *)
let initial_state : 'a zipper option = Some zbotleft in
(* construct the next_leaf function *)
let next_leaf : 'a zipper option -> ('a * 'a zipper option) option =
fun state -> match state with
| Some z -> (
(* extract label of currently-focused leaf *)
let Leaf current = z.in_focus in
(* create next_state pointing to next leaf, if there is one *)
let next_state : 'a zipper option = match move_right_or_up z with
| None -> None
| Some z' -> Some (move_botleft z') in
(* return saved label and next_state *)
Some (current, next_state)
)
| None -> (* we've finished enumerating the fringe *)
None in
(* return the next_leaf function and initial state *)
next_leaf, initial_state

Here's an example of make_fringe_enumerator in action:

# let tree1 = Leaf 1;;
val tree1 : int tree = Leaf 1
# let next1, state1 = make_fringe_enumerator tree1;;
val next1 : int zipper option -> (int * int zipper option) option = <fun>
val state1 : int zipper option = Some ...
# let Some (res1, state1') = next1 state1;;
val res1 : int = 1
val state1' : int zipper option = None
# next1 state1';;
- : (int * int zipper option) option = None
# let tree2 = Node (Node (Leaf 1, Leaf 2), Leaf 3);;
val tree2 : int tree = Node (Node (Leaf 1, Leaf 2), Leaf 3)
# let next2, state2 = make_fringe_enumerator tree2;;
val next2 : int zipper option -> (int * int zipper option) option = <fun>
val state2 : int zipper option = Some ...
# let Some (res2, state2') = next2 state2;;
val res2 : int = 1
val state2' : int zipper option = Some ...
# let Some (res2, state2'') = next2 state2';;
val res2 : int = 2
val state2'' : int zipper option = Some ...
# let Some (res2, state2''') = next2 state2'';;
val res2 : int = 3
val state2''' : int zipper option = None
# next2 state2''';;
- : (int * int zipper option) option = None

You might think of it like this: make_fringe_enumerator returns a little subprogram that will keep returning the next leaf in a tree's fringe, in the form Some ..., until it gets to the end of the fringe. After that, it will return None. The subprogram's memory of where it is and what steps to perform next are stored in the next_state variables that are part of its input and output.

Using these fringe enumerators, we can write our same_fringe function like this:

let same_fringe (t1 : 'a tree) (t2 : 'a tree) : bool =
let next1, initial_state1 = make_fringe_enumerator t1 in
let next2, initial_state2 = make_fringe_enumerator t2 in
let rec loop state1 state2 : bool =
match next1 state1, next2 state2 with
| Some (a, state1'), Some (b, state2') when a = b -> loop state1' state2'
| None, None -> true
| _ -> false in
loop initial_state1 initial_state2

The auxiliary loop function will keep calling itself recursively until a difference in the fringes has manifested itself---either because one fringe is exhausted before the other, or because the next leaves in the two fringes have different labels. If we get to the end of both fringes at the same time (next1 state1, next2 state2 matches the pattern None, None) then we've established that the trees do have the same fringe.

2. Test your implementations of move_botleft and move_right_or_up against some example trees to see if the resulting make_fringe_enumerator and same_fringe functions work as expected. Show us some of your tests.

3. Now we'll talk about another way to implement the make_fringe_enumerator function above (and so too the same_fringe function which uses it). Notice that the pattern given above is that the make_fringe_enumerator creates a next_leaf function and an initial state, and each time you want to advance the next_leaf by one step, you do so by calling it with the current state. It will return a leaf label plus a modified state, which you can use when you want to call it again and take another step. All of the next_leaf function's memory about where it is in the enumeration is contained in the state. If you saved an old state, took three steps, and then called the next_leaf function again with the saved old state, it would be back where it was three steps ago. But in fact, the way we use the next_leaf function and state above, there is no back-tracking. Neither do we "fork" any of the states and pursue different forward paths. Their progress is deterministic, and fixed independently of anything that same_fringe might do. All that's up to same_fringe is to take the decision of when (and whether) to take another step forward.

Given that usage pattern, it would be appropriate and convenient to make the next_leaf function remember its state itself, in a mutable variable. The client function same_fringe doesn't need to do anything with, or even be given access to, this variable. Here's how we might write make_fringe_enumerator according to this plan:

  let make_fringe_enumerator (t: 'a tree) =
(* create a zipper focusing the botleft of t *)
let zbotleft = move_botleft (new_zipper t) in
(* create a refcell, initially pointing to zbotleft *)
let zcell = ref (Some zbotleft) in
(* construct the next_leaf function *)
let next_leaf : unit -> 'a option = fun () ->
match !zcell with
| Some z -> (
(* extract label of currently-focused leaf *)
let Leaf current = z.in_focus in
(* update zcell to point to next leaf, if there is one *)
let () = zcell := match move_right_or_up z with
| None -> None
| Some z' -> _____ in
(* return saved label *)
_____
)
| None -> (* we've finished enumerating the fringe *)
None in
(* return the next_leaf function *)
next_leaf

let same_fringe (t1 : 'a tree) (t2 : 'a tree) : bool =
let next1 = make_fringe_enumerator t1 in
let next2 = make_fringe_enumerator t2 in
let rec loop () : bool =
match _____, _____ with
| Some a, Some b when a = b -> loop ()
| None, None -> true
| _ -> false in
loop ()

You should fill in the blanks.

Here's an example of our new make_fringe_enumerator in action:

  # let tree1 = Leaf 1;;
val tree1 : int tree = Leaf 1
# let next1 = make_fringe_enumerator tree1;;
val next1 : unit -> int option = <fun>
# next1 ();;
- : int option = Some 1
# next1 ();;
- : int option = None
# next1 ();;
- : int option = None
# let tree2 = Node (Node (Leaf 1, Leaf 2), Leaf 3);;
val tree2 : int tree = Node (Node (Leaf 1, Leaf 2), Leaf 3)
# let next2 = make_fringe_enumerator tree2;;
val next2 : unit -> int option = <fun>
# next2 ();;
- : int option = Some 1
# next2 ();;
- : int option = Some 2
# next2 ();;
- : int option = Some 3
# next2 ();;
- : int option = None
# next2 ();;
- : int option = None

## Same-fringe using streams

Now we'll describe a different way to create "the little subprograms" that we built above with make_fringe_enumerator. This code will make use of a data structure called a "stream". A stream is like a list in that it wraps a series of elements of a single type. It differs from a list in that the tail of the series is left uncomputed until needed. We turn the stream off and on by thunking it, nad by forcing the thunk.

We'll first show how to implement streams in OCaml, so that the types are manifest. But then we'll switch to a Scheme version to do the same-fringe problem. In part because that's how we have the code already written; but also because some later discussion will use the Scheme code as a starting point. In principle, though, we could have used OCaml throughout.

So here is a natural OCaml type for a stream. (You could also do things differently.)

type 'a stream = End | Next of 'a * (unit -> 'a stream);;

We have a special variant called End that encodes a stream that contains no (more) elements, analogous to the empty list []. Streams that are not empty contain a first element, paired with a thunked stream representing the rest of the stream. In order to get access to the next element in the stream, we must force the thunk by applying it to (). Watch the behavior of this stream in detail. This stream delivers the natural numbers, in order: 1, 2, 3, ...

# let rec make_int_stream i = Next (i, fun () -> make_int_stream (i + 1));;
val make_int_stream : int -> int stream = <fun>

# let int_stream = make_int_stream 1;;
val int_stream : int stream = Next (1, <fun>)         (* First element: 1 *)

# let tail = match int_stream with Next (i, rest) -> rest;;
val tail : unit -> int stream = <fun>                 (* Tail: a thunk *)

(* Force the thunk to compute the second element *)
# tail ();;
- : int stream = Next (2, <fun>)                      (* Second element: 2 *)

# match tail () with Next (_, rest) -> rest ();;
- : int stream = Next (3, <fun>)                      (* Third element: 3 *)

You can think of int_stream as a functional object that provides access to an infinite sequence of integers, one at a time. It's as if we had written [1;2;...] where ... meant "continue for as long as some other process needs new integers".

Okay, now armed with the idea of a stream, let's use a Scheme version of them to handle the same-fringe problem. This code is taken from http://c2.com/cgi/wiki?SameFringeProblem. It uses thunks to delay the evaluation of code that computes the tail of a list of a tree's fringe. It also involves passing "the rest of the enumeration of the fringe" as a thunk argument (tail-thunk below). Your assignment is to fill in the blanks in the code, and also to supply comments to the code, to explain what every significant piece is doing. Don't forget to supply the comments, this is an important part of the assignment.

This code uses Scheme's cond construct. That works like this;

(cond
((test1 argument argument) result1)
((test2 argument argument) result2)
((test3 argument argument) result3)
(else result4))

is equivalent to:

(if (test1 argument argument)
; then
result1
; else
(if (test2 argument argument)
; then
result2
; else
(if (test3 argument argument)
; then
result3
; else
result4)))

Some other Scheme details or reminders:

• #t is true and #f is false
• (lambda () ...) constructs a thunk
• there is no difference in meaning between [...] and (...); we just sometimes use the square brackets for clarity
• '(1 . 2) and (cons 1 2) are pairs (the same pair)
• (list) and '() both evaluate to the empty list
• (null? lst) tests whether lst is the empty list
• non-empty lists are implemented as pairs whose second member is a list
• '() '(1) '(1 2) '(1 2 3) are all lists
• (list) (list 1) (list 1 2) (list 1 2 3) are the same lists as the preceding
• '(1 2 3) and (cons 1 '(2 3)) are both pairs and lists (the same list)
• (pair? lst) tests whether lst is a pair; if lst is a non-empty list, it will also pass this test; if lst fails this test, it may be because lst is the empty list, or because it's not a list or pair at all
• (car lst) extracts the first member of a pair / head of a list
• (cdr lst) extracts the second member of a pair / tail of a list

1. Here is the Scheme code handling the same-fringe problem. You should fill in the blanks:

(define (lazy-flatten tree)
(letrec ([helper (lambda (tree tail-thunk)
(cond
[(pair? tree)
(helper (car tree) (lambda () (helper _____ tail-thunk)))]
[else (cons tree tail-thunk)]))])
(helper tree (lambda () _____))))

(define (stream-equal? stream1 stream2)
(cond
[(and (null? stream1) (null? stream2)) _____]
[(and (pair? stream1) (pair? stream2))
(and (equal? (car stream1) (car stream2))
_____)]
[else #f]))

(define (same-fringe? tree1 tree2)
(stream-equal? (lazy-flatten tree1) (lazy-flatten tree2)))

(define tree1 '(((1 . 2) . (3 . 4)) . (5 . 6)))
(define tree2 '(1 . (((2 . 3) . (4 . 5)) . 6)))

(same-fringe? tree1 tree2)

format: plain


text:
## Evaluation of elements in the residue class field of number/function field

Hi,

I have a question. Let $F|\mathbb{F}_q$ be algebraic function field and $P$ is a rational place of $F$. suppose $f \in F$ is regular at $P$. Is there any algorithm to evaluate the image of $f$ in the residue class field $\tilde{F}_p= O_p/P$?

I know Magma can do that, but I am trying to know how I can do it myself for specific example?

-
Depends on how $F,P,f$ are presented to you. – Felipe Voloch May 22 2012 at 13:07


text:
# Lebesgue's Density Theorem - intuition and weaker forms

Lebesgue's Density Theorem states that given a measurable set $E$ on the real line, then the set of points $E'$ for which $\lim_{h \to 0} \frac{m(E \cap (x-h,x+h) )}{2h} = 1$ is $E$ up to a nullset, i.e., $m(E \Delta E') = 0$. I interpret it as follows: a measurable set behaves locally as an interval, measure-wise.

1. Other, perhaps more correct interpretations
2. Intuition for why this is true? The only intuition I know is that the theorem is easy consequence in the case of an open set, and a measurable set is almost a $G_{\delta}$ set (intersection of countably many open sets).
3. Interesting proofs (I know a proof that uses the regulairy of Lebesgue measure, i.e., the ability to approximate the measure of set $E$ arbitrarily close by closed\open sets contained\containing $E$. Are there any different proofs?)
4. A direct consequence of the lemma is that for almost all $x\in E$ we have: $\forall h>0 m(E\cap (x-h,x+h)) > 0$. Is there a simple proof for this, not using Lebesgue's Density Theorem?
5. With my intuition, a nowhere dense closed set (closed set that doesn't contain an interval) of positive measure (say, 'thick' Cantor set) might contradict the theorem (but it doesn't). I know that topological denseness and positive measure don't imply each other, but still - it's not trivial for me to see how the theorem works in this case.
-
I'm not sure it's useful to try to convert the Lebesgue density theorem into a catchy slogan. But regarding question 4, certainly the (countable) union of the basic open sets which have null intersection with $E$ again has null intersection with $E$, and the points leftover satisfy your condition. (Also, your $H$ seems to play no role in the statement.) – user83827 Aug 17 '11 at 20:36
You're right, I simplified the statement now. Also, thanks for the solution, it is indeed trivial (I didn't give it enough thought) – Ofir Aug 17 '11 at 22:39
Well oftentimes these sorts of questions are difficult until you happen to rephrase it in a particular way that makes it almost immediate! Anyway, I apologize if the first sentence of my comment was overly blunt: I meant it as a simple, lighthearted statement of opinion (i.e., "I think the statement of the theorem is straightforward as is without need of a soft version"), but upon rereading it seems more negative than intended. And of course your opinion may differ. – user83827 Aug 17 '11 at 23:04
Catchy slogan: "The edge of a measurable set is negligible." – Marcel T. Jul 24 '14 at 18:10

I don't know if this is a more correct interpretation or a more interesting proof, but I always pictured the density theorem as a corollary of Lebesgue's Differentiation Theorem, which says that for $f \in L^1(\mathbb{R}^n)$, $Q_x \subset \mathbb{R}^n$ a cube centered at $x$, and $|\cdot|$ denoting Lebesgue measure,

$$\lim_{|Q_x|\searrow 0} \frac{1}{|Q_x|} \int_{Q_x} f(y) dy = f(x) \text{ for a.e. x \in \mathbb{R}^n}.$$

Since the indicator function $\chi_E(x)$ of a measurable set $E$ with finite measure is in $L^1$, we can use this theorem to say that

$$\lim_{|Q_x|\searrow 0} \frac{|E \cap Q_x|}{|Q_x|} = \chi_E(x) \text{ for a.e. x\in \mathbb{R}^n}.$$

By breaking up any measurable set into measurable pieces with finite measure, I think the result follows for general measurable sets as well.

In this sense, the density theorem basically says that the indicator function of a measurable set obeys the (Lebesgue version of the) fundamental theorem of calculus.

-

I like to think of it in terms of a probability: any event that can happen (i.e. any set with non-zero probability) has arbitrarily high probability of happening in some interval.

-
Which could be easily seen in the following formulation (which is a consequence of the one above): If $A$ is a Lebesgue measurable set and $\mu(A) > 0$ and if $0 < p < 1$, then there is an open interval $U = (a,b)$ such that $\mu(A\cap U) \ge p\mu(U) = p(b-a)$. Using the Lebesgue measure on $[0,1]$ and interpreting as a probability we see the it reads that the conditional probability on some interval could get arbitrary high. – Stefan Jul 16 '15 at 19:24


text:
# On the existence of the probability density of stochastic processes of the type $X_t = \sum_{i = 0}^{\infty} \phi^i W_{i-t}$.

Let the stochastic process $$\{X_t \}_{t \in \mathbb{N}}$$ be defined as $$X_t = \sum_{i = 0}^{\infty} \phi^i W_{i-t}$$

where $$\phi \in \mathbb{R}$$ and the $$W_i$$ are white noise with $$W_i\sim N(0, \sigma) \forall i \in \mathbb{Z}$$. It is well known in econometric theory that these kind of processes are solutions to auto-regressive (and ARMA) type equations.

It is quite natural to wonder if the associated distribution function to the process

$$F_t(x) = P(X_t \le x) \quad x \in \mathbb{R}$$

has a density (this type of question is asked all the time in the theory of stochastic differential equations, that is when we have continuous time $$t$$). So do we have a density in this case and can it be recovered explicitly?

• Is the following correct? For $|\phi| < 1$, $X_t = W_{-t}+\phi W_{1-t}+\ldots \sim \mathcal{N}\left(0,\sigma^2\sum_{i=0}^\infty \phi^{2i}\right) = \mathcal{N}\left(0,\frac{\sigma^2}{1-\phi^2}\right)$ – Timothy Hedgeworth Feb 28 at 17:24

To expand Hedgeworth's comments, we need two premises to hold in order to have $$X_t \sim N(0,\sigma^2 \sum_{i=0}^\infty \phi^{2i})$$ (Note that I use $$\sigma^2$$ to represent the variance of $$W_i$$).
We can easily prove the second premise by calculating that the $$cov(\phi^i W_{i-t}, \phi^j W_{j-t})=0$$.
The independence is a bit less obvious. Given your wording of the problem, my understanding is that the white noises are independent across the elements in the summation because you allow different subscripts for $$W_{i-t}$$ over different i.
So both premises hold and $$X_t \sim N(0,\sigma^2 \sum_{i=0}^\infty \phi^{2i})$$


text:
How to approach the design of a pixel circuit using MOSFETs

sijafae

Joined Nov 12, 2020
4
Good Day,

I am working on a school project in which we are supposed to decide the component sizes of transistors and capacitors in order for the 4-pixel camera to function properly. I have so far come up with some calculations which I am not very confident is the correct way to approach this problem and I hope I can get some feedback on what I have done wrong so far, and how I should approach the problem.

To begin with, it might be good with some background information as to what I am supposed to do:

The first picture shows the pixel circuit. Pins Expose, Erase, and NRE (or Read_n) are digital control signals that we are going to create in Verilog. How these signals, and the other control signals work, are shown in the waveform diagram above. The NRE signal is used to select between 2-pixel circuits in a column to be available at the ADC input pin. Erasing the charge of Cs by using the Erase pin has to occur before Expose can happen because no voltage should be present on Cs before allowing the diode current to be stored in it, otherwise it will seem as if a different lighting strength is present than there actually is. The amount of time of exposing can be anything between 2ms and 30ms, while the diode current is 50pA (very weak lighting) to 750pA (strong lighting). When there is little light, generally the amount of exposure time should be higher while much light should have a lower expose time. It is possible to set the exposure time manually, but the default is that 30ms corresponds to 50pA, while 2ms corresponds to 750pA.

The picture below shows a picture of the 2x2 pixel camera, where Mc1 and Mc2 are active loads each connected to a column of pixel circuits. When a pixel in each column is selected with the NRE signal, the ADC signal will go high the next clock cycle to digitize the values of the outputs of the pixel circuits in parallel.

I have attached the calculations I have performed so far, although those notes may be hard to read. I can in case write a short summary of what I have been thinking:
Constraints:
1. 0.36μm < L < 1.08μm
2. 1.08μm < W < 5.04μm
3. 50pA < $I_{pd}$ < 750pA
4. 2ms < t < 30ms
5. 180nm transistor technology
6. $V_{DD}$ = 1.8V
7. $C_c$ = 3pF (parasitic capacitance)
8. $C_s \leq 3pF$
9. f = 1kHz

Firstly, I tried to figure out how each transistor functioned in the circuit:
1. M1: When transistor M1 goes active, it allows the constant current from the photodiode to pass through and be stored in the capacitor, Cs. (Operates in the Active region).
2. M2: When transistor M2 goes active, Cs will be discharged through the transistor to the ground. (Using formula $i = C*\frac{dV}{dt}$ to find the amount of time required to discharge the capacitor and use the MOSFET size to decide current?) (Operates in the active region)
3. M3: This is a buffer stage (source follower). The amount of charge stored in Cs will decide the current through this transistor. When there is no light, no charge will be present in the capacitor, and the current through the transistor will be at its maximum. The current and exposure time combination is at the proper value, the capacitor charge will reach its maximum, which I can decide is the value in which transistor M3 is Off. Thus, when the exposure and the current combination is correct for the lighting, the value the ADC will see is a constant value. (Operates in the active region)
4. M4: Not-read transistor. When the NRE pin goes low, the transistor will become active. In my calculations, I figured that this transistor can not be in the active region, and must operate in the triode region due to the overdrive voltage always being higher than the source-drain voltage.
5. MC: This is the active load transistor connected to each column of pixel circuits. When the circuit is idle, capacitor Cc will be charged by the transistor, due to the overdrive voltage always being one threshold voltage below the source-drain voltage, making the transistor be in the active region. When the capacitor charge has reached Vdd - |Vtp|, the transistor will turn off due to the overdrive voltage reaching 0V.

Now that I have (hopefully) figured out the transistor functionalities, I begin looking for other constraints that will help me decide the component values. I decide to first figure out the lower boundary of the capacitor as well as finding the appropriate formula to later calculate the value. Since the current going through the photodiode is constant, the capacitor charging formula can be simplified to: $C_s = \frac{I_{pd}*t}{V_{cs}}$.
Figuring out the upper boundary of $V_{cs}$ will give the lower boundary of $C_s$. To figure out the upper boundary of $V_{cs}$, I can look at a few conditions for M3 and MC that must be met in order for them to always be conducting current:
1. $V_{DD} - V_{out} = V_{SG\_C} > |V_{tp}|$
2. $V_{S\_4} - V_{cs} = V_{SG\_3} > |V_{tp}|$
By setting $V_{cs} = V_{out}$, transistor M3 will still be conducting, but no current will be present due to M4 having $V_{SD} = 0$. Using this, and the conditions stated above, it can be shown that $V_{cs} < V_{DD} - 2*|V_{tp}|$. Placing this condition in for $V_{cs}$ in the capacitor formula, it can be shown that: $1.67pF < C_s \leq 3pF$.
Now that the range of $C_s$ is figured out, I decided to solve the size ratio between the transistors M3 and M4. Here, I looked at the transistors when the maximum current is present, that is when $V_{cs} = 0$. Knowing that transistor M4 is operating in the triode region, I decided that having $V_{SD\_4}|_{max} = |V_{tp}|$, that way I can approximate the current to be using the formula: $I_{s4} = \mu_p*C_{ox}*(\frac{W}{L}*k_4)*V_{eff\_4}*V_{SD\_4}$. By a few calculations (shown in the handwritten notes, however I calculated $V_{eff\_4}$ incorrectly, where it is supposed to be $V_{eff\_4} = V_{DD} - |V_{tp}| = 3*|V_{tp}|$, so the ratio should be $K_3 = \frac{3K_4}{2}$), the size relationship can be decided. Here I used the following values and relationships to solve for the size ratio:
1. $I_{S\_3} = I_{S\_4} \\ \frac{\mu_p*C_{ox}}{2}*(\frac{W}{L}*K_3)*V_{eff\_3}^2 = \mu_p*C_{ox}*(\frac{W}{L}*K_4)*V_{eff\_4}*V_{SD\_4}$
2. $V_{eff\_3} = |V_{tp}|$
3. $V_{eff\_4} = 3*|V_{tp}|$
4. $V_{SD\_4} = |V_{tp}|$
Using these formulas for the equal currents and the values for the transistor voltages, it can be shown that the size relationship between transistor M3 and M4 is: $k_3 = \frac{3*k_4}{2}$.

This is as far as I have gotten so far. I am unsure about how I should solve for the discharge of $C_c$, but maybe I can use the formula $i = C*\frac{dV}{dt}$ and then set the current going through transistors M3 and M4 equal to the sum of the capacitor discharge current and the current that will be produced in transistor MC when the voltage over the capacitor starts dropping? As I can see, the source-gate voltage of MC will increase as the capacitor is discharged, while the source-drain voltage will always be high enough to drive the transistor in the active region. Then, since the transistor only varies by the source gate voltage (exponential), the discharge current of the parasitic capacitor $C_c$ and the transistor current of MC are both exponentially changing currents. In the end, these currents will sum up to the current of $I_{S\_3}$. That is, $I_{S\_3} = I_c+I_{S\_c}$. Using this relationship, the lower boundary that the output voltage will reach before the voltage is stabilizing due to no more current being discharged from the parasitic capacitor, or rather, the charging current from the transistor outweighs the discharging of the capacitor.

I have attached my handwritten notes to this post in order to give a more thorough overview of what I have done (sorry for horrible handwriting).

Thank you for taking the time to read this. I hope someone could please help me get on the right path to solving this problem. Any help, tips, or suggestions would be very useful and I would greatly appreciate it.

Have a great day,
Sindre

Attachments

• 3.2 MB Views: 4


text:
## Geometry: Common Core (15th Edition)

According to theorem 6-18, if the diagonals of a parallelogram are congruent, then that parallelogram is a rectangle. In the diagram, the diagonal $\overline{SO}$ is congruent to the diagonal $\overline{TP}$; therefore, this parallelogram is a rectangle.


text:
Reaction forces for structure

1. Sep 23, 2013

Woopydalan

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

2. Relevant equations

3. The attempt at a solution
My problem here seems that I can't figure out how to solve for my unknowns. I have 3 equations for force, and I'm not sure if that equation I have for the moment around the origin counts as 3 new independent equations if I split them up into i, j, and k components. If that is allowed, then I have 6 equations and 6 unknowns, which should be solvable. Otherwise, I only have 4 equations and can't solve this thing.

Is my analysis of the moments ok as well as the forces present on my FBD? I figured the rings cannot oppose motion in the directions where the hole of the ring is perpendicular to the force, thus has no reaction force.

Attached Files:

File size:
334.8 KB
Views:
44
• 4.74.pdf
File size:
193 KB
Views:
68
2. Sep 23, 2013

SteamKing

Staff Emeritus
When writing moment equations, it's usually a good idea to pick one of the locations where an unknown reaction occurs. Look at it like solving a loaded beam on two supports: the sum of the forces involves both R1 and R2, but by writing a moment equation about one of the reaction points, say point 1, there is no moment produced by R1 but the resulting equation contains only R2, which can then be calculated. R1 is found by plugging R2 into the force equation.

You have your force equations involving the reactions. Instead of writing a moment equation about the origin, pick one of the reaction points and write a moment equation for it.

3. Sep 23, 2013

pongo38

Best to take moments about an axis rather than a point.

4. Sep 23, 2013

Woopydalan

If I take the moment about the X-axis, does that mean the moments caused by all forces that originate at points on the axis (A_y and A_z) go to zero, or is it that the moment caused by forces that act along the x-axis (B_x) go to zero, or is it something else??

5. Sep 23, 2013

Woopydalan

I did the moment around point B, I get Cy = -Ay, which if I put into the summation of F_x equation, I get 1200 = 0, so something is wrong here

Attached Files:

• 4.74 attempt 2.pdf
File size:
247.7 KB
Views:
40
Last edited: Sep 23, 2013
6. Sep 24, 2013

Woopydalan

Solved it. I just made one sign error in the matrix on one of my determinants!

Last edited: Sep 24, 2013


text:
# About fixing the potential on the surface of a conductor

In Purcell's Electricity and Magnetism, p.116 section 3.3, the author spoke about Laplace's equation and said that the boundary conditions for the potential$$\,\phi$$ on the surface of the conductor may be fixed:

In a real system the potentials may be fixed by permanent connections to batteries or other constant-potential "power supplies."

Does it mean then if I had a conductor like this one:

I would be able to set the potential $$\phi$$ on its surface to a given value by connecting it in the way below?


text:
# Re: accelerated movies

> What tool did you use to convert BMP files to a MPEG file?

bmp ? do i look like someone who would use micro$oft format ? considering pgm aint much different, i wont argue on that :) so : i save all my picts (a1, a2, a3, ...) in a directory. These are all in the pgm format (man pgm if u installed netpbm for more details). These picture are usually too dark so i run a /whatever_path/ImageMagick/convert -normalize$i $i.pgm;\rm$i
to expand to 255 colors the histograms of all these pictures.
Then just ask mpeg_encode to take all the a*.pgm files and it
will give u the .mpg in return.
Can be viewed using mpeg_play or xanim.
jmfriedt@ens-lyon.fr

Follow-Ups:


text:
Time dilation on accelerated reference frames

1. Aug 28, 2007

hover

Hey all,

I made a post earlier today on gravitational time dilation. It got me thinking that in any accelerated frame of reference there must also be a time dilation, due to the equivalence principle. This can simply be a elevator accelerating through space or it can be caused by centripetal force. I was wondering what these equations are. I've tried searching for these equations but every time i searched, i somehow end up with the equation for gravitational time dilation.

What are the equations?

2. Aug 28, 2007

Loren Booda

This reminds me of the twin paradox, which to solve one must somewhere apply acceleration to return to Earth. It requires more than just special relativity to reckon. Acceleration involves general relativity, but some simple solutions are known.

3. Aug 28, 2007

bernhard.rothenstein

acceleration time dilation

Could that be of interest arXiv:physics/0607288v1 [physics.class-ph]

4. Aug 29, 2007

SpitfireAce

I don't get the question... according to the equivalence principle an accelerated reference frame is equivalent to a uniform gravitational field... so the equations for time dilation in an accelerated reference frame would be the same as those for the corresponding uniform gravitational field, right?...

5. Aug 29, 2007

hover

The answer of both equations would be the same because both situations are the same/equivalent. But say for example someone is in a box accelerating in space. He knows he is truly accelerating through space because before he got in the box it was already accelerating. Since he knows he is accelerating and not in a gravitational field, he knows that there is no planet with mass pulling him. So no mass and no gravitational constant. In the gravitational time dilation equation, there is variable M for mass and G for gravitational constant-

t'=t/sqrt(1-(2GM/rc^2))

6. Aug 29, 2007

pervect

Staff Emeritus
Here's one simple way to get the answer. Do your work in an inertial frame, then there is no time dilation except for that due to velocity.

The rotating case is particularly well suited to this approach, where you probably have all the dimensions measured in the lab frame anyway.

In an inertial frame, there is no gravity, or gravity like forces, so there is no time dilation. It's important to realize this, to avoid false "double counting" of the effects.

In the elevator, it may be natural to use non-inertial coordinates (much more so in the rotating case).

I'm going to skip over a long discussion of how these non-inertial coordinates are defined, and skip on to the answer using them.

It turns out that the accelerating elevator can be described by the Rindler metric.

ds^2 = -(1+gz) dt^2 + dz^2

This metric actually completely defines the coordinate system used (this may not be obvious to you yet, unfortunately).

The result of this metric is that clocks tick at a rate of 1+gz in this coordinate system. You can think of the z coordinate as being the altitude, so "higher" clocks tick faster. Lower clocks tick slower, and eventually stop at some negative value of z. This marks the "rindler horizon", which is a topic in and of itself.

You should think of the coordinate system of an accelerated observer as being only defined for points "close enough" to the observer. Any point that is lower than the Rindler horizon is "too far away".

Note again that the very concept of time dilation depends on the coordinates you use, it's not a "geometric" concept. So in the elevator case, it's only when one uses non-inertial coordinates that one sees anything simlar to gravitational time dialtion, in an inertial coordinate system there is none.


text:
# Estimating a sinc period and width of a rectangular pulse

For a rectangular pulse that has as a fourier transform of a sinc function like so:

How can I estimate its sinc period in the frequency domain and compute the width of the rectangular pulse in the point-spread function?

I would like to do it in Matlab (if necessary), but I have no idea how to start off.

Let's look at what is generally known:

Time domain is generally inversely related to frequency domain in as much as if you have a narrow width in the time domain the frequency domain spectrum will be wider. and vice versa. When you have narrower time domain pulses you need higher frequencies to represent it in the frequency domain therefore more bandwidth -> wider plot.

Looking at your sinc function to find when the function will go to zero you simply need to look at the sin aspect of it. At what values does sin = $0$ ? Answer $0, \pi , 2\pi , 3\pi$ ... That means the value ${\tau}f = 0, \pi.2\pi,3\pi ...$

There is one exception here, as ${\tau}f$ -> $0$ sinc ->$1$ becasue the sin and the $\tau$ cancel in the limit. so you only need to look at $\pi,2\pi,3\pi, ...$

Lets pick $\tau=1$ then the rect function will run from $-\frac{\tau}{2}$ to $\frac{\tau}{2}$ and the first zero in the frequency domain will be at f=1.

Lets pick $\tau=2$ then the rect function will run from $-\frac{\tau}{1}$ to $\frac{\tau}{1}$ and the first zero in the frequency domain will be at f=$\frac{1}{2}$.

a wider time pulse means a narrower frequency band.

Lets pick $\tau=\frac{1}{2}$ then the rect function will run from $-\frac{\tau}{4}$ to $\frac{\tau}{4}$ and the first zero in the frequency domain will be at f=$\frac{2}{1}= 2$.

a narrower time pulse means a wider frequency band.

• Thanks! Just one thing, how do you determine where the rect function will from and to? Like with $\tau=2$, how did you know that the rect function will run from time $-\frac{\tau}{1}$ to $\frac{\tau}{1}$? – xenon Mar 14 '13 at 15:53
• From the definition ... substitute 2 into Tau in that defn. – placeholder Mar 14 '13 at 15:55
• Pardon my slowness. I am very new to this. I have been trying to substitute but I'm confuse with how after substituting I still get Tau. Just in case I substitute into the wrong part of the definition, which part of the equation should I use to sub Tau into? – xenon Mar 14 '13 at 17:15


text:
QUESTION

# ECE 353 Week 3 DQ 1 Phonological Acquisition

This file of ECE 353 Week 3 Discussion Questions 1 Phonological Acquisition includes:

Watch the talk by Patricia Kuhl, entitled The Linguistic Genius of Babies.

How do her findings support language acquisition in regard to phonology as it is described in Chapter Six of your text? Include a discussion of the need for a social context when developing phonological awareness and why. How did your own social context impact your phonological awareness?

Guided Response: Review several of your classmates

• @
Tutor has posted answer for $5.19. See answer's preview$5.19


text:
# The point loads are placed at the fixed positions shown in the figure and they are...

###### Question:

The point loads are placed at the fixed positions shown in the figure and they are live loads. A B D E k w C (centre) a b f' = 32 Mpa fas= 3 MPa fsy = 500 MPa E = 200 GPa E = 28600 MPa A a Cross section The following values are used for the question. • 1 = 3 m load before cracking • 12 = 2.5 m • P = 2 kN • a = 50 mm G = 2 kN/m . b = 500 mm • Q=0 kN/m • c = 450 mm • Asc = 620 mm • Ast = 3200 mm? load after cracking • P=15 kN . G = 15 kN/m • Q=15 kN/m Now, the live load increases gradually and the moment at the critical section just exceeds the cracking moment (Mc) but the compressive section of the concrete is still governed by the linear elastic region. It should be noted that in the flexural design, the cracking of the concrete section starts where the tensile stress reaches the tensile strength. It is assumed that the cracks then propagate rapidly up to the entire tension section (up to the neutral axis) and this cracked concrete section cannot resist the tension. In reality, concrete sections between the primary cracks can still resist some tensile stresses as shown in the figure below, which should be considered in the displacement design. However, in the flexure design, we design the critical section where the largest moment occurs and thus it is reasonable to have such primary cracks in this critical section. In the linear elastic region, the transformed area method do Por is still valid. However, we may solve the quadratic equation for calculating the neutral axis distance from the top surface. a A a A NA b NA. b A a Stress profile (concrete) Cross section before cracking cracks A, a Stress profile (concrete) Cross section just after cracking Longitudinal Section secondary crack Neutral axis primary cracks M M concrete in tension 1. The new neutral axis is in the cracked section under the elastic region on the compressive section. dn (distance to the top surface) = mm (3 decimals) (2 marks) 2. Cracked second moment of areas ler *106 mm (3 decimals) (2 marks)

#### Similar Solved Questions

##### Can someone explain how it can be both answers? 1:2025 SPRE 14. As long as there...
can someone explain how it can be both answers? 1:2025 SPRE 14. As long as there are multiple alleles of a gene in a population, why will the frequencies of the alleles always change over time? a. Because natural selection changes all allele frequencies in populations over time. b. Because geneti...
##### Onsider the following information about three stocks: State of Economy Probability of State of Economy Rate...
onsider the following information about three stocks: State of Economy Probability of State of Economy Rate of Return if State Occurs Stock A Stock B Stock C Boom 0.20 0.20 0.32 0.54 Normal 0.45 0.18 0.16 0.14 Bust 0.35 0.02 −0.34 −0.42 a-1. If yo...
##### Answer please Assume that the readings at freezing on a batch of thermometers are normally distributed...
answer please Assume that the readings at freezing on a batch of thermometers are normally distributed with a mean of 0°C and a standard deviation of 1.00°C. A single thermometer is randomly selected and tested. Find the probability of obtaining a reading less than 0.02°C. P(Z < 0....
##### [The following information applies to the questions displayed below.] Glasgow Enterprises started the period with 80...
[The following information applies to the questions displayed below.] Glasgow Enterprises started the period with 80 units in beginning inventory that cost $2.70 each. During the period, the company purchased inventory items as follows: Purchase No. of Items Cost 1 250$ 3.20 2 110 $3.30 3 60$ 3.7...
##### JavaFX Write a program that displays a histogram to show the number occurrences of each number...
JavaFX Write a program that displays a histogram to show the number occurrences of each number in an input sequence. There should be a text filed to accept users input numbers and button to show the histogram. Enter your student number (9 digit number) to generate your unique histogram. A student wi...
##### What is the rate and unit rate of 420 miles in 7 hours?
What is the rate and unit rate of 420 miles in 7 hours?...
##### A time study analyst observes a worker over a period of time and calculates that the worker requi...
A time study analyst observes a worker over a period of time and calculates that the worker required 60.0 minutes per unit produced. The worker was rated by the analyst at a 120 percent performance rating, and an 15 percent allowance for personal time and fatigue applies. (1) What is the standard ti...
##### What whole number does n stand for if the number sentence bleow is true
what whole number does n stand for if the number sentence bleow is true?[n+5]+[3times2]=18...
##### A function of human resource management that integrates the use of training and employees and career...
A function of human resource management that integrates the use of training and employees and career development efforts to improve individual, group, and organizational effectiveness, is best known as: a) Strategic human resource formation (SHRF) Ob) Strategic human resource orientation (SHRO) c) H...


text:
# Seminar by Erik Backelin (Universidad de los Andes, Colombia)

Title: Singular localization of g-modules

2013.11.01 | Christine Dilling

 Date Wed 13 Apr Time 15:15 — 16:15 Location Aud D3

Abstract:

(Joint with K. Kremnizer) Let $\mathfrak{g}$ be a complex reductive Lie algebra. We prove a Beilinson-Bernstein localization theorem for the category of $\mathfrak{g}$-modules at a singular central character using a sheaf of extended differential operators on a parabolic flagmanifold. This has earlier been done in finite characteristic by Bezrukavninkov, Mircovic and Rumynin. As an application we generalize an equivalence of Soergel between a regular block in category $\mathcal{O}$ and a category of Harish-Chandra bimodules to singular blocks.

In subsequent talks we will do singular localization for quantum groups using quantized versions of the constructions giving here.

Seminar


text:
# An Homage to Gauss and His Model of the Earth’s Magnetic Field

CSHPM Notes
June 2020 (Vol. 52, No. 3)

CSHPM Notes bring scholarly work on the history and philosophy of mathematics to the broader mathematics community. Authors are members of the Canadian Society for History and Philosophy of Mathematics (CSHPM). Comments and suggestions are welcome; they may be directed to either of the column’s co-editors:

Amy Ackerberg-HastingsIndependent Scholar (aackerbe@verizon.net)
Hardy GrantYork University [retired] (hardygrant@yahoo.com)

Carl Friedrich Gauss (1777–1855) was not only the prince of mathematicians, but also an applied mathematician who contributed to the development of the least squares method, numerical solutions of systems of linear equations, numerical solutions of integrals, the theory of interpolation, the Fast Fourier Transform (FFT), systems of ordinary differential equations, and more. He was also a mathematical physicist and a talented experimenter who conducted research in astronomy, geodesy, and geomagnetism [1, 2, 4, 6, 8]. In this study, we are interested in Gauss’s work on the terrestrial magnetic field. He was the first to model mathematically the magnetic field at the surface of the Earth, and to find numerical results. His model proved to be the most ambitious project of applied mathematics of its time. It is a perfect example of Joseph Fourier’s view of natural philosophy: “Profound study of nature is the most fertile source of mathematical discoveries” [3, p. 7]. However, this particular contribution by Gauss has been largely overlooked by the historians of mathematics, although it has been noted by geophysicists.

Erroneous explanations of the terrestrial magnetic field had existed for centuries. Christopher Columbus believed that the Polar star attracts a magnetic needle. Others thought that a magnetic mountain exists in the Arctic. William Gilbert, the physician of Queen Elizabeth I, destroyed all previous hypotheses when in 1600 he published De Magnete, theorizing that the Earth is a gigantic magnet with a north pole and a south pole.

Although Gauss became interested in the Earth’s magnetic field by 1803, it was not until 1839 that he published “Allgemeine Theorie des Erdmagnetismus” (on the mathematical modeling of the Earth‘s magnetic field). This appeared in a very obscure journal [5]—a fact that may help explain the historians’ neglect. Gauss examined William Gilbert’s hypothesis that the Earth is a magnet, and he further postulated that the magnetic potential obeys Laplace’s equation at the surface of the Earth as well as outside of the Earth. Gauss thus needed to solve the Laplace equation in spherical coordinates for a heterogeneous spherical terrestrial surface. Therefore, he required a magnetic map as a boundary condition at the surface of the Earth. Unfortunately for Gauss, the magnetic potential was not measurable. Only the components of the magnetic field were observable. The model is [6]:

$$\nabla^2V=0, \quad V(r\to +\infty) = 0,\quad {\partial V \over \partial r}(a, \theta, \phi) = Z,$$$${1 \over r} {\partial V \over \partial \theta}(a,\theta,\phi)=X, \quad-{1\over r\sin\theta}{\partial V\over \partial \phi}(a,\theta,\phi)=Y,$$$$V(r,\theta,\phi+2\pi)=V(r,\theta,\phi).$$

Here $V$  is the potential,  $r$ the radial distance, and $a$ the earth’s radius. Therefore the declination is given by $D=\arctan(Y/X)$, the inclination will be $I=\arctan(Z/H)$ and the horizontal intensity is $H=\sqrt{X^2+Y^2}$.

Gauss used the method of separation of variables for the Laplace equation. He wrote the equation for the spherical harmonics, the formula to obtain the associated Legendre polynomials $P_n^m$, and arrived directly—without any explanations or references—at the correct solution in the form of a series of trigonometric functions and associated Legendre polynomials:

$$Y^{(n)}=g^{n,0}P^{n,0}+(g^{n,1}\cos\phi + h^{n,1}\sin\phi)P^{n,1} + (g^{n,2}\cos2\phi + h^{n,2}\sin2\phi)P^{n,2}+\cdots\quad +(g^{n,n}\cos n\phi + h^{n,n}\sin n\phi)P^{n,n}.$$

This equation represents fundamental progress over Legendre’s and Laplace’s 18th-century results on the theory of gravitation. The coefficients $g$ and $h$ came to be called the Gauss coefficients. In modern notation, we have for the complete series of the potential [5]:
$$V(r,\theta,\phi) = a \sum_{n=0}^N \left({a\over r}\right)^{n+1}\sum_{m=0}^n P_n^m(\theta)[g_n^m\cos m\phi + h_n^m \sin m\phi].$$

Gauss limited himself to N = 4, or 24 Gauss coefficients for the derivatives of the potential. He already had observations from magnetic stations all around the terrestrial globe. By a technique of interpolation, he brought back the information to the nodes of a grid where the increments of $\theta$ and $\phi$ are constant [5, pp. 631–632]—creating one of the first examples of tessellation! This technique of bringing the information to the nodes of a grid is now called an objective analysis. Gauss selected 12 nodes on each of seven circles of latitude, for a total of 84 nodes. He then decomposed the problem by working from the colatitudes $\theta =$ constant, and by doing a harmonic analysis. For example, he had:

$$X = \sum_{n=1}^4 \sum_{m=0}^n (g_n^m \cos m\phi + h_n^m \sin m\phi) {dP_n^m\over d\theta}, \quad\rm$$$${i.e.,}\quad X=\sum_{m=0}^4(\alpha_{mx}\cos m\phi + \beta_{mx}\sin m\phi).$$

Consequently, he calculated nine Fourier coefficients $\alpha_m,\beta_m$ per circle of latitude. In $\alpha_{mx}$, the subscript $x$ refers to the component $X$ of the magnetic field. These calculations had to be repeated for the seven circles of latitude, for a total of 63 coefficients. But he had to do the computations for the three components of the magnetic field, for an overall total of 63 $\times$ 3 = 189 coefficients. By identification, he had [4]:

$$\begin{Bmatrix}\alpha_m \\ \\ \beta_{mx}\end{Bmatrix} =\sum_{n=m}^4 \begin{Bmatrix}g_n^m\\ \\h_n^m\end{Bmatrix}{dP_n^m(\theta) \over d\theta}.$$

Gauss used the 189 values to solve by least squares these systems of equations in order to find the 24 Gauss coefficients. In Figures 1 and 2 we have recreated, with a computer, Gauss’s model and his results for the declination and the total intensity for the year 1835. His map of the intensity corresponds well to modern numerical results.

In conclusion, Gauss’s memoir of 1839, by using spherical harmonics, gave a mathematical form to Gilbert’s hypothesis. It was also a major contribution to applied mathematics. Let us quote James Clerk Maxwell, who wrote in 1873 [7, p. viii]:

Gauss, as a member of the German Magnetic Union, brought his powerful intellect to bear on the theory of magnetism, and on the methods of observing it, and not only added greatly to our knowledge of the theory of attractions, but reconstructed the whole magnetic science as regards the instruments used, the methods of observation, and the calculation of the results, so that his memoirs on Terrestrial Magnetism may be taken as models of physical research by all those who are engaged in the measurement of any of the forces in nature.

Geomagnetism remains a very active field of observation and theory, not least for its continuing utility for orientation and navigation, but also due to interest in ongoing changes within the Earth. The wandering of the magnetic poles over recent decades is one example, and outside the scope of this article, but it is worth noting that geomagnetic field data continue to be assimilated and distributed in the form of spherical harmonic coefficients, so that creating any map from historical or modern data involves tables of harmonic coefficients in the manner that Gauss began.

The International Geomagnetic Reference Field (IGRF) presents its model using spherical harmonics of degree and order 13. NOAA also has a series of models extending in some cases to degree and order 720. The terrestrial and lunar gravitational fields are reported and modelled using spherical harmonics, as are such diverse data as the cosmic microwave background. Gauss was visionary in his identification and application of a methodology critical to so many modern endeavours.

Roger Godard is an emeritus professor at Royal Military College, Kingston, Ontario, who has contributed regularly to CSHPM since 1991. John de Boer is a retired Major from the Department of National Defence and a professor at RMC. His field is mainly numerical simulations of auroral events and mathematical physics.

### References

[1] Chapman, S., and J. Bartels. (1962) Geomagnetism. Vol. II. Oxford: Clarendon Press.

[2] Forbes, E. G. (1978) The astronomical work of Carl Friedrich Gauss (1777–1855). Historia Mathematica 5(2), 167–181.

[3] Fourier, J-B-J. (1822) The Analytical Theory of Heat. Translated by A. Freeman. Cambridge University Press, 1878.

[4] Garland, G. D. (1979) The contributions of Carl Friedrich Gauss to geomagnetism. Historia Mathematica 6(10), 5–29.

[5] Gauss, C. F. (1839) Allegemeine Theorie des Erdmagnetismus. In Resultate aus den Beobachtungen des magnetischen Vereins im Jahre 1838, 121–193. Leipzig. Reprint: Gauss-Werke, Vol. V, 119–193. Göttingen, 1876–1929.

[6] Godard, R., and J. de Boer. (2020) Gauss et le modèle du champ magnétique terrestre (Gauss and his model of the earth’s magnetic field). In M. Zack and D. Schlimm (eds), Research in History and Philosophy of Mathematics: The CSHPM 2018 Volume, 125–138. New York: Birkhäuser.

[7] Maxwell, J. C. (1873) A Treatise on Electricity and Magnetism. Vol. I. Oxford: The Clarendon Press.

[8] Sheynin, O. B. (1994) C. F. Gauss and Geodetic Observations. Archive for the History of Exact Sciences 46, 253–283.

Email the author: rgodard3@cogeco.ca