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Question: GENESIS: assocTestMM error for lack of sex chromosome info (not relevant for this species)
0
7 months ago by
naglemi0
naglemi0 wrote:

I'm working with a species that does not have sex chromosomes. The only part of the workflow in which I see I can note this is while building the GRM – by setting the option autosome.only to TRUE. I've looked through R documentation for assoctestMM and don't see a way to specify that there are no sex chromosomes in this dataset. How can I do this? Thanks for your time.

Here are relevant lines prior to the line causing the error, showing how I specify autosome-only to be TRUE:

grm <- snpgdsGRM(genofile, sample.id=keep, snp.id=NULL, autosome.only=TRUE, remove.monosnp=TRUE, maf=.05, missing.rate=.05, method="GCTA", num.thread=1L, with.id=TRUE, verbose=TRUE)
GRM=grm$grm row.names(GRM)=grm$sample.id
colnames(GRM)=grm$sample.id scanAnnotTRANS=ScanAnnotationDataFrame(mydat) LMMnullmod=fitNullMM(scanData=scanAnnotTRANS,outcome="SW3_TDZ",covMatList=GRM, family=gaussian) Here is the line that causes the errror, and the error: > assoc <- assocTestMM(genoData = genoData, nullMMobj = LMMnullmod, test = "Wald") Error in assocTestMM(genoData = genoData, nullMMobj = LMMnullmod, test = "Wald") : Sex values for the samples are required to compute MAF for X chromosome SNPs gwas genesis assoctestmm • 173 views ADD COMMENTlink modified 7 months ago by mconomos20 • written 7 months ago by naglemi0 Answer: GENESIS: assocTestMM error for lack of sex chromosome info (not relevant for thi 0 7 months ago by mconomos20 United States mconomos20 wrote: Hello, How many chromosomes does your species have? Also, are all chromosomes diploid for all individuals? The reason assocTestMM wants sex is so that chromosome 23 (which typically represents the X chromosome in humans) can be coded correctly for males vs. females. If all of your individuals are diploid at all chromosomes, then the easiest thing might be to make a fake sex column in your scanAnnot that says 'F' for everyone. Let me know if you have further questions. Best, Matt On Tue, Jul 24, 2018 at 7:59 PM, naglemi [bioc] <noreply@bioconductor.org> wrote: > Activity on a post you are following on support.bioconductor.org > > User naglemi <https: support.bioconductor.org="" u="" 16588=""/> wrote Question: > GENESIS: assocTestMM error for lack of sex chromosome info (not relevant > for this species) <https: support.bioconductor.org="" p="" 111361=""/>: > > I'm working with a species that does not have sex chromosomes. The only > part of the workflow in which I see I can note this is while building the > GRM – by setting the option autosome.only to TRUE. I've looked through R > documentation for assoctestMM and don't see a way to specify that there are > no sex chromosomes in this dataset. How can I do this? Thanks for your time. > > Here are relevant lines prior to the line causing the error, showing how I > specify autosome-only to be TRUE: > > grm <- snpgdsGRM(genofile, sample.id=keep, snp.id=NULL, autosome.only=TRUE, remove.monosnp=TRUE, maf=.05, missing.rate=.05, method="GCTA", num.thread=1L, with.id=TRUE, verbose=TRUE) > GRM=grm$grm > row.names(GRM)=grm$sample.id > colnames(GRM)=grm$sample.id > scanAnnotTRANS=ScanAnnotationDataFrame(mydat) > LMMnullmod=fitNullMM(scanData=scanAnnotTRANS,outcome="SW3_TDZ",covMatList=GRM, family=gaussian) > > Here is the line that causes the errror, and the error: > > > assoc <- assocTestMM(genoData = genoData, nullMMobj = LMMnullmod, test = "Wald") > Error in assocTestMM(genoData = genoData, nullMMobj = LMMnullmod, test = "Wald") : > Sex values for the samples are required to compute MAF for X chromosome SNPs > > ------------------------------ > > Post tags: GENESIS, assoctestmm, GWAS > > You may reply via email or visit https://support.bioconductor. > org/p/111361/ > -- Matthew P. Conomos, PhD. Research Scientist Department of Biostatistics Genetics Analysis Center University of Washington Seattle, WA 98105-1016, USA email: mconomos@uw.edu phone: (206) 685-8848

Thank you for this suggestion. I've gotten it to work with a modification of the approach you suggested, and am now trying to figure out how to avoid this roundabout method. When I create a column for sex and set them all to either M or F, I get the following:

> assoc <- assocTestMM(genoData = genoData, nullMMobj = LMMnullmod, test = "Wald")

Error in assocTestMM(genoData = genoData, nullMMobj = LMMnullmod, test = "Wald") :

Y chromosome SNPs should be analyzed with only males

Fortunately, there are some genotypes with NA phenotypes. If I change one of these so that one genotype is M or F and all the others are the other sex, it seems to work – no error. However, this is not ideal. In the future, we'll have datasets with no NA values, so this won't work. Can you suggest a good way to get around this?

Thanks!

ADD REPLYlink modified 5 months ago • written 5 months ago by naglemi0

In the next release of GENESIS, assocTestMM will be deprecated in favor of a new function called assocTestSingle that does not have such strict requirements about chromosome and sex coding. (If you have chromosomes called X and Y and supply a sex column it will do the right thing with frequency, if not, it will treat all chromosomes the same.) If you want to try out this function sooner, you can install the development versions of GWASTools and GENESIS. Instructions for using devel are here: http://www.bioconductor.org/developers/how-to/useDevel/

ADD REPLYlink written 5 months ago by Stephanie M. Gogarten570


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# Cross Price Elasticity Calculator

Before understanding the concept of cross price elasticity of demand we have to understand first the concepts of complementary and substitute goods or products.

Complementary products are those products that have used with each other. Without one product other product has no use. These products must be in combination form, without one other is useless or have minimum use. In simple words, both products are dependent on each other. Like Pen and Ink etc.

Substitute products are those products that have used instead of one another. Without one product other product is also useable. These products must not be in combination form, without one other have maximum or wide use. In simple words, both products are independent of each other. Like Flour and Rice etc.

The concept of cross price elasticity of demand is of different nature as compared to other concepts of elasticities. Cross price elasticity of demand shows the effect of demand in one commodity by changing or variations in the price of another commodity.

Let take an example to clear the concept. Let say we have two products pen and ink. We say the price of the pen is 5$and the price of ink is 2$. Market conditions changes and the demand for the pen increases, by following the Law of Demand it means the price of the pen has decreased. When the demand of the pen pulls up the demand for ink too. The demand for both products moves in the same manner which shows that both are complementary products.

## Cross Price Elasticity Calculator

In the below Cross Price Elasticity Calculator, you just have to put the price and demand of both products. The details of products at time point 1 and time point 2, Cross Price Elasticity Calculator will give you the results that whether the products are complementary or substitute for each other.

Cross Price Elasticity Calculator

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text: 167 10 g/unit cell u u V m d Mass of unit cell Volume of unit cell Density of unit cell. Density Functional theory, AlSb nanocrystals, Large unit cell, Ab-initio. Body-centered cell: 2 atoms/unit cell of bcc. 3) Number of Atoms Per Unit Cell:-Each Corner atom of hexagonal face is shared by 6 unit cells i. • There are two lattice parameters in HCP, a and c, representing the basal and height parameters respectively. Many of common metals (Cu,Ni,Pb. Option 2) 2 and 4. study electric charges. Results and Discussion In this section, we have discussed various simulation results such as overall unit cell structures and molecular. How writing services help students to achieve success in their life? Resolution of normal reaction in a smooth inclined plane; Hey there my name is Pranav, preparing for neet 2017 need study plan for one month for class 11&12. (d ) Number of carbon atoms in an unit cell of diamond is '4' 18. Calculate the density, if 200 g of this element contains 5 × 1024 atoms. (b) The actual packing of atoms within the unit cell. the volume of a cube. D= M/V The usual units of density are grams per cubic centimeter or. For covalently bonded molecules and transition metal complexes, chemists derive the coordination number from the chemical formula. Number of Atoms (n) and Number Density (N) The number of atoms or molecules ( n ) in a mass ( m ) of a pure material having atomic or molecular weight ( M ) is easily computed from the following equation using Avogadro's number ( N A = 6. Each level or layer of atoms, ions, or molecules in a unit cell can be represented by a two-dimensional base which is a square, hexagon, parallelogram, or other packable shape. There are 8 Au atoms on the corners: each contributes 1/8 to the unit cell so the net number of Au atoms is 8 × 1/8 = 1. From this figure, is one-half of the bond angle or = 109. NUMBER OF ATOMS PER UNIT CELL IN ACUBIC LATTICE• Simple cubic cell: 1atom/unit cell of sce. One of the two atoms is sitting on the lattice point and the other one is shifted by $\frac{1}{4}$ along each axes. Each of the corner atoms is the corner of another cube, so the corner atoms are shared among eight unit cells. Use the three unit cells shown below to answer the questions. (6) number of atoms. The unit cell • The atoms in a crystal are in a regular repeating pattern called the crystalline lattice. Table 1 and Figure 4 indicate that the cohesive energy decreased with higher energy cutoff, converged to an accuracy of 0. Cl- is located in the center of the unit cell. To a crystallographer however, cubic cells can be "face centered" or "body centered". of atoms in 1unit cell = (1/8 x 8) = 1 atom Radius of each atom = a / 2 where a is the lattice constant Volume of 1 atom (sphere) = 4/3 π r3 = 4/3 x π x (a / 2)3 = π /6 x a3 Total volume taken up by atoms in 1 unit cell = No. The BCT lattice has a non-primitive unit cell containing two lattice points, so the 2 atom motif accounts for the 4 atoms per unit cell. For now, let us take the chemical symbol of an element to stand for one atom of that element. Number of Atoms in a Cubic Unit Cell The body-centered cubic unit cell of a particular crystalline form of iron is 0. • S atoms (0,0,0) ( ½ ,½ ,0). Therefore, the 8 corner lattice points account for. As each unit cell consists of two carbon atoms, therefore, the total number of carbon atoms in the overall SWCNT unit cell can be calculated as N T (SWCNT unit cell) = 2 × N (SWCNT unit cell) = (27) 3. Body-centered unit cell 11. The mass of a unit cell is equal to the product of number of atoms in a unit cell and the mass of each atom in unit cell. The blue cell is referred to as a centred unit cell; the rest containing only one lattice point are all. The cubic unit cell is the smallest repeating unit when all angles are 90 o C and all lengths are equal (figure 12. study electron flow and its causes. How many atoms are there in a simple cubic unit cell? in a bcc unit cell? in a fcc unit cell? in the unit cell characterizing the diamond lattice? b) In terms of the lattice constant a, what is the distance between nearest-neighbor atoms in a simple cubic lattice? Assuming a cubic crystal system, make a sketch of the following planes. Number of Atoms in a Cubic Unit Cell The body-centered cubic unit cell of a particular crystalline form of iron is 0. The unit cell is the cube connecting the individual lattice points. QUESTIONS. volume bcc cell The factor is 1/2 because there are two atoms per unit cell. 52 APF = a3 4 3 1 π (0. Number of atoms per unit cell in this case = 2- Body Centered Cubic Unit Cell This is a cube with an atom at each of the 8 corners, and another atom at the centre of the body diagonals. 34 % Iron contracts upon heating. This is called a body-centered cubic (BCC) solid. The crystal basis is defined by the type, number, and arrangement of atoms inside the unit cell. The density of a metal and length of the unit cell can be used to determine the type for packing. Close-Packed Structures. Calculate the density, if 200 g of this element contains 5 × 1024 atoms. The unit cell of CsCl has one formula unit in the cell. Crystal structure - Simple Cubic, Face Centered Cubic, Body Centered Cubic, Hexagonal Close Packed, Diamond, NaCl, Zincblende, CsCl, Wurzite, Perovskite Atoms are organized in a three-dimensional regular model in a crystal in straight lines. The correct structure of CClCsCl N f No of Cl--i it ll 8 1/8 1ions per unit cell = 8 x 1/8 =1 ONE CsCl unit is present per unit cell The number of atoms in. Number of atoms per unit cell = 8 Atoms touch each other along the cross diagonal of the cube. The number of atoms per unit cell and the number of slip systems respectively, for a face centered cubic (FCC) crystal are a) 3,3 b) 3,12 c) 4,12 d) 4,48. How many atoms are there in a simple cubic unit cell? in a bcc unit cell? in a fcc unit cell? in the unit cell characterizing the diamond lattice? b) In terms of the lattice constant a, what is the distance between nearest-neighbor atoms in a simple cubic lattice? Assuming a cubic crystal system, make a sketch of the following planes. This is obvious if we compare the CsCl unit cell with the simple cubic unit cell showing the interstitial site. The tetragonal unit cell is distinguished by an axis of fourfold symmetry, about which a rotation of the cell through an angle of 90° brings the atoms into coincidence with their initial positions. An element crystallizes in a structure having FCC unit cell of an edge 200 pm. Recall that the simple cubic lattice has large interstitial sites between each 8 atoms. Keeping the following points in mind we can calculate the number of atoms in a unit cell. The unit cell is that unique part of the crystal structure such that when translated along parallel lines, generates the entire crystal. • The crystalline lattice can be reproduced by translation of the unit cell in three dimensions. How Many Atoms Per Unit Cell Is There For Gold? 2. In CsCl, however, the metal ions, instead of being surrounded by six chloride ions, are shifted into the center of each cubic element of the Cl –-ion lattice. This difference should be sufficient to avoid any possible confusion. (d ) Number of carbon atoms in an unit cell of diamond is ‘4’ 18. (a) List the unit cell(s) from the small volume A-F of the crystal. This is called a body-centered cubic (BCC) solid. This calculates the mass of 1 molecule and that is all we are interested in because that is the broad definition of a unit cell (1 atom or 1 molecule of a substance). An atom at centre of cube belongs only to this unit cell and there is only one body centre in the unit cell. Please, notice that the corner atoms are not touching one another. For a cubic unit cell, a=b=c and α=β=γ=90°. Table 3032b. Determine the number of lattice atoms per cell and the coordination number in the cubic crystal systems SC -one point at each corner i. 1) There are 4 atoms per face-centered cubic unit cell. Density of a unit cell is given as the ratio of mass and volume of unit cell. Because of the easy movement of dislocations. \n \n \n \n\n \n. Cubic Unit Cell. The unit cell of caesium chloride has one Cs + ion and one Cl- ion as calculated below. NaCl Anion forms fcc unit cell, cation occupies half of tetrahedral voids, CN is 6 for both. If the length of the body diagonal for CsCl which crystallises into a cubic structure with Cl - ions at the corners and Cs + ions at the centre of the unit cells is 7 Å and the radius of the Cs + ion is 1. Table 1 and Figure 4 indicate that the cohesive energy decreased with higher energy cutoff, converged to an accuracy of 0. This tutorial will walk you through an exploration of the sphalerite structure. APF, the atomic packing factor, which is the fraction of the volume of the cell actually occupied by the hard spheres. • Each point represents one or more atoms in the actual crystal, and if the points are connected by lines, a crystal lattice is formed. The unit cell. If one considers the lattice form of the ccp structure shown, we can see that the unit cell contains 8 atoms at the corners, and 6 atoms in the faces, specifically one in the center of each face. If you consider as hexagonal structure, which includes 3 boxes (from method 1), number of atoms inside and volume of the unit cell need to be reconsidered. The simple cubic lattice contains 1 lattice point per unit cell. Inorganic Chemistry with Doc M. (b) List the primitive cell(s) from the small volume A-F of the crystal. The unit cell edge of a substance (1 monatomic cation:1 monatomic anion) having a crystal lattice identical to that of sodium chloride has a length of 586. 172 A/atom in the rocksalt structure and 13. The elements boron and tin can crystallize in tetragonal form,…. By saying primitive unit cell does it means that the unit cell contain only 1 atom per unit cell ? Boron contains 57 atoms per unit cell. Chapter 2 - 2 Atomic Structure (Freshman Chem. • Each unit cell has eight 1/8 atom at corners and 1 full atom at the center. Because this hole is equidistant from all eight atoms at the corners of the unit cell, it is called a cubic hole The hole located at the center of the simple cubic lattice. Introduction. Calculate the number of atoms per unit cell in each type of cubic unit cell 1. Answer (4) Option (4) is incorrect because number of atoms in an unit. Simple cubic unit cell 9. The unit cell depicted below shows chloride ions surrounding a cesium ion. Both the HCP and FCC crystal structures are close-packed structure. Just as for the diamond primitive unit cell, this MIL47/53 primitive unit cell is not a nice cubic cell. 3 (a) In order to determine the number of grams in one amu of material, appropriate manipulation of the amu/atom, g/mol, and atom/mol relationships is all that is necessary, as. At first glance you might think that it is body-centered, but this would be true only if the atom at the body center was the same kind of atom as those on the corners of the cells. The basis keyword is only used when atoms (not molecules) are being created. S1 unit cell S2 unit cell S3 unit cell One hexagonal unit cell might look like any of the following. The unit cell size ( a ) is directly proportional to the atom size ( r ), so one can be defined in terms of the other and then canceled in the above expression. The sodium chloride structure is an arrangement in which each positive ion is surrounded by six negative ions, and vice versa. NUMBER OF ATOMS PER UNIT CELL IN A CUBIC LATTICE. As each unit cell consists of two carbon atoms, therefore, the total number of carbon atoms in the overall SWCNT unit cell can be calculated as N T (SWCNT unit cell) = 2 × N (SWCNT unit cell) = (27) 3. expected for a fcc unit cell? 12. Unit cell: the repeating unit in a lattice. A "body centered" crystal has a central atom and 4 atoms neatly arranged around it. By saying primitive unit cell does it means that the unit cell contain only 1 atom per unit cell ? Boron contains 57 atoms per unit cell. with $$E_{total}$$is the total energy of the unit cell, $$E_{atom}$$ is the total energy of the atom, and N as the number of atoms in the unit cell. Note: Ions are not drawn to full size. Calculate the interionic distance in CsCl. This is the problem: The substance rubidium is found to crystallize in a cubic lattice, with an edge length of 562. For a cubic unit cell, a=b=c and α=β=γ=90°. The difference between these is in the number of lattice points per unit cell. Zinc atoms are violet spheres and are located that there are four sulphur atoms around each zinc atom. Cesium chiloride, CsCl CsCl: simple cubic Cl with Cs in all cubic sites. In the unit cell of CsCl, each ion is at the center of a cube of ions of the opposite kind, so the co - ordination number is eight. structure of the CsCl-type, containing 944 atoms per unit cell. 2 nm ØThe most common types of unit cells are the faced-centered cubic (FCC), the body-centered cubic (BCC) and the hexagonal close-packed (HCP). "There are 6 atoms in the hcp unit cell. Hence The packing fraction in this case is = iii) Face-Centred Cubic Unit Cell: Besides atoms present at the corners of the cube, there are atoms at the centres of six faces, each of which is shared between two unit cells. Be the first to write the explanation for this question by commenting below. d) What is the total number of atoms in the cubic close-packed unit cell? The type of packing adopted by a metal actually has some influence on the physical properties of the metal. The coordination number for the central atom is an indicator of how many atoms or molecules form bonds with it, and that is a determinant of the molecular shape and ultimately the properties of the solid. Number of Atoms per Unit Cell. Then each atom at the corner is only an 1/8 atom. 7 and X Cl well as the number and probability distributions of the constituent electrons. of face centered atoms = 6 (no. calculate the number of atoms in a cubic based unit cell having one atom on each corner and two atoms on each body diagonal 0fm3u1kk -Chemistry - TopperLearning. ∴ 1 n 8 1 8 =×= (b) The body centred cubic cell consists of 8 atoms at the corners and one atom. If the length of the body diagonal for CsCl which crystallises into a cubic structure with Cl - ions at the corners and Cs + ions at the centre of the unit cells is 7 Å and the radius of the Cs + ion is 1. 023 x 10 23 atoms or molecules V is the volume of the unit cell expressed in cubic centimeters (cm 3 ). CsCl packing NaCl packing CsCl has a cubic unit cell. The first thing we need do is to determine the unit cell edge length from the given C—C distance. Relation between radius of cation and anion and edge length of the cube, r CS to r Cl = a√3/2; Other common examples of. Radius rule and coordination number for ionic crystals:. Also note that the cubic arrangement of cesium ions (or chloride ions) does not constitute a closest-packed structure. Examples of how to use “unit cell” in a sentence from the Cambridge Dictionary Labs. 046307 – 2 (0. Calculate the number of carbon atoms in the unit cell of diamond. Atoms on the corners, edges, and faces of a unit cell are shared by more than one unit cell, as shown in the figure below. Unit cell representation of Wurtzite structure. Calculate the number atoms present in 1 g of the crystal. For CsCl, X Cs = 0. Match box is an example of (a ) Tetragonal unit cell (b ) Orthorhombic unit cell (c) Monoclinic unit cell (d ) Triclinic unit cell 19. Solve (a) There are four copper ions and one oxygen ion located completely inside the unit cell. Density of a unit cell is given as the ratio of mass and volume of unit cell. The unit cell of CsCl has one formula unit in the cell. There are two such hexagonal faces i. For example, the molecule might possess mirror symmetry, so that one half of the atoms are related to the other half across a crystallographic mirror plane. waht is the total number of ions (Cs+ and Cl- ions) that lie within a unit cell of CsCl. 516 x 10 22 Solving the numerical problem using the log table, the rules applied are - Log (m x n) = Log m + Log n. InAs is a III-V semiconductor that. Hence The packing fraction in this case is = iii) Face-Centred Cubic Unit Cell: Besides atoms present at the corners of the cube, there are atoms at the centres of six faces, each of which is shared between two unit cells. 6 kJ/mol to f E calculate the lattice energy of. The atom at the center of the hexagonal faces is shared by two cells each. This is called a body-centered cubic (BCC) solid. in a unit cell of NaCl there are 4 formula units of NaCl moles of Nacl in 1 gm=1/58. Crystallographers also choose unit cells which contain the smallest number of lattice points. Some examples are shown in figure 1. Lithium crystallizes in BCC crystal structure. The set of directions that are related to these by symmetry are the <110> set. (The number of atoms per unit cell is given in parentheses. The unit cell is a parallelepiped defined by the lengths of its edges a, b, c and angles between them α, β, γ as shown in the figure below. In relation to the cubic unit cell, the two atoms of the basis are at (0,0,0) and along the cube diagonal. Here we consider the "stuff" in the unit cells. Distance between two nearest atoms placed at the centre of the bcc. (i) Explain Linde’s method for liquefaction of gas. But a lattice node can correspond to any number of. to enroll in courses, follow best educators, interact with the community and track your progress. To calculate the density we must divide the mass of the atoms per unit cell by the volume of the unit cell. Thereby the number of atoms per conventional unit cell is doubled from 4 to 8. The tetragonal α-FeSe structure with coordination number four can be thought of as the CsCl structure with an elongated c axis and coordination number of eight. You should have completed the CsCl and NaCl tutorials first. Contribution towards One Unit Cell. This topic briefly explains about the number of atoms in a unit cell and the types of unit cell. see different ways to represent electron flow in electric circuits. of atoms in a unit cell 1. unit cell 4 atoms 22 23 u u m u u V a3 4. 023 x 10 23) / 27 atoms of Aluminium can be obtained as, (8. •Simple cubic structure:. ) Packing Efficiency of Face Centred Cubic Crystal Lattice (FCC): In a face centred cubic lattice the eight atoms are located on the eight corners of the cube and one at the centre of the cube. The maximum packing density occurs when the atoms have a radius which equals half of the side of the unit cell. BCC describes the structure and atoms needed to form a single unit cell. 5 Crystal Lattices and Unit Cells - Bravais Lattices - Primitive and Centered Unit Cells 1. • These three-dimensional configurations of points used to describe the orderly arrangement of atoms in a crystal. Gallium atoms are shown in black and arsenic atoms are shown in red. Using the atomic radius of 1Å and your data on the nearest neighboring atoms, calculate the dimensions of the unit cell (this should include the lengths of the unit cell axes and the total volume). restriction on numbers or positions of nearest-neighbor atoms Þ large number of nearest neighbors and dense atomic packing ØAtom (hard sphere) radius, R, defined by ion core radius- typically 0. 232 A/atom for the CsCl structure. The set of directions that are related to these by symmetry are the <110> set. As a principal rule, any given atom can be transferred to one or more atoms, of the same type and with the same surroundings, within the unit cell by symmetry operations. When you do this, you need to remember that the unit cell repeats. Since the diamond lattice is face-centered cubic, the equation can easily be solved by making substitutions. Question: Identify the types of lattice and number of atoms per unit cell for CsCI and BaTiO3 (above 120 C) CsCl crystallizes in a cubic unit cell. Since ordered CsCl-type systems have no nearest neighbor atoms of the same kind, in. Note that although different materials may have the. CsCl crystallizes in a unit cell that contains a Cs+ ion at the center of the cube and a Cl- ion at each corner. Number of Atoms per Unit Cell Unit Cell Type Net Number Atoms SC 1 BCC FCC 2 4 Comparing NaCl and CsCl • Even though their formulas have one cation and one. In a _____ cubic, 4 atoms lie within each unit cell 2,1 CsCl crystallizes in a unit cell that contains the Cs+ ion in the center of a cube that has a Cl- at each corner. Cubic unit cells of metals show (in the upper figures) the locations of lattice points and (in the lower figures) metal atoms located in the unit cell. Each cubic cell has 8 atoms in each corner of the cube, and that atom is shared with 8 neighboring cells. why z no of atoms present per unit cell for cscl is 1 - Chemistry - TopperLearning. Directly edit the coordinates and/or offsets of any selected atoms. [The space lattice of diamond is FCC] Q. In addition there are eight oxygen ions on the corners of the unit cell. In the face-centered unit cell, there are also atoms or ions in the center of each of the six faces of the unit cell. 795 å Atomic positions: Ti at (0, 0, 0) Ca at (1/2, 1/2, 1/2) O at (1/2, 0, 0) Back to top. It can be easily seen that each ion on the edge is shared by four unit cells and its contribution per unit cell is ¼ , while the ion at the body centre contributes only to one unit cell. " This would mean that the total parts of "A" would add to ____ atom(s). 5 Crystal Lattices and Unit Cells - Bravais Lattices - Primitive and Centered Unit Cells 1. Calculate the number of atoms in a cubic unit cell having one atom on each corner and two atoms on each body diagonal. Each sphere represents an atom or an ion. 02 x 10^23, and you will get your number of atoms. In FCC unit cell atoms are present in all the corners of the crystal lattice. 75 , which means that = 90 54. CsCl, CsBr. In a unit cell, every constituent particle( atom, molecule or ion ) has a specific and fixed position called lattice site. In a body-centered cubic (bcc) unit cell, the atoms are present in the body-center besides the ones that are at its corners that wholly belongs to the unit cell in which it is present. This is incorrect. Because there are a total of eight one eighth volume spheres in the cell the simple cubic unit contains one net particle. If more than two ‘ ion pairs ‘ come together, ‘a three-dimensional lattice is built up. CsCl has the bcc arrangement and its unit cell edge length is 400 pm. The primitive GaN unit cell contains 4 atoms, in the case of the wurtzite structure (space group ), and 2 atoms, in the case of the zinc blende structure (space group ). Density of a unit cell is given as the ratio of mass and volume of unit cell. In a _____ cubic, 4 atoms lie within each unit cell 2,1 CsCl crystallizes in a unit cell that contains the Cs+ ion in the center of a cube that has a Cl- at each corner. Calculate the density of CsCl given that the unit cell dimension is 4. In a simple cubic lattice, the unit cell that repeats in all directions is a cube defined by the centers of eight atoms, as shown in Figure 10. For the purposes of these initial calculations we will define the structures as follows:. atoms hexagonal unit. This misconception is easy to make, since there is a center atom in the unit cell, but CsCl is really a Non-closed packed structure type. Be the first to write the explanation for this question by commenting below. For example if you have a primitive cubic cell with atoms on the corners only. a)On the model, identify a unit cell. structure is shown below in Fig. In the unit cell, each of the constituent particle comprising of a molecule, ion, or atom carries a fixed and specific position, known as the lattice site. Hence The packing fraction in this case is = iii) Face-Centred Cubic Unit Cell: Besides atoms present at the corners of the cube, there are atoms at the centres of six faces, each of which is shared between two unit cells. 6 x 1/2 = 3. Since the sodium ion is 52. Each particle in the unit cell is called a lattice point. This is called a body-centered cubic (BCC) solid. Non-Primitive Unit Cells. For example, sodium has a density of 0. so, number of atoms per unit cell = 1 + 3 = 4 Actually, A slip system is composed of a slip plane and a slip direction. Body-centered unit cell 11. The B atoms can be found at the 12 edges and at the center of the unit cell as shown below. For face centered cubic (FCC), z=4. For body centred cubic unit cell, the coordination number is 8:8. Contents of a Unit Cell. 5% the size of the chloride ion, we can predict that the chloride ions will occupy the lattice points that define the face centered cubic unit cell, with the sodium ions occupying the octahedral holes. Examples of unit cell structures. An important feature of the unit cell is the number of lattice points it contains. Problem #8: Sodium crystallizes in body-centered cubic system, and the edge of the unit cell is 430. You are given two identical looking cubes of a metal alloy. "There are 6 atoms in the hcp unit cell. 19 g/cm3 and a unit cell side length a of 3. volume bcc cell The factor is 1/2 because there are two atoms per unit cell. Each of these crystalline solids consists of a unit cell which is a small group of atoms that contain unique features. 023 x 1023 atoms 152 g 1 mol = 5. How many number of molecules per unit cell which crystallizes in the form of end face centred (monoclinic) lattice with a molecule at each lattice. 8 points per unit cell. There are 1 Cs and 1 Cl in each unit cell. 75 , which means that = 90 54. This tutorial will walk you through an exploration of the sphalerite structure. Draw planes of each and identify the Bravaice lattice, P, C, I or F a) One atom per unit cell located at (x’, y’, z’) b) Two atoms per unit cell of the same type located at (0, ½, 0) and (1/2, 0, ½). Silicon crystallizes in the same pattern as diamond , in a structure which Ashcroft and Mermin call "two interpenetrating face-centered cubic. The set of directions that are related to these by symmetry are the <110> set. In addition there are eight oxygen ions on the corners of the unit cell. Each atom is surrounded by four nearest neighbors that form the vertices of a regular tetrahedron. The maximum packing density occurs when the atoms have a radius which equals half of the side of the unit cell. Which of the following alternatives are correct (A) it forms a base centered cubic unit cell (B) it forms a face centered cubic unit cell (C) its coordination number is 8 (D) its coordination number is 6 24. Number Of Atoms In A Unit Cell - Solid State video for Chemistry is made by best teachers who have written some of the best books of Chemistry. LD 110 = 1 atoms/2√2 R LD 100 = 1 atoms/2R. NUMBER OF ATOMS PER UNIT CELL IN ACUBIC LATTICE• Simple cubic cell: 1atom/unit cell of sce. nanocrystal drop to a very small value (0. Atoms at adjacent corners of this unit cell contact each other, so the edge length of this cell is equal to two atomic radii, or one atomic diameter. Contents of a Unit Cell. Cubic unit cells of metals show (in the upper figures) the locations of lattice points and (in the lower figures) metal atoms located in the unit cell. 3D-view of the bcc-structure: Conventional unit cell, primitive unit cell and Wigner-Seitz cell. The Seven Crystal Systems. There are 2 atoms per unit cell as CsCl crystallises in body centered cubic crystalline structure. 4 and its unit cell has 8 atoms. An important feature of the unit cell is the number of lattice points it contains. center for a total of TWO atoms in the unit cell -Each layer is offset from the layer before. 046307 – 2 (0. CsCl has the bcc arrangement and its unit cell edge length is 400 pm. • Cations surround themselves with as many anions as possible and vice-versa. In o-La 2 Ni 2 In m = n = 4, but intergrowing of the LaIn and LaNi 2 slabs are in the new mode. In order to describe all. •Then all you need to know is what types of atoms, ions or molecules lie on each point. The atoms on the periphery of the top and bottom are each shared by six cells, and. atoms, for example, around the center of an FCC unit cell. CsCl has the bcc arrangement and its unit cell edge length is 400 pm. Number of atoms per unit cell : Body Centered Cubic Unit Cell. To calculate the density we must divide the mass of the atoms per unit cell by the volume of the unit cell. Body-centered cell: 2 atoms/unit cell of bcc. 3D-view of the bcc-structure: Conventional unit cell, primitive unit cell and Wigner-Seitz cell. The unit cell • The atoms in a crystal are in a regular repeating pattern called the crystalline lattice. Number of atoms per unit cell •If a metal crystallized in a simple cubic unit cell, there would be a metal atom on each of the eight corners of the cell. CsCl has a simple cubic lattice. of atoms present in 200gm of the element. Atoms/ions/molecules are often located at lattice points. Each particle in the unit cell is called a lattice point. CaTiO 3 Perovskite Space group: Pm-3m Number: 221 Pearson symbol: cP5 Unit cell dimensions: a = 3. The size and shape of a unit cell is described, in three dimensions, by the lengths of the three edges (a, b, and c) and the angles between the edges (α, β, and γ). To calculate the density we must divide the mass of the atoms per unit cell by the volume of the unit cell. Option 4) 1. The cell looks the same whether you start with anions or cations on the corners. How many formula units of KBr are there in a unit cell? Does the unit cell have a NaCl or CsCl structure? The number of atoms in a unit cell having f. This is correct. Cesium chiloride, CsCl CsCl: simple cubic Cl with Cs in all cubic sites. Therefore, the 8 corner lattice points account for. The Seven Crystal Systems. of the unit cell is known? 3 A ZM * d a N Hint: 10. CsCl has a simple cubic lattice. Volume 6 atoms per unit cell. The volume of the unit cell then is the edge cubed (edge 3). 125 nm n = 2 atoms/unit cell. how to calculate the number of atoms in a hexagonal unit cell. (a) How many aluminum atoms are in a unit cell? (b) What is the coordination number of each aluminum atom? (c) Estimate the length of the unit cell edge, a, from the atomic radius of aluminum (1. BCC describes the structure and atoms needed to form a single unit cell. center for a total of TWO atoms in the unit cell -Each layer is offset from the layer before. n = number of atoms/unit cell or n’= number of formula units/u. NaCl Anion forms fcc unit cell, cation occupies half of tetrahedral voids, CN is 6 for both. These quantities are referred to as the lattice parameters of the unit cell. a) 1,8 b) 2,1 c) 1,1 d) 2,2 e) 2,4 Now from what i could tell it is a body center cubic unit cell. Calculate the density of CsCl given that the unit cell dimension is 4. Each unit cell has 1 formula unit of CsCl structure. Plan We can calculate the volume taken up by atoms by multiplying the number of atoms per unit cell by the volume of a sphere, 4 r3/3. Assuming the shape of an atom to be a Assuming the shape of an atom to be a sphere, any type of packing of the atoms within a unit cell would leave space in between the atoms. Using the data given below and the heat of formation ()H ) of -1214. A unit cell consists of a cube in which there are A atoms are at the corners and B atoms at the face centres and A atoms from 2 corners are missing what is the simplest formula of compound Calculate no. 25g of an organic compound; Which of the following is low boiling liquid. Also note that the cubic arrangement of cesium ions (or chloride ions) does not constitute a closest-packed structure. Note that although different materials may have the.


text:
# Nagaoka ferromagnetism observed in a quantum dot plaquette

## Abstract

Engineered, highly controllable quantum systems are promising simulators of emergent physics beyond the simulation capabilities of classical computers1. An important problem in many-body physics is itinerant magnetism, which originates purely from long-range interactions of free electrons and whose existence in real systems has been debated for decades2,3. Here we use a quantum simulator consisting of a four-electron-site square plaquette of quantum dots4 to demonstrate Nagaoka ferromagnetism5. This form of itinerant magnetism has been rigorously studied theoretically6,7,8,9 but has remained unattainable in experiments. We load the plaquette with three electrons and demonstrate the predicted emergence of spontaneous ferromagnetic correlations through pairwise measurements of spin. We find that the ferromagnetic ground state is remarkably robust to engineered disorder in the on-site potentials and we can induce a transition to the low-spin state by changing the plaquette topology to an open chain. This demonstration of Nagaoka ferromagnetism highlights that quantum simulators can be used to study physical phenomena that have not yet been observed in any experimental system. The work also constitutes an important step towards large-scale quantum dot simulators of correlated electron systems.

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## Data availability

The datasets obtained from the measurements described in this work are available in the repository Zenodo with the identifier https://doi.org/10.5281/zenodo.3258940.

## Code availability

The code used to plot the datasets and implement the models used to reproduce all the figures in the main manuscript is available in the repository Zenodo with the identifier https://doi.org/10.5281/zenodo.3258940.

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## Acknowledgements

We acknowledge input and discussions with M. Chan, S. Philips, Y. Nazarov, F. Liu, L. Janssen, T. Hensgens, T. Fujita and all of the Vandersypen team, as well as experimental support by L. Blom, C. van Diepen, P. Eendebak, F. van Riggelen, R. Schouten, R. Vermeulen, R. van Ooijik, H. van der Does, M. Ammerlaan, J. Haanstra, S. Visser and R. Roeleveld. L.M.K.V. thanks the NSF-funded MIT-Harvard Center for Ultracold Atoms for its hospitality. This work was supported by grants from the Netherlands Organisation for Scientific Research (FOM projectruimte and NWO Vici) (J.P.D., U.M., L.M.K.V.), the European Research Council (ERC-Synergy) (V.P.M., L.M.K.V.), the Postdoctoral Fellowship in Quantum Science of the Harvard-MPQ Center for Quantum Optics and AFOSR-MURI Quantum Phases of Matter (grant number FA9550-14-1-0035) (Y.W.), the Swiss National Science Foundation (C.R., W.W.) and The Villum Foundation (M.S.R.).

## Author information

B.W., M.S.R., E.D. and L.M.K.V. had equal contribution in conceptualization. Experimental investigation and methodology was performed with equal contribution from J.P.D. and U.M. Theoretical investigation was led by J.P.D., V.P.M. and B.W. (extended Hubbard models), and Y.W. (ab initio model). J.P.D. led the data curation and software, with support from U.M. and V.P.M. J.P.D. and U.M. had equal contribution in the formal analysis, with support from V.P.M. L.M.K.V. led the funding acquisition and supervision. U.M. led the resources (device fabrication) with support from J.P.D. C.R. and W.W. led the resources (heterostructure growth). J.P.D. led the writing—original draft, with review and editing support from U.M., V.P.M., Y.W., M.S.R., E.D. and L.M.K.V. Please refer to the Casrai Credit Taxonomy for definitions of each of these roles.

Correspondence to L. M. K. Vandersypen.

## Ethics declarations

### Competing interests

The authors declare no competing interests.

Peer review information Nature thanks Joe Salfi, Igor Zutic and the other, anonymous, reviewer(s) for their contribution to the peer review of this work.

Publisher’s note Springer Nature remains neutral with regard to jurisdictional claims in published maps and institutional affiliations.

## Extended data figures and tables

### Extended Data Fig. 1 Charge stability diagram of the relevant voltage regions.

a, Measured charge stability diagram showing both point N and point M, as highlighted in Fig. 2a. b, Measured charge stability diagram focusing on the 2001:1101 charge transition, where spin measurements are performed (point M).

### Extended Data Fig. 2 Tuning the gate voltages to the Nagaoka condition using charge stability diagrams.

a, Sample charge stability diagram where we have highlighted the visible interdot transitions, where the electrochemical potentials of two dots become resonant (that is, an electron is allowed to tunnel between the two dots). b, Charge stability diagram similar to a, where we have modified gates P1 and P3 such that the interdot transitions appear at different locations in the diagram. Dashed black lines delimit the regions with a fixed total electron occupation in the system. c, In this diagram, gates P1 and P3 have been tuned to observe the Nagaoka condition, where the three visible interdot transitions are aligned in the three-electron configuration. The intersite interaction in the system provides an effective isolation from the reservoirs for a narrow range of gate voltages, such that the system can remain stable with three electrons in the resonant configuration.

### Extended Data Fig. 3 Effects of spin-coupling mechanisms.

Calculated spectra of the system in the region of pε close to the level crossing of the s = 1/2 and s = 3/2 energies, comparing the effects of different mechanisms for spin coupling. a, Spectrum without any spin coupling effects. b, Spectrum including only spin–orbit coupling (SOC) effects. c, Sample spectrum with both spin–orbit and hyperfine induced Overhauser field gradients, using a single combination of hNa fields (as defined in the Supplementary Methods) selected from a normal distribution with standard deviation δN = 73 neV. The Supplementary Methods describe the implementations of these spin-coupling terms in the theoretical model.

### Extended Data Fig. 4 Characterization of the Nagaoka condition.

a, Average PT in the detuning region 1.00 < pε < 1.01 for 40 values of τramp within the same range shown in Fig. 3. Solid lines are fits using the time evolution simulations described in the Supplementary Methods, for different values of distance $$\ell$$ between neighbouring dots. Inset shows the unscaled results of the time-evolution simulations, where the probability of s = 3/2 is the sum of the lowest four eigenstate probabilities from the final evolved state. b, Thermal relaxation measurements. PT is measured for increasing wait times at point N, for diabatic (blue) and adiabatic (red) passages. Solid lines are exponential fits as guide to the eye.

### Extended Data Fig. 5 Ab initio simulations from 2D to 1D.

a, Schematic of the methodology used in the ab initio simulations to reproduce the effect of the four-dot system transition from a 2D plaquette to a 1D chain. We gradually vary the angle θ, which effectively varies the distance between two of the dots. b, c, The ground-state energy and spin configuration (b) and the ferromagnetic to low-spin energy gap ΔE as a function of θ (c). The ground state soon becomes a low-spin state for the rotating angle at 0.3°.

### Extended Data Fig. 6 Ab initio simulations for local energy offsets.

a, Schematic of the methodology used in the ab initio simulations to reproduce the effect of a local energy offset. The amplitude of the potential V of one of the quantum wells is changed by an amount dV. The variation of the single-well potential by positive or negative dV gives unbalanced site energies. Besides, with the change of eigenstate basis, the hybridization and interaction parameters are also affected in the ab initio calculation. b, c, The ground-state energy and spin configuration (b) and the ferromagnetic to low-spin energy gap ΔE as a function of dV (c). When the potential detuning is dV = 0.11 meV or dV = −0.07 meV, the system undergoes a transition to a low-spin ground state. The transitions at these two directions have a different nature, as drawn in the insets. For dV > 0, the particular quantum dot is deeper and tends to trap more electrons. However, a negative dV raises the energy cost on the particular quantum well and leads to a lower probability of occupation in a three-electron system. Without the ‘mobile’ hole in the ‘half-filled’ system, the ground state becomes a low-spin state instead of a Nagaoka ferromagnetic state.

### Extended Data Fig. 7 Local energy offsets on all dots.

Same measurement as in Fig. 6, applying the ±50 μeV offset on each of the four dots. Panels correspond to offsets in dots 1 to 4, clockwise from the top left. Note that the asymmetry in the plots is related to the fact that the local energies at point M are in an asymmetric detuning configuration and we pulse linearly from this configuration to point N. As expected, the simulated energies of the different spin states at point N (pε = 1), are the same in all four plots.

### Extended Data Fig. 8 Large local offsets.

Each pair of panels show experimental measurements (left) and simulated spectra (right), where point N has been redefined such that the chemical potential of dot 1 is offset by the amount shown on the top right of each panel. Green crosses highlight the detuning points used to obtain the values in Fig. 6b. For experimental plots, these points where obtained using a peak-finding algorithm (local maxima by simple comparison with neighbouring values); for simulated plots, the points correspond to the energy-level crossings.

## Supplementary information

### Supplementary Information

This file contains Extended Fermi-Hubbard models used to simulate different experiments in the main text, Ab initio exact diagonalisation simulations of the 2×2 plaquette and Mapping 3-spin states onto 2-spin measurements.

## Rights and permissions

Reprints and Permissions

Dehollain, J.P., Mukhopadhyay, U., Michal, V.P. et al. Nagaoka ferromagnetism observed in a quantum dot plaquette. Nature 579, 528–533 (2020). https://doi.org/10.1038/s41586-020-2051-0

• Accepted:

• Published:

• Issue Date:

• ### Discrete time crystal in the gradient-field Heisenberg model

• Bikun Li
• , John S. Van Dyke


text:
# Computing the Euclidean norm: Numerical aspects

The Euclidean norm (also known as the 2-norm) is arguably the most widely used norm on vector spaces. It is also very simple conceptually: The length of a vector $$\mathbf{x} = \left[ x_1, \ldots, x_n \right]^\mathrm{T} \in \mathbb{R}^n$$ is given by its Euclidean norm $$\|\mathbf{x}\|_2$$ .

In this post, I want to talk about computing the Euclidean norm numerically, i.e. using floating-point arithmetic. As I recently found out, there is more than meets the eye to it. So let’s dive in…

# The naïve approach

The formula for the Euclidean norm can be found in any textbook. It doesn’t look very complicated, right? $$\|\mathbf{x}\|_2 = \sqrt{x_1^2 + \ldots + x_n^2} = \sqrt{\sum_{i = 1}^n x_i^2}$$ We just square all elements of $$\mathbf{x}$$ , add them up and take the square root. Sounds easy enough! One can instantly write a C function for it:

/* Uses ANSI C */
#include <stddef.h>
#include <math.h>

double euclidean_norm_naive(const double x[], const size_t n) {
size_t i = 0;
double acc = 0.0;
for (i = 0; i < n; i++) { acc += x[i]*x[i]; }
return sqrt(acc);
}

So far, so good. But we’ve all heard that floating-point math is hard. For instance, we should take rounding errors, under- and overflow into account. And indeed, if you use our naïve function to compute the Euclidean norm of a vector with very large elements, you might be unpleasantly surprised:

const double foo[2] = {1.0e200, 1.0e200};

printf("%f", euclidean_norm_naive(foo, 2)); /* inf */

Oops. This is what we call overflow. Input vector elements $$10^{200}$$ and the correct result $$\sqrt{2} \cdot 10^{200}$$ can be represented in double-precision floating-point arithmetic, but the intermediate value $$\left( 10^{200} \right)^2 = 10^{400}$$ cannot. So we get an inf instead of the correct result.

Similarly, our naïve implementation will underflow if vector $$\mathbf{x}$$ has very small elements. Hence, this function is unusable for a considerable range of inputs. How can we solve this problem?

# What would LAPACK do?

Let’s take a look at how LAPACK does this, ultimately it is the reference for doing numerical linear algebra. Here is the DNRM2 function:

      DOUBLE PRECISION FUNCTION DNRM2(N,X,INCX)
*     .. Scalar Arguments ..
INTEGER INCX,N
*     ..
*     .. Array Arguments ..
DOUBLE PRECISION X(*)
*     ..
*
*  Purpose
*  =======
*
*  DNRM2 returns the euclidean norm of a vector via the function
*  name, so that
*
*     DNRM2 := sqrt( x'*x )
*
*  Further Details
*  ===============
*
*  -- This version written on 25-October-1982.
*     Modified on 14-October-1993 to inline the call to DLASSQ.
*     Sven Hammarling, Nag Ltd.
*
*  =====================================================================
*
*     .. Parameters ..
DOUBLE PRECISION ONE,ZERO
PARAMETER (ONE=1.0D+0,ZERO=0.0D+0)
*     ..
*     .. Local Scalars ..
DOUBLE PRECISION ABSXI,NORM,SCALE,SSQ
INTEGER IX
*     ..
*     .. Intrinsic Functions ..
INTRINSIC ABS,SQRT
*     ..
IF (N.LT.1 .OR. INCX.LT.1) THEN
NORM = ZERO
ELSE IF (N.EQ.1) THEN
NORM = ABS(X(1))
ELSE
SCALE = ZERO
SSQ = ONE
*        The following loop is equivalent to this call to the LAPACK
*        auxiliary routine:
*        CALL DLASSQ( N, X, INCX, SCALE, SSQ )
*
DO 10 IX = 1,1 + (N-1)*INCX,INCX
IF (X(IX).NE.ZERO) THEN
ABSXI = ABS(X(IX))
IF (SCALE.LT.ABSXI) THEN
SSQ = ONE + SSQ* (SCALE/ABSXI)**2
SCALE = ABSXI
ELSE
SSQ = SSQ + (ABSXI/SCALE)**2
END IF
END IF
10     CONTINUE
NORM = SCALE*SQRT(SSQ)
END IF
*
DNRM2 = NORM
RETURN
*
*     End of DNRM2.
*
END

Wait, what? SCALE? Branching? Why?

Let’s try to understand what happens in DNRM2. First I want to translate it into C, because, frankly, I cannot read fixed format FORTRAN without overheating my brain1). Here is a cleaned-up C version without some fine implementation details:

double euclidean_norm_scaled(const double x[], const size_t n) {
size_t i = 0;
double absxi = 0.0;
double scale = 0.0;
double ssq = 1.0;

if (n < 1) {
return 0.0;
} else if (n == 1) {
return fabs(x[0]);
} else {
/* Inlined DLASSQ */
for (i = 0; i < n; i++) {
if (x[i] != 0.0) {
absxi = fabs(x[i]);
if (scale < absxi) {
ssq = 1.0 + ssq*(scale/absxi)*(scale/absxi);
scale = absxi;
} else {
ssq += (absxi/scale)*(absxi/scale);
}
}
}
return scale*sqrt(ssq);
}
}

The most interesting part is the inlined DLASSQ subroutine. What does it do? It took me some time to understand it, but actually it does one very simple thing: It iterates through an array once and keeps track of the largest value in it. It also calculates the sum of squares scaled with the current largest value. In other words, it computes $$s = \max \left\{|x_1|, \ldots, |x_n|\right\}$$ and $$a = \sum_{i = 1}^n \left(\frac{x_i}{s}\right)^2$$ in one iteration through the array. That’s it.

And now we see the trick used in DNRM2: We ‘extract’ the largest element of the vector and put it before the square root, replacing the textbook formula with this one: $$\|\mathbf{x}\|_2 = s \, \sqrt{\sum_{i = 1}^n \left(\frac{x_i}{s}\right)^2}$$ with $$s = \max\left\{|x_1|, \ldots, |x_n|\right\}$$ . By doing this, we avoid problems with under- and overflow we discussed earlier.

# Some examples

Let’s test these functions and see how they perform on over- and underflow. We use the following inputs2): \begin{aligned} \mathbf{x}_\mathrm{of} &= \big[ 3 \cdot 10^{200} \ \ {-4} \cdot 10^{200} \big]^\mathrm{T} & \|\mathbf{x}_\mathrm{of}\|_2 &= 5 \cdot 10^{200} \phantom{\frac{1}{1}} \\ \mathbf{x}_\mathrm{uf} &= \big[ 3 \cdot 10^{-200} \ \ 4 \cdot 10^{-200} \big]^\mathrm{T} & \|\mathbf{x}_\mathrm{uf}\|_2 &= 5 \cdot 10^{-200} \phantom{\frac{1}{1}} \\ \mathbf{x}_\mathrm{ac} &= \big[ 10^{-7} \quad 2 \quad 2 \cdot 10^7 \big]^\mathrm{T} & \|\mathbf{x}_\mathrm{ac}\|_2 &= 10^{-7} + 2 \cdot 10^7 \phantom{\frac{1}{1}} \end{aligned}

Overflow test:
exact:  5.00000000000000019e+200
naive:                       inf
scaled: 4.99999999999999951e+200

Underflow test:
exact:  4.99999999999999991e-200
naive:   0.00000000000000000e+00
scaled: 4.99999999999999991e-200

Accuracy test:
exact:   2.00000000000001006e+07
naive:   2.00000000000001006e+07
scaled:  2.00000000000000969e+07

We observe that the scaled version is much less prone to under- and overflow, allowing us to use the whole floating-point range. On the other hand, it certainly requires more operations and is sometimes less accurate, probably due to the division during the scaling process.

# Conclusion

So, what did I learn from this simple example?

First, numerical linear algebra and floating-point arithmetic is hard. Even the simplest formulæ often have to be rearranged for better numerical stability or floating-point accuracy. I think that it is often the art of numerical linear algebra to find this transformation.

Second, open-source is important not only because it’s free. It is important because anyone can peek under the hood and see how it’s done in real world.


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## Discussion Tag Cloud

Vanilla 1.1.10 is a product of Lussumo. More Information: Documentation, Community Support.

• CommentRowNumber1.
• CommentAuthorscott
• CommentTimeFeb 1st 2012
• (edited Feb 9th 2012)

I’ve had a look at Scholastica, which seems quite good. They’re run by a bunch of UChicago associated people, and seem very responsive to comments. (I went through most of their interface, using their convenient builtin feedback form to comment on rough edges. I’m sure they would appreciate others doing the same.)

I would really like to see them allow stronger arxiv integration — primarily, submitting manuscripts simply by providing an arXiv URL.

Have other people looked at Scholastica? What other systems are out there?

It would be great if we could collectively prepare a summary of the options available for review management — I think this would be a necessary part of making the case to editorial boards that it’s time to make the jump away from Elsevier.

• CommentRowNumber2.
• CommentAuthorscott
• CommentTimeFeb 9th 2012

I’m talking to the Scholastica folks on Friday Feb 10, as they want to show me the preliminary arXiv integration they’ve implemented over the last week. I’ll report back here.

• CommentRowNumber3.
• CommentTimeFeb 9th 2012
I believe that MSP wrote and now uses EditFlow -- http://msp.org/services.html -- talk to Rob Kirby to find out more.
• CommentRowNumber4.
• CommentAuthorCschaffner
• CommentTimeFeb 9th 2012
I've also been talking to the developers at Scholastica. They showed me their prelim arXiv integration today and it looks smooth to me. They seem eager to provide whatever service is required by the community and they are really fast responding! They do not seem to have enough clients yet to make concrete promises about their pricing policy.
• CommentRowNumber5.
• CommentAuthorCschaffner
• CommentTimeFeb 9th 2012
There's an open-source software called Open Journal Systems at http://pkp.sfu.ca/?q=ojs
Open Journal Systems (OJS) is a journal management and publishing system that has been developed by the Public Knowledge Project through its federally funded efforts to expand and improve access to research.

I do not have any personal experience with running or using this software. Does anybody else?
• CommentRowNumber6.
• CommentAuthorScott Morrison
• CommentTimeFeb 10th 2012

I actually talked to Rob last week about EditFlow. I’ll try to blog about what I learnt, soon.

• CommentRowNumber7.
• CommentAuthorAndy Putman
• CommentTimeFeb 10th 2012
The big problem with Scholastica is that in their model, authors pay to submit their papers. I would be very opposed to going down this road with mathematics.
• CommentRowNumber8.
• CommentAuthorNoah Snyder
• CommentTimeFeb 10th 2012
Andy could you explain that further? When I look at things like PLoS ONE it seems that the price point is higher than makes sense for math, but at a lower price point it seems like a totally reasonable sustainable model. The ArXiv's cost is $7 per submission, say we go up an order of magnitude and charge$70 per paper on average (with some sort of price discrimination: more if you have a grant, less if you're unemployed, or a grad student, or in a developing country). I'm not sure if that's enough, but if it is it seems like a totally great model to me. The only real downside I see is that it stops us from getting some money from universities library budgets that we get now. But given the decreases in library budgets I don't know how significant that loss is in the long run.
• CommentRowNumber9.
• CommentAuthorNoah Snyder
• CommentTimeFeb 10th 2012
Hrm, trying to do some back-of-the-envelope calculations with AGT. At $300 a year for an electronic subscription with say 300 subscribers that's$90,000/year. They publish roughly 100 papers per year. So that's $900 per paper. I'm not sure what their acceptance rate is, but say it's 50%, that would give you a price of$450 submission. So maybe I'm wrong in thinking you can get by with one order of magnitude above the arXiv, maybe it's two. That does make that funding model less appealing. I really would like to get a better handle on what a journal's budget looks like.
• CommentRowNumber10.
• CommentAuthorHenry Cohn
• CommentTimeFeb 10th 2012
My second-hand understanding is that for a small math journal publisher, you seem to need roughly $40-50/page for high-quality journal publishing, and that's if you're pretty efficient (which some journals aren't). You might be able to cut costs a little if you compromise on things like careful copyediting or formatting, or from amortizing fixed costs over many journals, but you won't save a lot without some real innovations. The main reason the arXiv's costs are so low is that submitting an arXiv paper hardly requires any human intervention; another is that they have huge economies of scale. • CommentRowNumber11. • CommentAuthorTerence Tao • CommentTimeFeb 10th 2012 The Journal of the AMS (and, I think, some other AMS journals) switched to Editflow a few years back. I quite like the system as an editor (it is certainly superior to Centrack, which was AMS’s in-house system), but I don’t know how smooth it is to use on the production side of things. • CommentRowNumber12. • CommentAuthorSam Nead • CommentTimeFeb 10th 2012 @Scott - I'll be interested to hear about EditFlow. Thanks! 1. I would like to mention Annotum It’s a wordpress.org based solution, coming out of the BeyondPDF workshop last year. This is not really interesting for mathematics since we’re too used to LaTeX based solutions, but it’s useful to have in mind when trying to get other sciences on board. • CommentRowNumber14. • CommentAuthorScott Morrison • CommentTimeFeb 11th 2012 • (edited Feb 11th 2012) @nsnyder, @Henry, my conversations with Rob confirm this sort of lower bound — MSP doesn’t think they could produce something like Geometry & Topology at under$50/page, and that’s after the investment in developing their own software (editFlow), and a lot of volunteer labour. Copy-editing (dubious importance, mostly it has added negative value to my own papers so far) comes in at about $30/page, so is essentially prohibitive for “low budget” journals. • CommentRowNumber15. • CommentAuthorScott Morrison • CommentTimeFeb 11th 2012 @Noah, I’ve heard that the acceptance rate at G&T is about 1 in 3. I agree this makes “pay on submission” look very unlikely. To add some context regarding Scholastica’s intentions to charge on submission, currently the biggest journals they have on board are law reviews, with about 1500 submissions and 15 acceptances per year. They think$5 per submission looks fine there, and it sounds believable, but very far from our situation.

• CommentRowNumber16.
• CommentAuthorScott Morrison
• CommentTimeFeb 11th 2012

Further, @Noah (referring to the comment above about not getting money from university budgets), I think it actually makes sense to think about journal subscriptions and author fees as coming out of closely connected pots of money. UC Berkeley, for example, will pay up to $6000 per year (max$3000 per paper) to subsidize open access costs for faculty, and my understanding is that this money passes through the library system.

Rob Kirby is currently proposing that libraries make this official policy — they promise to cover all their faculty’s submission fees, and then subscribe to journals out of whatever remains in that pot.

This may end up being a necessary approach — the big commercial publishers are very much looking forward to charging us at both ends, submitting and subscribing. See, for example, the language in the investment report quoted from Gower’s blog post:

Open Access is unlikely to hurt Elsevier in the next five years and the longer term risk is more than priced in, in our view.

• CommentRowNumber17.
• CommentAuthorNoah Snyder
• CommentTimeFeb 11th 2012
Do I understand right that the $50/page includes that$30/page? It's shocking to me that anyone thinks copy editing is worth that much.
• CommentRowNumber18.
• CommentAuthorNoah Snyder
• CommentTimeFeb 11th 2012
Hrm, looking over it again probably since they're "low cost" you're saying that they don't do copy-editing. But probably that \$50 includes printing? Looking at their pricing it appears that the costs of printing and distribution are probably only a third of the total, so electronic only would still be a lot if money.
2. This recent addition to journals of the LMS: http://education.lms.ac.uk/the-journal/ might be interesting.

It’s entirely online, open-access, combining a blog with a journal, allowing commenting via wordpress etc. Very simple, using established, reliable technology. Should be worth asking for advice.

• CommentRowNumber20.
• CommentAuthorScott Morrison
• CommentTimeFeb 12th 2012

@Noah, I’m actually a little unclear on this. I think that G&T does spend significant money on copyediting. I’ll interrogate Rob next week.

• CommentRowNumber21.
• CommentAuthorHenry Cohn
• CommentTimeFeb 12th 2012
I don't know how much they spend, but in my experience MSP has some of the best copyeditors I've ever dealt with. (AK Peters did too, before Alice and Klaus were terminated by Taylor & Francis.) This makes a real difference - at most journals I find dealing with copyeditors to be a frustrating exercise in keeping them from introducing typos or inconsistencies. However, the two times I've published in G&T I've been really impressed, so whatever they are doing, it's working.

One of my fears, as a perfectionist, is that in the end there will be no market for high-quality journals, and it will turn out that putting author-typeset papers on a web site is all the community is willing to pay for. That wouldn't be so terrible, but something would be lost.


text:
# Gravitation

## Free Fall

When an object falls from any height under the influence of gravitational force only, it is known as free fall. In the case of free fall no change of direction takes place but the magnitude of velocity changes because of acceleration.

This acceleration acts because of the force of gravitation and is denoted by ‘g’. This is called acceleration due to gravity.

### Expression for acceleration due to gravitation ‘g’.

Let mass of the object put under free fall = m.

And acceleration due to gravity = g.

Therefore, according to Newton’s Second Law of Motion which states that Force is the product of mass and acceleration,

F = m x g -----------------(i)

Now, according to Universal Law of gravitation,

F= G*(M*m)/d^2 ---(ii)

Thus, from above two expressions, we get

mg = G*(M*m)/d^2
=>g = (GMm)/(d^2*m)
=>g=(GM)/d^2 ---(iii)

Where, g is acceleration due to gravity,

G is the Universal Gravitational Constant.

M is the mass of earth.

And d is the distance between object and centre of earth.

#### When object is near the surface of earth

When an object is near the surface of earth, the distance between object and centre of the earth will be equal to the radius of earth because the distance of object is negligible in comparison of the radius of earth.

Let the radius of earth is equal to R.

Therefore, after substituting ‘R’ at the place of ‘d’ we get,

g = (GM)/R^2 ----(iv)

Since, earth is not a perfect sphere rather it has oblique shape. Therefore, radius at the equator is greater than at the poles.

Since, value of ‘g’ is reciprocal of the square of radius of earth, thus, the value of ‘g’ will be greater at the poles and less at the equator.

And the value of ‘g’ will decrease with increase of distance of object from earth.

## Calculation of value of g

We know that,

The accepted value of G is 6.673xx10^(-11)Nm^2kg^(-2)

The mass of earth, M=6xx10^(24)kg

The radius of earth, R = 6.4xx10^6m

Therefore, by using expression g = (GM)/R^2, the value of g can be calculated.

Therefore, after substituting the value of G, M and R in the expression for ‘g’ we get.

g=(6.673xx10^(-11)Nm^2kg^(-2)xx6xx10^(24)kg)/(6.4xx10^6m)^2

=>g= 9.8\ m\ s^(-2)

Motion of an object under the influence of gravitational force of earth

The expression for ‘g’ is written as g=(GM)/R^2

Since, the value of ‘g’ does not depend upon the mass or distance of an object, therefore, all objects fall over the earth with the same rate.

The equations for motion are as follows:

v=u+at ---(i)
s=ut+1/2at^2 ---(ii)
v^2=u^2+2as ---(iii)

Therefore, the equations of motion are also applied to calculate the velocity, distance, etc by replacing ‘a’ by ‘g’. After substituting ‘g’ at the place of ‘a’ we get above equations as follows:

v=u+g\t ---(i)
s=ut+1/2g\t^2 ---(ii)
v^2=u^2+2gs ---(iii)

In the calculation; initial velocity (u), final velocity (v), time taken (t), or distance covered (s), the value of ‘g’ is taken as positive in the case of object moving towards earth and taken as negative in the case of object is thrown in opposite direction of earth.


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## A note on the large sample properties of linearization, jackknife and balanced repeated replication methods for stratified samples.(English)Zbl 0745.62008

Summary: D. Krewski and J. N. K. Rao [ibid. 9, 1010-1019 (1981; Zbl 0474.62013)] considered inference for a (nonlinear) function of a vector of finite population means $$\theta = g(\bar Y)$$. For a sequence of finite populations with increasing number of strata, they demonstrate that $$\hat \theta = g(\bar y)$$ is asymptotically normal, where $$\bar y$$ is the usual unbiased stratified estimator of $$\bar Y$$. Additionally, they demonstrate that $$(\hat \theta - \theta)/v^{1/2}(\hat \theta )$$ is asymptotically a standard normal distribution, where $$v(\hat \theta)$$ is a variance estimator obtained using linearization, jackknife or balanced repeated replication (BRR) methods.
We extend their results to when the partial first derivatives $$(g_ 1(\mu ),g_ 2(\mu ),\dots ,g_ p(\mu))\equiv 0$$, where $$\mu$$ is the limit of $$\bar Y$$ with increasing number of strata. We explore the asymptotic distribution of $$(\hat \theta - \theta)/v^{1/2}(\hat \theta )$$ and show (1) that it is no longer normal and (2) that it depends upon which variance estimator is used. We describe an application of these results to hypothesis testing using complex survey data.

### MSC:

 62D05 Sampling theory, sample surveys 62F12 Asymptotic properties of parametric estimators

Zbl 0474.62013
Full Text:


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# Extreme and Mean Ratio

## The Golden Ratio is also a Golden Mean

$\large \frac{a}{b} = \frac{b}{a+b}$

The Golden ratio is a geometric mean

a geometric mean is a relation between 3 values $$a, b & c$$

in a golden mean, we set the value of $$c = a + b$$


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# Answers From The Book Information Content in the Limit Order Book for Crude Oil Futures (WTI)

### Motivation

Order book imbalance strategies have been a big alpha source in automated market making. Tick by tick observations provide important information about general market sentiment and direction, and high frequency trading firms (HFTs) have been very efficient at trading on this information at very low latency intervals.

In their recent paper on HFT strategies, Goldstein, Kwan and Philip analyzed six months of full order book and trade data for stocks in the S&P/ASX 100 and showed that order book imbalances are strong predictors of future prices. They found that HFT firms increase price efficiency in markets by trading in the same direction as the imbalance but could also use order book information to detect institutional buying or selling pressure. In addition, their results also show that HFT firms trade more aggressively during periods of high market volatility and more successful a picking off stale orders from institutional and retail investors.

This post aims to expand on the work by Goldstein et al. by investigating the predictive significance of order book imbalance on price changes, short term intraday volatilities, and realized spreads for the NYMEX WTI Crude Oil futures market to assess its implications on execution.

In addition to finding strong evidence that WTI futures prices move in the direction in the direction of order balance, we found a more effective definition to the predictive relationship between order imbalance and price changes. We also discovered an interesting causality between order book imbalance and realized spreads.

### Data and Metric Definitions

Data

Our study uses one week’s worth of tick data for WTI futures which was obtained from Bloomberg. This comprises of order book and trade data with second based time stamps that shows prices and volumes for every trade and change to best bid and ask prices from the period of 7th to 14th July 2017:

###### Fig 1 – Tick Data for NYMEX WTI Crude Oil Futures

Order Book Imbalance

Order book imbalance (OBI) is defined as the difference between the volume available at the best bid and ask prices, as a proportion of the total volume available at the best bid and ask prices. This is calculated at each point in time by:

$OBI_{t} = \frac{VolBestBid_{t}-VolBestAsk_{t}}{VolBestBid_{t}+VolBestAsk_{t}}$

Where VolBestBid and VolBestAsk are the quote volumes available at the top bid and ask levels respectively. This results in a normalized indicator for the shape of the order book, and we obtain the following distribution of values for OBI for our dataset:

###### Fig 2 – Histogram and Descriptive Statistics for Distribution of Order Book Imbalance Values

Mid Price

The mid price is defined as the average of the current best bid and ask prices being quoted for the security:

$Mid\:Price_{t} = \frac{BestBidPrice_{t}-BestAskPrice_{t}}{2}$

Short Term Volatility

In their paper, Goldstein et al. calculate volatility as the difference between the log of the highest  ask price and the log of the lowest bid price within each interval. We use their measure for calculating volatility within short intraday time frames for our study:

$Volatility_{t,t+1} = log(max(BestAskPrice_{t,t+1}))-log(max(BestBidPrice_{t,t+1}))$

In market microstructure, the realized half spread is the difference between the transaction price and the mid-price over an interval of time. This measure incorporates price impact into trading costs and is calculated using the following formula:

$Realized\:Spread_{t,t+1} = \frac{Executed Price_{t}-Mid Price_{t+1}}{Mid Price_{t}}$

### Empirical Results

Overview

In investigating the statistical properties of order book imbalance relative to the other metrics we have defined, we first focus on analyzing the information content of order book imbalance on WTI crude oil futures by using Goldstein et al.’s restricted regression model, which uses the book imbalance for the best bid and offer:

$Return=\beta _{0}+\beta _{1} OBI _{TopLevel}+\epsilon$

They calculated return as the log of the difference between the bid-ask midpoint 10 trades in the future and the midpoint price just prior to the trade. OBI is the order book imbalance immediately before the trade, based on the equation provided in the previous section. Besides applying their return definition for a range of future trade sequences, we also explore the predictability of OBI on midpoint price returns over a range of time intervals as well as the next successive midpoint price change.

We then extend this regression analysis to analyze the predictive power of order imbalance on realized spreads and short term volatility which are of relevance to price impact and execution. In the case of the volatility model, we use average values for OBI over the previous interval in order to smoothen out noise. We expect to find that order imbalance of the limit order book contains information on future realized spreads and short term volatility. Following the literature by Foucault et al. (2007) wherein they developed a theoretical limit order market in which traders had different views of future volatility given their private information. Given limit orders have options like features, they would need to be priced using the trader’s private volatility information. Upon order submission into the limit order book also disseminate their volatility information into the market place.

$Realized\:Spread=\beta _{0}+\beta _{1} OBI _{TopLevel}+\epsilon$

$Volatility=\beta _{0}+\beta _{1} Average\:OBI _{TopLevel}+\epsilon$

Price Changes

###### Fig 3 – Regression Analysis of Order Book Imbalance vs. Various Mid Price Return Definitions

Although all regressions were significant at 5%, we see that based on R-squares most of them do not provide much incremental information about future price movements. However, the model which uses the next midpoint change as a return definition stands out as the best for prediction. In their study, Goldstein et al. split their data by day and stock and obtained mean and median R-squares of 12.02% and 10.96% respectively before regressing OBI against the change in mid-price 10 trades in the future. Surprisingly, we obtained an R-square of 10.9%, which is very close to their results. Overall, this proves that the order book shape contains statistically significant information about future price movements for WTI futures.

###### Fig 4 – Regression Analysis of Order Book Imbalance vs. Various Realized Spread Definitions

Poor results were obtained from using order book imbalance to predict realized spreads. Regressions on time based definitions of realized spreads yielded insignificant results across different short term intervals (regression results on OBI vs. 5-second realized spreads shown in table). On the other hand, although using imbalance to predict realized spreads over 10 future trades yielded significant results, R-square was small.

We were interested in finding out whether OBI and realized spreads had any implications on each other at all. Granger Causality tests on OBI values and realized spreads over intervals of 10 trades suggest that instead of OBI being a significant predictor of realized spreads, the reverse was true.

###### Fig 6 – Granger Causality Analysis of Order Imbalance Values vs. Successive Realized Spreads over 10 Trades

Fitting the model the other way round showed a highly significant regression with R-square of 11.5%. In addition, realized spreads and lagged imbalance values had a correlation of 33.9%. This implies that the previous spread size causes an imbalance. A possible economic reasoning behind this could be that when realized spread delta is large, orders are not able to refresh fast enough in order to rebalance the book, which leaves a liquidity imbalance.

###### Fig 7 – Regression Analysis of Realized Spreads over 10 Future Trades vs. Lagged Order Imbalance Values

Short Term Volatility

###### Fig 8 – Regression Analysis of Short Term Volatility over Different Intervals vs. Average Order Imbalance Values

Results suggest that order imbalance does not contain much information about future short term volatility. In addition, fitting the regression on standard deviations of mid prices over the different intervals did not yield better results. In this case, Granger causality tests were not in favor of the opposite relationship either.

On the other hand, we found a relationship between 30 min volatility and corresponding trade volumes over the same interval to be highly significant and have correlation of 85.4%.

### Conclusion

Our study reinforces the previous findings of Goldstein et al. that order book imbalances are strong predictors of future prices, in particular the direction of the next price change. This could have economic implications for strategies based on order book shape. In addition, we find evidence that realized spreads contain information about future imbalances. However, we acknowledge that in this complex realm of microstructure, simple linear models may not be optimal in dealing with the significant noise within the data. In future, we seek to explore these relationships with more advanced techniques for dealing with noisy tick data such as Levy processes and Kalman filters.

### References

• Goldstein, Michael A. and Kwan, Amy and Philip, Richard, High-Frequency Trading Strategies (May 23, 2017). Available at SSRN: https://ssrn.com/abstract=2973019
• Foucault, T., Moinas, S., Theissen, E., 2007. Does Anonymity Matter in Electronic Limit Order Markets? Review of Financial Studies 20, 1707-1747.


text:
Chapter 29

Hand injuries are one of the more common reasons for emergency department visits. Hand injuries may be isolated or part of multiple trauma. These injuries are seldom life-threatening but can result in significant disability, threatening a patient's livelihood and lifestyle. These injuries can have significant economic impact with possible loss of occupation and lost time from work. Systematic examination and initial care has a direct effect on the ultimate consequence of any hand injury. Overall, it is far better to under treat and refer than to over treat and cause avoidable iatrogenic injury or disability. The course of recovery depends on initial management.

### Position the Patient

A hand-injured patient may be examined while they are sitting but the environment should be calm and controlled. That does not apply to a patient with hand lacerations. It is best to examine the injured patient with a hand laceration while supine. The hand should be extended on a hand table next to the bed. The area should be clean, sterile, and have excellent lighting and all essential instruments available. It is best to have an assistant if possible for helping with lighting, positioning, and repair. There is a tendency in EM to see hand patients in hallways, chairs, and less than ideal circumstances. This must be avoided at all costs.

### Control Bleeding

Bleeding is controlled by direct pressure with sterile gauze packs, elevation, and, if necessary, an arterial tourniquet (eg, blood pressure cuff inflated above systolic blood pressure). Do not use clamps unless all other measures fail. Blind clamping of vessels can lead to further injury.

### Obtain History

#### Current Injuries

Ascertain the mechanism of injury by questioning the patient or others. An exact description of how the injury occurred will help determine the need for X-ray, antibiotics, urgent consultation in the ED, or in a follow-up visit. It is important to document the mechanism of injury, the time at which it occurred, and the environment in which it occurred. Ask about the following details associated with the mechanism of injury:

• Type of injury—crush, exploding, or simple amputation.
• Lacerations—exactly what device was involved and how the lacerations were caused.
• Position of the hand at the time of the injury (ie, fingers extended or flexed).
• Presence of pain, numbness, paresthesias, weakness, discoloration, coldness, clumsiness or poor coordination, or crepitus.
• Circumstances surrounding open wounds—environment, whether inflicted in a dirty environment (ie, sewer or barnyard) or in a specific situation (ie, fight-clenched fist).

#### Relevant History

• History of prior or existing hand or upper extremity injuries or disorders. Dupuytren's fasciitis, arthritis, and benign tumors are the most common nontraumatic problems noted.
• Bleeding disorders likely to influence hemostasis (eg, hemophilia).
• Factors that might impair wound healing (eg, use of corticosteroids).
• Tetanus immunization status (Chapter 30).
• Any allergies.
• General state ...

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text:
# Solve a system of linear equations?

GROUPS:
 Hi!I have a pretty easy problem where I have 6 equations and 6 unknown of the form m*x = 0, where m is a 6x6 matrix and x is a vector of length 6 with the 6 unknown variables. I'm looking for a nontrivial solution to this problem so I use the Nullspace command.My code is the following m = {{E^(-I*x*b/2), E^(I*x*b/2), -E^(-I*y*b/2), -E^(I*y*b/2), 0, 0}, {I*x*E^(-I*x*b/2), -I*x*E^(I*x*b/2), -I*y*E^(-I*y*b/2), I*y*E^(I*y*b/2), 0, 0}, {0, 0, -E^(I*y*b/2), -E^(-I*y*b/2), E^(I*x*b/2), E^(-I*x*b/2)}, {0, 0, -I*y*E^(I*y*b/2), I*y*E^(-I*y*b/2), I*x*E^(I*x*b/2), -I*x*E^(-I*x*b/2)}, {E^(-I*x*(a + b/2)), E^(I*x*(a + b/2)), 0, 0, 0, 0}, {0, 0, 0, 0, E^(I*x*(a + b/2)), E^(-I*x*(a + b/2))}} ns = NullSpace[m] This gives me the output { }. What does this result mean? All the coefficients x,y,a,b are real and greater than zero. Is that something that I should specify in the code? I wonder what is wrong with my code? I've attached my code to this post.Thanks in advance! Attachments:
Answer
10 months ago
3 Replies
 Perhaps MatrixForm[RowReduce[m]] and NullSpace[IdentityMatrix[6]] might give a hint
Answer
10 months ago
 For m *x == 0 you must have Det[ m ] == 0, but In[17]:= Det[m] // FullSimplify Out[17]= E^(-I (2 a x + b y)) (-(x + E^(2 I a x) (x - y) + y)^2 + E^(2 I b y) (x - y + E^(2 I a x) (x + y))^2)
Answer
10 months ago
 No nulls is good nulls?
Answer
10 months ago


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# No Natural Group Structure on Conjugacy Classes

Is it possible to show that there is no natural group structure on conjugacy classes in a group?

Alternatively, for a path connected space $X$, the set $[S^1,X]$ of free (unpointed) homotopy classes of loops in $X$ is in bijection with the set of conjugacy classes of $\pi_1(X)$. Is it possible to show that this has no natural group structure?

I interpret your question to mean the following: there is a functor, which I'll write $G \mapsto LG$, sending a group $G$ to its set of conjugacy classes. Is it possible to lift this functor to a group-valued functor?
The answer is no. To prove this, it suffices to find an inclusion $H \to G$ of groups inducing an inclusion $LH \to LG$ of conjugacy classes, but such that the order of $LH$ doesn't divide the order of $LG$ (so it can't be a group homomorphism with respect to any group structure on $LH$ or $LG$). A very small example suffices: the inclusion $C_2 \to S_3$ induces an inclusion $LC_2 \to LS_3$, but the former has order $2$ and the latter has order $3$.
(However, $LG$ is naturally a groupoid-valued functor; namely, it should really be thought of as returning the action groupoid $G/G$ of $G$ acting on itself by conjugation. The geometric realization of this thing is, among other things, the free loop space of $BG$.)