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# for an onto function range is equivalent to the codomain

All elements in B are used. The term surjective and the related terms injective and bijective were introduced by Nicolas Bourbaki,[4][5] a group of mainly French 20th-century mathematicians who, under this pseudonym, wrote a series of books presenting an exposition of modern advanced mathematics, beginning in 1935. 3. is one-to-one onto (bijective) if it is both one-to-one and onto. So. The French word sur means over or above, and relates to the fact that the image of the domain of a surjective function completely covers the function's codomain. Solution : Domain = All real numbers . Its domain is Z, its codomain is Z as well, but its range is f0;1;4;9;16;:::g, that is the set of squares in Z. The domain is basically what can go into the function, codomain states possible outcomes and range denotes the actual outcome of the function. X Let’s take f: A -> B, where f is the function from A to B. For example consider. Function such that every element has a preimage (mathematics), "Onto" redirects here. Range vs Codomain. X Any morphism with a right inverse is an epimorphism, but the converse is not true in general. Equivalently, a function f with domain X and codomain Y is surjective, if for every y in Y, there exists at least one x in X with {\displaystyle f (x)=y}. A right inverse g of a morphism f is called a section of f. A morphism with a right inverse is called a split epimorphism. In mathematics, a function f from a set X to a set Y is surjective (also known as onto, or a surjection), if for every element y in the codomain Y of f, there is at least one element x in the domain X of f such that f(x) = y. Equivalently, a function Two functions , are equal if and only if their domains are equal, their codomains are equal, and = Ὄ Ὅfor all in the common domain. 2. is onto (surjective)if every element of is mapped to by some element of . Any function induces a surjection by restricting its codomain to its range. De nition 64. Further information on notation: Function (mathematics) § Notation A surjective function is a function whose image is equal to its codomain. De nition 65. We know that Range of a function is a set off all values a function will output. Definition: ONTO (surjection) A function $$f :{A}\to{B}$$ is onto if, for every element $$b\in B$$, there exists an element $$a\in A$$ such that $f(a) = b.$ An onto function is also called a surjection, and we say it is surjective. A surjective function is a function whose image is equal to its codomain. Both Codomain and Range are the notions of functions used in mathematics. Range can be equal to or less than codomain but cannot be greater than that. Functions, Domain, Codomain, Injective(one to one), Surjective(onto), Bijective Functions All definitions given and examples of proofs are also given. The codomain of a function sometimes serves the same purpose as the range. The "range" is the subset of Y that f actually maps something onto. While codamain is defined as "a set that includes all the possible values of a given function" as wikipedia puts it. These properties generalize from surjections in the category of sets to any epimorphisms in any category. Practice Problems. Onto Function. Your email address will not be published. The proposition that every surjective function has a right inverse is equivalent to the axiom of choice. Let A/~ be the equivalence classes of A under the following equivalence relation: x ~ y if and only if f(x) = f(y). Right-cancellative morphisms are called epimorphisms. g is easily seen to be injective, thus the formal definition of |Y| ≤ |X| is satisfied.). However, the domain and codomain should always be specified. {\displaystyle f(x)=y} Your email address will not be published. This page was last edited on 19 December 2020, at 11:25. By definition, to determine if a function is ONTO, you need to know information about both set A and B. Another surjective function. Y In previous article we have talked about function and its type, you can read this here.Domain, Codomain and Range:Domain:In mathematics Domain of a function is the set of input values for which the function is defined. and codomain This video introduces the concept of Domain, Range and Co-domain of a Function. [2] Surjections are sometimes denoted by a two-headed rightwards arrow (.mw-parser-output .monospaced{font-family:monospace,monospace}U+21A0 ↠ RIGHTWARDS TWO HEADED ARROW),[6] as in {\displaystyle f} the range of the function F is {1983, 1987, 1992, 1996}. The range of a function, on the other hand, can be defined as the set of values that actually come out of it. A function is said to be onto if every element in the codomain is mapped to; that is, the codomain and the range are equal. When this sort of the thing does not happen, (that is, when everything in the codomain is in the range) we say the function is onto or that the function maps the domain onto the codomain. Onto functions focus on the codomain. Here, codomain is the set of real numbers R or the set of possible outputs that come out of it. When you distinguish between the two, then you can refer to codomain as the output the function is declared to produce. For instance, let A = {1, 2, 3, 4} and B = {1, 4, 9, 25, 64}. In a 3D video game, vectors are projected onto a 2D flat screen by means of a surjective function. [1][2][3] It is not required that x be unique; the function f may map one or more elements of X to the same element of Y. {\displaystyle X} Codomain of a function is a set of values that includes the range but may include some additional values. (This one happens to be an injection). 1.1. . The purpose of codomain is to restrict the output of a function. Specifically, if both X and Y are finite with the same number of elements, then f : X → Y is surjective if and only if f is injective. For example, in the first illustration, above, there is some function g such that g(C) = 4. This function would be neither injective nor surjective under these assumptions. If (as is often done) a function is identified with its graph, then surjectivity is not a property of the function itself, but rather a property of the mapping. If you have any doubts just ask here on the ask and answer forum and our experts will try to help you out as soon as possible. x A function f : X → Y is surjective if and only if it is right-cancellative:[9] given any functions g,h : Y → Z, whenever g o f = h o f, then g = h. This property is formulated in terms of functions and their composition and can be generalized to the more general notion of the morphisms of a category and their composition. For every element b in the codomain B, there is at least one element a in the domain A such that f(a)=b.This means that no element in the codomain is unmapped, and that the range and codomain of f are the same set.. Then if range becomes equal to codomain the n set of values wise there is no difference between codomain and range. The term range, however, is ambiguous because it can be sometimes used exactly as Codomain is used. R n x T (x) range (T) R m = codomain T onto Here are some equivalent ways of saying that T … If A = {1, 2, 3, 4} and B = {1, 2, 3, 4, 5, 6, 7, 8, 9} and the relation f: A -> B is defined by f (x) = x ^2, then codomain = Set B = {1, 2, 3, 4, 5, 6, 7, 8, 9} and Range = {1, 4, 9}. In this article in short, we will talk about domain, codomain and range of a function. www.differencebetween.net/.../difference-between-codomain-and-range On the other hand, the whole set B … Three common terms come up whenever we talk about functions: domain, range, and codomain. y {\displaystyle x} The function f: A -> B is defined by f (x) = x ^3. Example In mathematics, the codomain or set of destination of a function is the set into which all of the output of the function is constrained to fall. Here, x and y both are always natural numbers. in Co-domain … f Both the terms are related to output of a function, but the difference is subtle. There is also some function f such that f(4) = C. It doesn't matter that g(C) can also equal 3; it only matters that f "reverses" g. Surjective composition: the first function need not be surjective. The function f: A -> B is defined by f (x) = x ^2. A function is bijective if and only if it is both surjective and injective. In simple terms: every B has some A. In order to prove the given function as onto, we must satisfy the condition Co-domain of the function = range Since the given question does not satisfy the above condition, it is not onto. {\displaystyle f\colon X\twoheadrightarrow Y} Using the axiom of choice one can show that X ≤* Y and Y ≤* X together imply that |Y| = |X|, a variant of the Schröder–Bernstein theorem. f(x) maps the Element 7 (of the Domain) to the element 49 (of the Range, or of the Codomain). Example 2 : Check whether the following function is onto f : R → R defined by f(n) = n 2. By knowing the the range we can gain some insights about the graph and shape of the functions. The range can be difficult to specify sometimes, but larger set of values that include the entire range can be specified. The function may not work if we give it the wrong values (such as a negative age), 2. Thus, B can be recovered from its preimage f −1(B). An onto function is such that every element in the codomain is mapped to at least one element in the domain Answer and Explanation: Become a Study.com member to unlock this answer! As prepositions the difference between unto and onto is that unto is (archaic|or|poetic) up to, indicating a motion towards a thing and then stopping at it while onto is upon; on top of. with Every surjective function has a right inverse, and every function with a right inverse is necessarily a surjection. In native set theory, range refers to the image of the function or codomain of the function. Let fbe a function from Xto Y, X;Ytwo sets, and consider the subset SˆX. The composition of surjective functions is always surjective. Y For e.g. Any surjective function induces a bijection defined on a quotient of its domain by collapsing all arguments mapping to a given fixed image. When working in the coordinate plane, the sets A and B may both become the Real numbers, stated as f : R→R . ↠ Any function induces a surjection by restricting its codomain to the image of its domain. The range is the square of A as defined by the function, but the square of 4, which is 16, is not present in either the codomain or the range. Problem 1 : Let A = {1, 2, 3} and B = {5, 6, 7, 8}. It’s actually part of the definition of the function, but it restricts the output of the function. The range is the subset of the codomain. In fact, a function is defined in terms of sets: in Every function with a right inverse is necessarily a surjection. For example, let A = {1, 2, 3, 4, 5} and B = {1, 4, 8, 16, 25, 64, 125}. inputs a function is defined by its set of inputs, called the domain; a set containing the set of outputs, and possibly additional elements, as members, called its codomain; and the set of … Equivalently, A/~ is the set of all preimages under f. Let P(~) : A → A/~ be the projection map which sends each x in A to its equivalence class [x]~, and let fP : A/~ → B be the well-defined function given by fP([x]~) = f(x). Conversely, if f o g is surjective, then f is surjective (but g, the function applied first, need not be). In context|mathematics|lang=en terms the difference between codomain and range is that codomain is (mathematics) the target space into which a function maps elements of its domain it always contains the range of the function, but can be larger than the range if the function is not surjective while range is (mathematics) the set of values (points) which a function can obtain. If f : X → Y is surjective and B is a subset of Y, then f(f −1(B)) = B. More precisely, every surjection f : A → B can be factored as a projection followed by a bijection as follows. In other words, g is a right inverse of f if the composition f o g of g and f in that order is the identity function on the domain Y of g. The function g need not be a complete inverse of f because the composition in the other order, g o f, may not be the identity function on the domain X of f. In other words, f can undo or "reverse" g, but cannot necessarily be reversed by it. Codomain = N that is the set of natural numbers. ( The cardinality of the domain of a surjective function is greater than or equal to the cardinality of its codomain: If f : X → Y is a surjective function, then X has at least as many elements as Y, in the sense of cardinal numbers. However, in modern mathematics, range is described as the subset of codomain, but in a much broader sense. This is especially true when discussing injectivity and surjectivity, because one can make any function an injection by modifying the domain and a surjection by modifying the codomain. Any function can be decomposed into a surjection and an injection. [8] This is, the function together with its codomain. 1. The prefix epi is derived from the Greek preposition ἐπί meaning over, above, on. These preimages are disjoint and partition X. Theimage of the subset Sis the subset of Y that consists of the images of the elements of S: f(S) = ff(s); s2Sg We next move to our rst important de nition, that of one-to-one. From this we come to know that every elements of codomain except 1 and 2 are having pre image with. This post clarifies what each of those terms mean. {\displaystyle y} So here. But not all values may work! Before we start talking about domain and range, lets quickly recap what a function is: A function relates each element of a set with exactly one element of another set (possibly the same set). For example: Then f = fP o P(~). with domain The “codomain” of a function or relation is a set of values that might possibly come out of it. 2.1. . (This one happens to be a bijection), A non-surjective function. While codomain of a function is set of values that might possibly come out of it, it’s actually part of the definition of the function, but it restricts the output of the function. (The proof appeals to the axiom of choice to show that a function Unlike injectivity, surjectivity cannot be read off of the graph of the function alone. 0 ; View Full Answer No. See: Range of a function. ) Difference Between Microsoft Teams and Zoom, Difference Between Microsoft Teams and Skype, Difference Between Checked and Unchecked Exception, Difference between Von Neumann and Harvard Architecture. And knowing the values that can come out (such as always positive) can also help So we need to say all the values that can go into and come out ofa function. We want to know if it contains elements not associated with any element in the domain. March 29, 2018 • no comments. That is the… He has that urge to research on versatile topics and develop high-quality content to make it the best read. Given two sets X and Y, the notation X ≤* Y is used to say that either X is empty or that there is a surjection from Y onto X. The codomain of a function can be simply referred to as the set of its possible output values. . The range of T is equal to the codomain of T. Every vector in the codomain is the output of some input vector. In modern mathematics, range is often used to refer to image of a function. A function is said to be a bijection if it is both one-to-one and onto. : Most books don’t use the word range at all to avoid confusions altogether. In the above example, the function f is not one-to-one; for example, f(3) = f( 3). The set of actual outputs is called the rangeof the function: range = ∈ ∃ ∈ = ⊆codomain We also say that maps to ,and refer to as a map. The term range is often used as codomain, however, in a broader sense, the term is reserved for the subset of the codomain. f Any function with domain X and codomain Y can be seen as a left-total and right-unique binary relation between X and Y by identifying it with its function graph. Here are the definitions: 1. is one-to-one (injective) if maps every element of to a unique element in . So the domain and codomain of each set is important! Range (f) = {1, 4, 9, 16} Note : If co-domain and range are equal, then the function will be an onto or surjective function. Sagar Khillar is a prolific content/article/blog writer working as a Senior Content Developer/Writer in a reputed client services firm based in India. For example the function has a Domain that consists of the set of all Real Numbers, and a Range of all Real Numbers greater than or equal to zero. So here, set A is the domain and set B is the codomain, and Range = {1, 4, 9}. this video is an introduction of function , domain ,range and codomain...it also include a trick to remember whether a given relation is a function or not For other uses, see, Surjections as right invertible functions, Cardinality of the domain of a surjection, "The Definitive Glossary of Higher Mathematical Jargon — Onto", "Bijection, Injection, And Surjection | Brilliant Math & Science Wiki", "Injections, Surjections, and Bijections", https://en.wikipedia.org/w/index.php?title=Surjective_function&oldid=995129047, Short description is different from Wikidata, Creative Commons Attribution-ShareAlike License. The Range can also mean all the output values of a function. In other words no element of are mapped to by two or more elements of . In mathematical terms, it’s defined as the output of a function. In mathematics, a surjective or onto function is a function f : A → B with the following property. Range is equal to its codomain Q Is f x x 2 an onto function where x R Q Is f x from DEE 1027 at National Chiao Tung University Notice that you cannot tell the "codomain" of a function just from its "formula". Math is Fun That is, a function relates an input to an … . I could just as easily define f:R->R +, with f(x)= e x. X Thanks to his passion for writing, he has over 7 years of professional experience in writing and editing services across a wide variety of print and electronic platforms. The set of all the outputs of a function is known as the range of the function or after substituting the domain, the entire set of all values possible as outcomes of the dependent variable. The composition of surjective functions is always surjective: If f and g are both surjective, and the codomain of g is equal to the domain of f, then f o g is surjective. Y x A function f from A to B is called onto if for all b in B there is an a in A such that f (a) = b. Domain is also the set of real numbers R. Here, you can also specify the function or relation to restrict any negative values that output produces. Let N be the set of natural numbers and the relation is defined as R = {(x, y): y = 2x, x, y ∈ N}. A surjective function with domain X and codomain Y is then a binary relation between X and Y that is right-unique and both left-total and right-total. Function, but the converse is not one-to-one ; for example, f n. In short, we will talk about domain, codomain and range of the function f a. Factored as a projection map, and consider the subset SˆX 3D video game, are... Bijective ) if it contains elements not associated with any element in the above example, the term “ ”... Easily seen to be injective, thus the formal deﬁnition range ” sometimes is used refer. Function g such that g ( C ) = x ^2 show that a function is onto ( )... 3D video game, vectors are projected onto a 2D flat screen by means of a function just its... 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Notes on Introduction to Chemical Arithmetic | Grade 11 > Chemistry > Chemical Arithmetic | KULLABS.COM

Notes, Exercises, Videos, Tests and Things to Remember on Introduction to Chemical Arithmetic

Please scroll down to get to the study materials.

• Note
• Things to remember

Atomic Number: It is the number of protons present in the nucleus of an atom. For example Carbon, an atom contains 6 protons. Therefore, the atomic number of carbon is 6. It is also the number of electron present in a neutral atom.

Atomic mass number: It is the sum of a number of protons and neutrons present in an atom of an element. For example Atomic mass number of Carbon = No. of protons + No. of neutrons = 6 + 6 = 12 The proton and neutron component of the nucleus are collectively known as nucleons i.e. the total number of nucleons is a mass number. The atomic number and atomic mass number of an atom can be denoted symbolically as

$$_A^XZ$$

where A Mass number ( neutrons + protons)

Z = Atomic number ( Number of protons or electrons)

X = symbol of element

For Example: $$_6^{12}C$$

Isotopes: Atoms of the same element having different mass number are called isotopes. It arises due to the difference in the number of protons.

Example: 11H, 21H,31H (Isotopes of hydrogeN)

126C, 136C,146C ( Isotopes of carbon)

Isobars: Atoms of different elements having the different atomic number but the same atomic mass number are called isobars.

Example: 4018Ar, 4019K,4020Ca

Isotones: Atoms of different elements having the same number of neutrons are called isotones.

Example: 146C, 157N, 168O ( 8 neutrons each)

Atomic mass/ Atomic weight: Atomic mass of an element is defined as the number which shows how many times an atom of an element is heavier than one – twelfth of the mass of an atom of a C – 12 isotope. For example, The atomic mass of Magnesium is 24 amu means that one atom of Magnesium is 24 times heavier than one – twelfth of the mass of an atom of a C – 12 isotope.

Atomic mass = $$\frac {Mass \: of\: an\: atom\: of\: element}{ \frac{1}{12}\: th \: of \: mass\: of\: an\: atom\: of\: C-12\: isotope}$$

Atomic mass unit ( AMU): The mass of one-twelfth of the mass of an atom of a C – 12 isotopes is called one atomic mass unit ( 1 amu). The masses of all atoms and molecules are expressed with reference to the mass of 1 AMU.

Absolute mass of 1 amu

1 gram atomic mass of carbon = 12 gm of carbon

Mass of 6.023 x 1023 atoms of carbon = 12 gm of carbon

Mass of 1 atom of carbon = $$\frac {12}{6.023 x 10^{23}}$$ gm of carbon

$$\frac {1}{12}th$$ of Mass of 1 atom of carbon = $$\frac {12}{6.023 x 10^{23}}$$ gm of carbon = $$\frac {1}{12}$$$$\frac {12}{6.023 x 10^{23}}$$ gm of carbon = 1.66 x 10-24 gm of carbon

∴ 1 amu = 1.66 x 10-24 gm

Average atomic mass

There are many cases where different atoms of the same element are naturally abundant and posses different mass number. Such atoms of the same element having different mass number are called isotopes. Such isotopes having different relative masses as well. In such case, the atomic mass of the element is the average relative masses of isotopes of the element.

Atomic mass = $$\frac {Mass \:of\: an\: atom\: of\: element\: }{ \frac{1}{12}th\: of \: mass\: of\: an\: atom\: of\: C-12\: isotope}$$

Atomic mass being an average mass, it is fractional in number. For example, 2 isotopes of chlorine having relative masses 35 and 37 are present in the chlorine in the ratio 3: 1 respectively. Therefore, average relative atomic mass of chlorine could be

$$\frac {35× 3 + 37 × 1}{3 + 1}$$

= 35.5

Hence, the fractional atomic mass of an element is due to the presence of different isotopes of elements in a natural mixture in different percentage or ratio.

Therefore, Average atomic mass =$$\frac {m_1 R_1 + m_2R_2}{R_1 + R_2}$$ OR =$$\frac {m_1 R_1 + m_2R_2}{100\%}$$

Gram atomic mass ( Gram – atom): Atomic mass expressed in gram is known as gram atomic mass or simply gram atom or mole of an atom.

For example, 1 gm atom of Magnesium means 24 gm of Magnesium. It is also 1 mole of Magnesium.

1 gm atom of Carbon = 12gm of Carbon

1 gm atom of Sodium = 23 gm of Sodium

1 gm atom of Sulphur = 32gm of Sulphur

1gm atom of Oxygen = 16gm of Oxygen

No. of gram atom = $$\frac {Mass \:of\: an\: element\: in\: gram }{Atomic\: mass\: of\: an\: element }$$

Molecular mass: it is the mass of a molecule of a substance with respect to the one-twelfth of the mass of an atom of the C-12 isotope. A molecular mass is a relative term and it is unit less number. For example Molecular mass of sulphuric acid is 98 means that a molecule of sulphuric acid is 98 times heavier than one-twelfth of the mass of an atom of a C-12 isotope. A molecular mass can be calculated by the summation of atomic masses of all the atoms present in the given molecule.

For example: Molecular mass of C6H12O6 = 12 x 6 + 12 x 1 + 16 x 6 = 180 amu.

Gram molecular mass ( Gram molecule): When the molecular mass is expressed in gram, then it is called 1 gram molecular mass or 1 gram mole.

No. of gram molecule = $$\frac {Mass\: in\: gram }{Molecular\:Mass}$$

Mole concept: It is found that a number of atoms present in 1gram atom of the element are equal to the number of molecules present in a 1gm mole of any substance. This number is experimentally determined and found to be equal to ( 6.023 x 1023 ). This number is called Avogadro’s number or Avogadro’s constant.

Avogadro’s number = 6.023 x 1023

The amount of substance containing Avogadro’s number of atoms or molecules is called mole or simply written as mol.

Thus, a mole can be defined as the amount of substance which contains the same number of elementary particles ( atoms / molecules/ ions) as the number of atoms present in 12 gm of C – 12 isotope. Simply, a mole is the collection of Avogadro’s number of particles i.e. number of atoms/ molecules/ions/electrons/protons/neutrons, etc.

Significance of mole

1. It gives a number of particles.

1 mole of the substance contains Avogadro's number of particles. 1 mole of H2 molecule = 6.023 x 1023 molecules of H2

1 mole Hydrogen atom = 6.023 x 1023 H atom

1 mole of Na+ ion = 6.023 x 1023 ions of Na+

1 mole of electrons = 6.023 x 1023 electrons

2. It gives the mass of substance in gram.

1 mole of atoms = 6.023 x 1023 atoms = gram atomic mass of element

Eg: 1 mole of O atom = 6.023 x 1023 atos of O = 16gm of oxygen

1 mole of Na = 6.023 x 1023 atoms of Na = 23 grams of Na

1 mole of Sulphur atom = 6.023 x 1023 atoms of S = 32 gm of S

1 mole of molecules = 6.023 x 1023 molecules = gram molecular mass

Eg: 1 mole of O2 = 6.023 x 1023 atos of O = 32gm of oxygen

1 mole of CO2 = 6.023 x 1023 atoms of Na = 44 grams of CO2

1 mole of CaCO3 = 6.023 x 1023 atoms of CaCO3 = 100 gm of CaCO3

3. It gives the volume of gaseous substances at NTP.

The volume occupied by 1 mole of any gaseous substance is known as gram molecular volume (GMV) which is equal to 22.4liter.

•  Atomic mass expressed in gram is known as gram atomic mass or simply gram atom or mole of an atom.
• Molecular mass of C6H12O6 = 12 x 6 + 12 x 1 + 16 x 6 = 180 amu.
• The amount of substance containing Avogadro’s number of atoms or molecules is called mole or simply written as mol.
• The volume occupied by 1 mole of any gaseous substance is known as gram molecular volume (GMV) which is equal to 22.4liter.
.

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text:
Conditional moments tests where proposed by Tauchen (1985) and Newey (1985). As these tests are quite alike Lagrange multiplier (or score) tests, which is, along with the likelihood ratio and the Wald tests one of the three “classical” tests. These three tests have in common to test a set of hypothesis which enables to go from a unconstrained (or larger) model to a constrained (or smaller) model:

• the Wald test is based on the values of the vector implied the restrictions evaluated for the constrained model,
• the likelihood ratio is based on the comparison of the fits of both models (measured by the value of the log-likelihood function),
• the score test is based on the value of the score (ie the vector of the first derivatives) of the log-likelihood of the unconstrained models using the estimates of the constrained model.

# 1 The score (or Lagrange multiplier) test

For the special case where the hypothesis can be writen as a linear combination of the coefficients, the unconstrained model can always be parametrized in a way such that the likelihood function of the larger model is: $$\ln L (\beta_1, \beta_2)$$, with under H0: $$\beta_2 = 0$$ and the larger model reduces to the smaller one. For the score test, the statistic is based on the gradient of the log-likelihood computed with the parameters obtained by estimating the constrained model: $$\hat{\beta}_1 ^c, 0$$, which is:

$\frac{\partial \ln L}{\partial \beta}(\hat{\beta}_1^c, 0) = \left( \begin{array}{c} \frac{\partial \ln L}{\partial \beta_1}(\hat{\beta}_1^c, 0) \\ \frac{\partial \ln L}{\partial \beta_2}(\hat{\beta}_1^c, 0) \end{array} \right) = \left( \begin{array}{c} 0 \\ \frac{\partial \ln L}{\partial \beta_2}(\hat{\beta}_1^c, 0) \end{array} \right)$

As an example, consider a linear or log-linear gaussian model for which one wishes to test the normality hypothesis. For the smaller model, we have:

$y_n = \alpha + \beta x_n + \epsilon_n \mbox{ with } \epsilon_n \sim N(0, \sigma_\epsilon)$

for the linear model and:

$\ln y_n = \alpha + \beta x_n + \epsilon_n \mbox{ with } \epsilon_n \sim N(0, \sigma_\epsilon)$

for the log-linear model. A potential larger model consists on taking a Box-Cox transformation of $$y$$, denoted $$y^{(\lambda)}$$:

$y^{(\lambda)} = \alpha + \beta x_n + \epsilon_n \mbox{ with } \epsilon_n \sim N(0, \sigma_\epsilon)$

with:

$y^{(\lambda)} = \left\{ \begin{array}{lcl} \frac{y_n^\lambda - 1}{\lambda} & \mbox{if} & \lambda \neq 0 \\ \ln y_n & \mbox{if} & \lambda = 0 \end{array} \right.$

which obviously reduce to the smaller log-linear model for $$\lambda = 0$$. For $$\lambda = 1$$, $$y ^ {(\lambda)} = y - 1$$ and the larger model reduce to the linear model except that $$1$$ should be substracted from the left hand side of the equation, which means that the intercept of the linear model fitted by ordinary least squares should be reduced by 1.

The log-likelihood of the larger model is:

$\ln L (\beta, \sigma, \lambda) = -\frac{N}{2}\ln 2\pi - N \ln \sigma - (1 - \lambda) \sum_{n=1}^N \ln y_n - \frac{1}{2\sigma ^ 2}\sum_{n=1}^N\left(y^{(\lambda)} - \beta^\top x_n\right) ^ 2$

The gradient of the log-likelihood is:

$\frac{\partial \ln L}{\partial \theta}(\theta) = \left( \begin{array}{c} \frac{\partial \ln L}{\partial \beta}\\ \frac{\partial \ln L}{\partial \sigma}\\ \frac{\partial \ln L}{\partial \lambda}\\ \end{array} \right) = \left( \begin{array}{c} \frac{1}{\sigma ^ 2} \sum_{n=1}^N \left(y^{(\lambda)} - \beta^\top x_n\right) x_n\\ -\frac{N}{\sigma} + \frac{1}{\sigma^3}\sum_{n=1}^N \left(y^{(\lambda)} - \beta^\top x_n\right) ^ 2\\ \sum_{n=1} ^ N \ln y_n - \frac{1}{\sigma ^ 2}\sum_{n=1}^N \left(y_n ^ {(\lambda)} - \beta^\top x_n\right) \left(y_n^{(\lambda)} \ln y_n - \frac{y_n ^ {(\lambda)} - \ln y_n}{\lambda}\right) \end{array} \right)$

For the constrained linear ($$\hat{\lambda}_{i} = 1$$) and log-linear ($$\hat{\lambda}_o = 0$$) models, we have respectively:

$\frac{\partial \ln L}{\partial \theta}(\hat{\theta}_i) = \left( \begin{array}{c} 0 \\ 0 \\ \sum_{n=1} ^ N \ln y_n - \frac{1}{2\hat{\sigma}_c ^ 2}\sum_{n=1}^N \left(\ln y_n - \hat{\beta}_c^\top x_n\right) \left(1 + y(\ln y - 1)\right) \end{array} \right)$

$\frac{\partial \ln L}{\partial \theta}(\hat{\theta}_o) = \left( \begin{array}{c} 0 \\ 0 \\ \sum_{n=1} ^ N \ln y_n - \frac{1}{2\hat{\sigma}_c ^ 2}\sum_{n=1}^N \left(\ln y_n - \hat{\beta}_c^\top x_n\right) \ln y_n^2 \end{array} \right)$

The following function returns the log-likelihood for a box-cox model, given a vector of parameters coef, a matrix of covariates X and a vector of response y. It has further arguments:

• sum: if FALSE, invidual contributions to the log-likelihood (a vector of length $$N$$) and to the gradient (a matrix of dimensions $$N\times K$$ is returned,
• gradient: if TRUE the analytical gradient is computed and returned as an attribute,
• hessian: if TRUE the analytical hessian is computed and returned as an attribute.
lnL_bc <- function(coef, X, y, sum = FALSE, gradient = FALSE, hessian = FALSE){
y <- y
N <- length(y)
K <- length(coef) - 3
beta <- coef[1:(K + 1)]
sigma <- coef[K + 2]
lambda <- coef[K + 3]
bX <- as.numeric(X %*% beta)
if (lambda == 0){
bc <- log(y)
bc_l  <- 1 / 2 * log(y) ^ 2
bc_ll <- 1 / 3 * log(y) ^ 3
}
else{
bc <- (y ^ lambda - 1) / lambda
bc_l <- bc * log(y) - (bc - log(y)) / lambda
bc_ll <- bc_l * log(y) - (bc_l * lambda - (bc - log(y))) / lambda ^ 2
}
e <- bc - bX
lnL <- - 1 / 2 * log(2 * pi) - log(sigma) - (1 - lambda) * log(y) -
1 / (2 * sigma ^ 2) * e ^ 2

g_beta <- 1 / sigma ^ 2 * e * X
g_sigma <- - 1 / sigma + 1 / (sigma ^ 3) * e ^ 2
g_lambda <- log(y) - 1 / (sigma ^ 2) * e * bc_l
G <- cbind(g_beta, g_sigma,  g_lambda)
}
if (hessian){
h_bb <- - 1 / sigma ^ 2 * crossprod(X)
h_bs <- apply(- 2 / sigma ^ 3 * X * e, 2, sum)
h_bl <- apply( 1 / sigma ^ 2 * bc_l * X, 2, sum)
h_ss <- sum(1 / sigma ^ 2 - 3 / sigma ^ 4 * e ^ 2)
h_sl <- sum(2 / sigma ^ 3 * e * bc_l)
h_ll <- sum(- 1 / sigma ^ 2 * (e * bc_ll + bc_l ^ 2))
H <- rbind(cbind(h_bb, sigma = h_bs, lambda = h_bl),
sigma = c(h_bs, h_ss, h_sl),
lambda = c(h_bl, h_sl, h_ll))
}
if (sum){
lnL <- sum(lnL)
if (gradient) G <- apply(G, 2, sum)
}
if (hessian) attr(lnL, "hessian") <- H
lnL
}

Consider as an example the cars data set which is provided by base R. The response is dist (the stoping distance) and the only covariate is speed (the speed of the car). The smaller linear and log-linear models can be computed using lm:

lin_lm <- lm(dist ~ speed + I(speed ^ 2), cars)
log_lm <- update(lin_lm, log(.) ~ .)

as stated previously, the linear model is a special case of the box-cox model if one is substracted from the intercept.

coefs_lin <- coef(lin_lm)
coefs_lin[1] <- coefs_lin[1] - 1

To use the lnL_bc function, we need to extract the response vector and the covariates matrix:

X <- model.matrix(lin_lm)
y <- model.response(model.frame(lin_lm))

Finally we can evaluate the log-likelihood of the box-cox model for the two special cases of the linear and the log-linear models using the coefficients obtained using lm ($$\sigma$$ is a parameter in the log-likelihood, it is extracted from the lm models using the provided sigma2 function which differs from the sigma function by the fact that the deviance is divided by the number of observations and not the number of degrees of freedom of the regression).

sigma2 <- function(x) sigma(x) * sqrt(df.residual(x) / nobs(x))
coefs_lin <- c(coefs_lin, sigma = sigma2(lin_lm), lambda = 1)
coefs_log <- c(coef(log_lm), sigma = sigma2(log_lm), lambda = 0)
lnL_lin <- lnL_bc(coefs_lin, X, y, sum = TRUE, gradient = TRUE, hessian = FALSE)
lnL_log <- lnL_bc(coefs_log, X, y, sum = TRUE, gradient = TRUE, hessian = FALSE)
lnL_lin
## [1] -205.386
##   (Intercept)         speed    I(speed^2)       g_sigma      g_lambda
##  1.041376e-15  1.951564e-14  4.263673e-13  6.175616e-16 -2.184459e+01
lnL_log
## [1] -202.3903
##   (Intercept)         speed    I(speed^2)       g_sigma      g_lambda
## -2.936210e-13 -3.541334e-12 -4.704148e-11  2.620126e-13  2.956768e+01

According to the log-likelihood, the log-linear model fits fairly better the data than the linear one. As expected, all the elements of the gradient vector except the last one is 0. The derivative of the log-likelihood with respect with $$\lambda$$ is negative for the linear model and positive for the log-linear model. This means that the log-likelihood should increase for some values of $$\lambda$$ between 0 and 1. To fit the box-cox log-likelihood model, we use the maxLik::maxLik function. It is a good general practise to provide “good” (and different) starting values to unsures that the non-linear optimization process converges, and that it converges to a global maximum. We therefore run twice maxLik, with starting values corresponding to the linear and to the non-linear model:

library("maxLik")
bc_lin <- maxLik(lnL_bc, start = coefs_lin, X = X, y = y,
sum = TRUE, gradient = TRUE, hessian = TRUE)
bc_log <- maxLik(lnL_bc, start = coefs_log, X = X, y = y,
sum = TRUE, gradient = TRUE, hessian = TRUE)

We first check that the fitted model is the same for the two sets of starting values:

cbind(coef(bc_lin), coef(bc_log))
##                     [,1]         [,2]
## (Intercept)  0.213367880  0.213370726
## speed        0.582112891  0.582111088
## I(speed^2)  -0.005530642 -0.005530664
## sigma        1.365042140  1.365033776
## lambda       0.372872290  0.372870677

and we then summarize the fitted model:

summary(bc_lin)
## --------------------------------------------
## Maximum Likelihood estimation
## Newton-Raphson maximisation, 15 iterations
## Log-Likelihood: -197.3795
## 5  free parameters
## Estimates:
##              Estimate Std. error t value Pr(> t)
## (Intercept)  0.213368   1.351837   0.158 0.87459
## speed        0.582113   0.232726   2.501 0.01237 *
## I(speed^2)  -0.005531   0.006108  -0.905 0.36524
## sigma        1.365042   0.650610   2.098 0.03590 *
## lambda       0.372872   0.131795   2.829 0.00467 **
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
## --------------------------------------------

As expected, the fitted value of $$\lambda$$ lies between 0 and 1 and is statistically different from 0 and from 1. Note also that the log-likelihood of the box-cox model (-197.38) is larger than the one of the log-linear model (-202.39).

If the hypothesis corresponding to the smaller model are true, the distribution of the score vector is:

$g(\tilde{\theta}) \sim N(0, I(\theta))$

Were $$I$$ is the information matrix. Therefore if H0 is true:

$g(\tilde{\theta}) ^ \top I(\theta) ^ {-1} g(\tilde{\theta})$

is a $$\chi^2$$ with $$J$$ degrees of freedom ($$J$$ being the number of hypothesis, one in our example). $$I(\theta)=-\mbox{E}\left(\frac{\partial^2 \ln L}{\partial \theta \partial \theta^\top}(\theta)\right)$$ needs to be estimate and two natural estimators are:

• the empirical variance of the gradient: $$\sum_{n=1} ^ N g(\tilde{\theta}, x_n, y_n) g(\tilde{\theta}, x_n, y_n)^\top = G(\tilde{\theta}, X, y) ^ \top G(\tilde{\theta}, X, y)$$ (also called the outer-product of the gradient), where $$G$$ is an $$N\times K$$ matrix for which each line is the contribution of an observation to the gradient,
• the opposite of the hessian: $$-H(\tilde{\theta}) = -\frac{\partial^2 \ln L}{\partial \theta \partial \theta^\top}(\tilde{\theta})$$.

The $$G$$ matrix is obtained using the lnL_bc function with sum = FALSE and gradient = TRUE and the $$H$$ matrix using hessian = TRUE.

lnL_lin <- lnL_bc(coefs_lin, X, y, sum = FALSE, gradient = TRUE, hessian = TRUE)
lnL_log <- lnL_bc(coefs_log, X, y, sum = FALSE, gradient = TRUE, hessian = TRUE)
H_lin <- attr(lnL_lin, "hessian")
H_log <- attr(lnL_log, "hessian")
g_lin <- apply(G_lin, 2, sum)
g_log <- apply(G_log, 2, sum)

The test statistics can then be computed using any of the two estimators of the information matrix:

as.numeric(g_lin %*% solve(- H_lin) %*% g_lin)
## [1] 22.48473
as.numeric(g_lin %*% solve(crossprod(G_lin)) %*% g_lin)
## [1] 18.08425
as.numeric(g_log %*% solve(- H_log) %*% g_log)
## [1] 8.797853
as.numeric(g_log %*% solve(crossprod(G_log)) %*% g_log)
## [1] 9.428458

The two variants of the same test statistics are very close whatever the approximation of the information matrix used. The critical value of a $$\chi^2$$ with one degree of freedom at the 5% level being 3.84, the two hypothesis of linear and log normality are strongly rejected.

Note that $$g(\tilde{\theta}) = G(\tilde{\theta})^\top \iota$$ where $$\iota$$ is a vector of one of length $$N$$. Therefore the outer product variant of the test is:

$\iota^\top G(\tilde{\theta}) \left[G(\tilde{\theta}) ^ \top G(\tilde{\theta})\right] ^ {-1} G(\tilde{\theta}) ^ \top \iota = \iota^\top M_G(\tilde{\theta}) \iota$

where $$M_G(\tilde{\theta}) = G(\tilde{\theta})\left[G(\tilde{\theta}) ^ \top G(\tilde{\theta})\right] ^ {-1} G(\tilde{\theta}) ^ \top$$ is the projection matrix on the columns of $$G(\tilde{\theta})$$. $$M_G(\tilde{\theta}) \iota$$ is therefore the fitted values of the regression of a column of ones with the columns of $$G(\tilde{\theta})$$ and, as $$M_G(\tilde{\theta})$$ is idempotent, the test statistic is the sum of the fitted values of this regression. The (uncetered as there is no intercept in this regression) $$R^2$$ is this sum of the fitted values divided by the total variation of the response, which is equal to $$N$$ (the response being a vector of ones of length $$N$$). Therefore, the test statistic can be obtained as $$N$$ times the $$R^2$$ of the regression of $$1$$ one on the columns of $$G$$:

iota <- rep(1, length(y))
lm_ones_lin <- lm(iota ~ G_lin - 1)
lm_ones_log <- lm(iota ~ G_log - 1)
summary(lm_ones_lin)$r.squared * nobs(lm_ones_lin) ## [1] 18.08425 summary(lm_ones_log)$r.squared * nobs(lm_ones_log)
## [1] 9.428458

# 2 The conditional moment test

The conditional moment test approach consists on constructing a vector of moments which are 0 under H0. Denoting $$\mu_n$$ such a vector, we have, for the normality hypothesis:

$\mu_n(\theta) = \left( \begin{array}{c} \mu_{3n}\\ \mu_{4n} \end{array} \right)= \left(\begin{array}{c} \mbox{E}(\epsilon_n ^ 3) \\ \mbox{E}(\epsilon_n ^ 4 - 3 \sigma ^ 4) \end{array}\right) = 0$

Denoting $$\hat{\epsilon}_n = y_n - \tilde{\beta} ^ \top x_n$$ or $$\hat{\epsilon}_n = \ln y_n - \tilde{\beta} ^ \top x_n$$ the residuals of the linear or the log-linear model, the empirical moments are:

$m_n(\tilde{\theta}) = \left( \begin{array}{c} \hat{\epsilon}_n ^ 3 \\ \hat{\epsilon}_n ^ 4 - 3 \sigma ^ 4 \end{array} \right)$

Define : $$M (\tilde{\theta}) ^ \top = \left(m_1, m_2, \ldots, m_N\right)$$ an $$N\times r$$ matrix for which each line is the moment vector for on observation and $$W(\theta)= \frac{1}{N}\sum_n \mbox{E}\left(\frac{\partial m_n}{\partial \theta}(\theta)\right)$$ a $$K \times r$$ matrix containing the derivatives of the empirical moments with the parameters of the model. Defining:

$Q = \left[M(\tilde{\theta}) - G(\tilde{\theta}) I(\theta) ^ {-1} W(\theta)\right]^\top \left[M(\tilde{\theta}) - G(\tilde{\theta}) I(\theta) ^ {-1} W(\theta)\right]$

the test statistic is:

$m(\tilde{\theta}) ^ \top Q ^ {-1} m(\tilde{\theta})$

and under Ho follows a $$\chi^2$$ with $$r$$ (the length of $$m$$) degrees of freedom.

$$Q$$ plays the role of the information matrix for the score test. As previously, different flavours of the test can be obtained depending on how $$Q$$ is estimated:

• numerically, using as previously $$\hat{I}_N = G(\tilde{\theta}) ^ \top G(\tilde{\theta})$$ and $$\hat{W}_N = G(\tilde{\theta}) ^ \top M(\tilde{\theta})$$,
• using the analytical derivatives, using as previously $$\hat{I}_A = - H(\tilde{\theta})$$ and $$\hat{W}_A = - \frac{\partial M}{\partial \theta}(\tilde{\theta})$$

In this latter case, the test statistic can also be obtained as $$N$$ times the $$R^2$$ of the regression of a column of one on the columns of $$G$$ and $$M$$. The following function performs the test.

cm_norm <- function(x, type = c("analytical", "opg", "reg")){
type <- match.arg(type)
X <- model.matrix(x)
y <- model.response(model.frame(x))
N <- length(y)
K <- length(coef(x))
coefs <- c(coef(x), sigma = sigma2(x))
beta <- coefs[1:K]
sigma <- coefs[K + 1]
epsilon <- as.numeric(y - X %*% beta)
G <- cbind(1 / sigma ^ 2 * X * epsilon,
sigma = - 1  / sigma + 1 / sigma ^ 3 * epsilon ^ 2)
M <- cbind(asym = epsilon ^ 3, kurt = epsilon ^ 4 - 3 * sigma ^ 4)
m <- apply(M, 2, sum)
if (type == "analytical"){
I <- - rbind(cbind(- 1 / sigma ^ 2 * crossprod(X),
sigma = - 2 / sigma ^ 3 * apply(X * epsilon, 2, sum)),
sigma = c(- 2 / sigma ^ 3 * apply(X * epsilon, 2, sum),
1 / sigma ^ 2 - 3 / sigma ^ 4 * sum(epsilon ^ 2)))
W <- - rbind(cbind(asym = - 3 * apply(epsilon ^ 2 * X, 2, sum),
kurt = - 4 * apply(epsilon ^ 3 * X, 2, sum)),
sigma = c(0, - 12 * N / sigma ^ 3))
}
if (type == "opg"){
I <- crossprod(G)
W <- crossprod(G, M)
}
if (type != "reg"){
Q <- crossprod(M - G %*% solve(I) %*% W)
stat <- as.numeric(m %*% solve(Q) %*% m)
}
else stat <- summary(lm(rep(1, N) ~ G + M - 1))\$r.squared * N
stat
}

For the linear model, we get:

cm_norm(lin_lm, "analytical")
## [1] 12.2857
cm_norm(lin_lm, "opg")
## [1] 13.41273
cm_norm(lin_lm, "reg")
## [1] 13.41273

and the normality hypothesis is therefore strongly rejected. For the log-linear model:

cm_norm(log_lm, "analytical")
## [1] 0.07940664
cm_norm(log_lm, "opg")
## [1] 0.6255081
cm_norm(log_lm, "reg")
## [1] 0.6255081

on the contrary, the normality hypothesis is not rejected.

# References

Newey, Whitney K. 1985. “Maximum Likelihood Specification Testing and Conditional Moment Tests.” Econometrica 53 (5): 1047–70. https://www.jstor.org/stable/1911011.

Tauchen, George. 1985. “Diagnostic Testing and Evaluation of Maximum Likelihood Models.” Journal of Econometrics 30 (1): 415–43. https://doi.org/https://doi.org/10.1016/0304-4076(85)90149-6.


text:
International
Tables for
Crystallography
Volume B
Reciprocal space
Edited by U. Shmueli

International Tables for Crystallography (2010). Vol. B, ch. 3.4, pp. 452-453   | 1 | 2 |

## Section 3.4.5. Extension of the method to a composite lattice

D. E. Williamsa

aDepartment of Chemistry, University of Louisville, Louisville, Kentucky 40292, USA

### 3.4.5. Extension of the method to a composite lattice

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Define a general lattice sum over direct-space points which interact with pairwise coefficients , where :where the prime indicates that when the self-terms with are omitted. For convenience the terms may be divided into three groups: the first group of terms has , where j is unequal to k; the second group has d not zero and j not equal to k; and the third group had d not zero and . (A possible fourth group with and is omitted, as defined.)By expanding this expression we obtainThis expression for V has nine terms, which are numbered on the right-hand side. Term (3) can be expressed in terms of Γ rather than γ:It is seen that cancellation occurs with term (1) so thatwhich is the , j unequal to k portion of the treated direct-lattice sum. The d unequal to 0, j unequal to k portion corresponds to term (2) and the d unequal to 0, portion corresponds to term (6). The direct-lattice terms may be consolidated asNow let us combine terms (4) and (8), carrying out the h summation first:Terms (5) and (9) may be combined:The final formula is shown below. The significance of the four terms is: (1) the treated direct-lattice sum; (2) a correction for the difference resulting from the removal of the origin term in direct space; (3) the reciprocal-lattice sum, except ; and (4) the term of the reciprocal-lattice sum.


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# What is your Time Table ??

Hi guys...just thought of sharing my time table and I also want to know yours. how to manage time,how to give quality time for studying,are topics that we should always think while writing a time table.

Do share yours. $\boxed{6:00 A.M-7:00 A.M - wake up get freshed and revise concepts 7:00 A.M-8:00 A.M -go to Gym 8:30A.M-11:30 A.M-study new concepts 11:30A.M-1:00P.M P.M-Brilliant* 1:00P.M-3:30P.M-study 4:P.M-8:30 P.M-class 9:00-11:00 P.M-study 11:00-6;00 A.M- RIP. :D }$

thnkx

Note by Max B
6 years, 8 months ago

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sorry for the latex that i used..

- 6 years, 8 months ago

you study a lot....good

- 6 years, 8 months ago

what's ur timetable

- 6 years, 8 months ago

My time table is not a ideal time table which could be followed....I really need to work on my time table... I am not going to tell my time table ,but it's how I spend my whole day-

Get up at 5 am

From 6:30 am to 11:30 am I got my school

Then,I work on Brilliant...!till 2:30 pm ,about...

get a nap...

From 3:30 to 5:30 -I do some reading work,like newspapers,Editorial ,other than course ,et cetra

From 7:00 Do some school work till 9:30 or less.

Ate some ,then ,aaaaaaaammmmmm(drowsy)

ZZZzzzzz....

- 6 years, 8 months ago

Oh My god! Which school do you go to? 6.30 is when we wake up here in Chennai...and you have only 5 hours of schooling. o.O? Lucky!!!!!! We have school from 8.15 -4 :(...and by the time I come home , my head goes for a six and I barely have time left for homework which really dampens my progress prohibiting me from using Brilliant...U.ppl are so lucky

- 6 years, 8 months ago

@Krishna Ar Really we have similar daily routine in US I got to school from 6.30- to 2.30.Then I sleep, do my homework, study concepts and then use brilliant and Wikipedia to learn how to solve problems

- 6 years, 8 months ago

Wow! But you have school from 6.30am to 3.30 pm?

- 6 years, 8 months ago

sorry it is 2.30

- 6 years, 8 months ago

Hey @Max B

You can use Markdown to make the table, like this:

 This is a markdown table
 1 2 || This|| is|| || a markdown || table.||

I really want to see your timetable!

Staff - 6 years, 8 months ago

$\|| 1 || 2 || 3 || 4 || || a || b || c || d ||$

- 6 years, 8 months ago

I need practice....u know why it takes time.....thankx by the way

- 6 years, 8 months ago


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What is 1.5 pair cable?

I've come across a spec for 1.5 pair cable and wondered if it's just three conductor cable. It would be nice to have this confirmed, as well as have an explanation for why it's called in such a roundabout manner. Is the word pair a specific terminology in this context which somehow conveys something "(un)shielded 3-conductor cable" does not?

• Link? It may be 1 twisted pair and a loose conductor, which would have some uses. "Pair" does sort of imply twisted to improve balance. Dec 1 '20 at 20:12
• I think it is just a special case of a multipair cable. So in n-pair you substitute n with 1.5. The difference with 3-conductor cable would be that in 1.5-pair, you can have a twisted pair bundled separately from the additional conductor. Dec 1 '20 at 20:14


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# Area between $y=2+|x-1|$ and $y=-\frac{1}{5}x+7$

Question 17, page 448, from Anton 8th. The question asks for the area, and the answer is 24.

Now, I did draw the graph, found the points where both functions touch each other by $f(x) = g(x)$:

(-5,8) and (5,6);

I'm having problems solving the integration. I've tried so many times to solve it, but it comes to the same integral:

$$\int_{-5}^{5}-\frac{1}{5}x+7-2+|x-1|$$ then: $$\int_{-5}^{5}-\frac{1}{5}x+5+|x-1|$$

$$-\frac{x^2}{10} + 5x + |\frac{x^2}{2}-x|$$

Replacing values: 5:

$$-\frac{5^2}{10}+5.5+\frac{5^2}{2}-5 = \color{blue}{-\frac{25}{10}+25+\frac{25}{2}-5 = 30}$$

For -5:

$$-\frac{(-5)^2}{10}+5.(-5)-\frac{(-5)^2}{2}+(-5) = \color{blue}{-\frac{25}{10}-25-\frac{25}{2}-5} = -45$$

then the difference between these:

$$30 - (-45) = 75$$

I can't understand what I'm doing wrong, since the $|\frac{x^2}{2}-x|$ need to behave different on each value, for +5 = $\color{blue}{\frac{5^2}{2}-5}$ and for -5 = $\color{red}{-\frac{(-5)^2}{2}+(-5)}$

I searched by a integration calculator and found this: integral-calculator.com, but I've never seen anything like $$\frac{-(x+1).|x-1|}{2}$$

What am I doing wrong?

Updating with graph picture:

• I haven't looked your work but as a general tip, I find these kinds of problems much easier if you draw a picture (so that you know what the proper bounds of integration are). – Zubin Mukerjee Oct 22 '14 at 0:02
• @ZubinMukerjee I've put a image from WinPlot, as I stated, the points are correct so the bounds are -5 and 5. I'm having trouble to solve the integration. – Fabiano Araujo Oct 22 '14 at 0:06

It is not the case that: $$\int |g(x)| \, dx = \left|\int g(x) \, dx\right|$$ Instead, to get rid of the absolute values, we use the fact that: $$|x - 1| = \begin{cases} x - 1 &\text{if } x \geq 1 \\ 1 - x &\text{if } x < 1 \end{cases}$$ We thus split the integral into two separate ones: $$\int_{-5}^1 [(\tfrac{-1}{5}x + 7) - (2 + (1 - x))] \, dx + \int_{1}^5 [(\tfrac{-1}{5}x + 7) - (2 + (x - 1))] \, dx$$
• Well, that was an adventure! Finally! Thank you so much. I was considering the absolute values as $x >= 0$ and $x < 0$. That was confusing everything. – Fabiano Araujo Oct 22 '14 at 0:38


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?

Free Version
Moderate

# Kirchoff's Laws

APPHEM-1G4KQL

In the circuit above, a current of magnitude $I_1$ travels to the right through $R_1$ and a current of magnitude $I_2$ travels from top to bottom through resistor $R_2$.

Which of the following could be used to analyze the circuit shown in the figure above?

Select all that apply.

A

$\mathit{2V – I_1 R1 – I_2 R_2=0}$

B

$\mathit{2V – I_1 R_1 – I_2 R_2-V – I_2 R_3=0}$

C

$\mathit{2V – I_1 R_1 – (I_1 – I_2) R_3 – V=0}$

D

$\mathit{2V – I_1 R_1 – (I_1 – I_2) R_3 + V=0}$

E

$\mathit{2V – I_1 R_1 –I_2 R_3 – V=0}$

F

$\mathit{V – I_2 R_2 + I_2 R_3 =0}$

G

$\mathit{-V – (I_1-I_2) R_3 + I_2 R_2 =0}$


text:
# How to solve $\frac{\text{d}y}{\text{d}\tau} = -\varepsilon y/\left(1+\frac{k\Delta}{(ky+1)^2}\right)$

If we take

$$\frac{\text{d}y}{\text{d}\tau} = -\varepsilon y/\left(1+\frac{k\Delta}{(ky+1)^2}\right) \qquad\qquad (1)$$

with $y(0)=y_0$ then we can solve it as such: \begin{align} -\varepsilon\tau + C_0 &= \int \frac {1+\frac{k\Delta}{\left(ky+1\right)^2}} {y} \ \text{d}y \\ %% &= \int \frac {1} {y} % + % \frac {k\Delta} {y\left(ky+1\right)^2} \ \text{d}y\\ %% &= \log{y} % + % \frac{\Delta}{k}\log{\left(\frac{y}{k+y}\right)} % + % \frac{\Delta}{k+y} \end{align}

with \begin{align} C_0 &= \log{y_0} % + % \frac{\Delta}{k}\log{\left(\frac{y_0}{k+y_0}\right)} % + % \frac{\Delta}{k+y_0} \end{align}

Therefore we can write: \begin{align} \tau= - \frac{1}{\varepsilon} \left( \log{y} % + % \frac{\Delta}{k}\log{\left(\frac{y}{k+y}\right)} % + % \frac{\Delta}{k+y} -C_0\right) \qquad\qquad (2) \end{align}

I have solved $(1)$ numerically, using an ODE solver in MATLAB, with the following parameters: $$\Delta=0.4351$$ $$\varepsilon=4.4261\times 10^{-5}$$ $$k=22.9806$$ $$y_0=0.5945$$ $$\tau\in [0,200000]$$

This is plotted by the black line in the following figure. I then solved $(2)$ for $\tau$ for $y\in[y_\text{min},y_0]$ where $y_\text{min}$ was the smallest value for y output from the numerical simulation. I then plotted this on the same figure, in dashed magenta. They don't match! I have checked this over and over, I do not know what I have done wrong here. Any help would be really appreciated.

• How did you get $y + k$ terms from $ky +1$ terms? I have not gone through your calculations - maybe if you could update with the integrals you performed. – Chinny84 Mar 9 '17 at 14:51
• @Chinny84: Shit. One second. – Freeman Mar 9 '17 at 14:56
• @Chinny84: That fixed it. The integral is wrong. – Freeman Mar 9 '17 at 14:59
• Been there and done it - good thing about getting old you have seen alot of short cuts for checking your result ;)! – Chinny84 Mar 9 '17 at 15:02
• @Chinny84: You sound like my supervisor xD – Freeman Mar 9 '17 at 15:11

Well, as you said we can solve it:

$$\text{y}'\left(\tau\right)=-\frac{\epsilon\cdot\text{y}\left(\tau\right)}{1+\frac{\text{k}\cdot\Delta}{\left(1+\text{k}\cdot\text{y}\left(\tau\right)\right)^2}}\space\Longleftrightarrow\space\int\frac{\text{y}'\left(\tau\right)}{\frac{\text{y}\left(\tau\right)}{1+\frac{\text{k}\cdot\Delta}{\left(1+\text{k}\cdot\text{y}\left(\tau\right)\right)^2}}}\space\text{d}\tau=\int-\epsilon\space\text{d}\tau\tag1$$

Now, we get:

• $$\int-\epsilon\space\text{d}\tau=-\epsilon\int1\space\text{d}\tau=\text{C}_1-\epsilon\cdot\tau\tag2$$
• Substiute $\text{u}=\text{y}\left(\tau\right)$: $$\int\frac{\text{y}'\left(\tau\right)}{\frac{\text{y}\left(\tau\right)}{1+\frac{\text{k}\cdot\Delta}{\left(1+\text{k}\cdot\text{y}\left(\tau\right)\right)^2}}}\space\text{d}\tau=\int\frac{1}{\frac{\text{u}}{1+\frac{\text{k}\cdot\Delta}{\left(1+\text{k}\cdot\text{u}\right)^2}}}\space\text{d}\text{u}=\int\frac{1+\frac{\text{k}\cdot\Delta}{\left(1+\text{k}\cdot\text{u}\right)^2}}{\text{u}}\space\text{d}\text{u}=$$ $$\int\frac{1}{\text{u}}\space\text{d}\text{u}+\text{k}\cdot\Delta\int\frac{1}{\left(1+\text{k}\cdot\text{u}\right)^2}\cdot\frac{1}{\text{u}}\space\text{d}\text{u}\tag3$$
• $$\int\frac{1}{\text{u}}\space\text{d}\text{u}=\ln\left|\text{u}\right|+\text{C}_2\tag4$$
• $$\int\frac{1}{\left(1+\text{k}\cdot\text{u}\right)^2}\cdot\frac{1}{\text{u}}\space\text{d}\text{u}=\ln\left|-\text{k}\cdot\text{u}\right|-\ln\left|1+\text{k}\cdot\text{u}\right|+\frac{1}{1+\text{k}\cdot\text{u}}+\text{C}_3\tag5$$

So, we get:

$$\ln\left|\text{y}\left(\tau\right)\right|+\text{k}\cdot\Delta\cdot\left\{\ln\left|-\text{k}\cdot\text{y}\left(\tau\right)\right|-\ln\left|1+\text{k}\cdot\text{y}\left(\tau\right)\right|+\frac{1}{1+\text{k}\cdot\text{y}\left(\tau\right)}\right\}=\text{C}-\epsilon\cdot\tau\tag6$$

We can simplify a very little:

• $$\left|-\text{k}\cdot\text{y}\left(\tau\right)\right|=\left|-\text{k}\right|\cdot\left|\text{y}\left(\tau\right)\right|=\left|\text{k}\right|\cdot\left|\text{y}\left(\tau\right)\right|\tag7$$
• $$\ln\left|-\text{k}\cdot\text{y}\left(\tau\right)\right|-\ln\left|1+\text{k}\cdot\text{y}\left(\tau\right)\right|=\ln\left(\frac{\left|\text{k}\right|\cdot\left|\text{y}\left(\tau\right)\right|}{\left|1+\text{k}\cdot\text{y}\left(\tau\right)\right|}\right)\tag8$$

So, when we use the values:

$$\ln\left(0.5945\right)+9.99885906\ln\left(0.931796325795775\right)+0.6819589257422061=\text{C}\tag9$$

• Thank you for helping! Very much appreciated, it's all sorted now! – Freeman Mar 13 '17 at 13:29


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# TikZ picture: two arrow; one over the other

I have the following code:

\documentclass{article}

\usepackage{tikz}
\usetikzlibrary{matrix,arrows}

\begin{document}
$\begin{tikzpicture} \matrix (m) [matrix of math nodes,row sep=3em,column sep=1em,minimum width=2em]{ A & & G \\ & S & \\ }; \path[-stealth] (m-1-1) edge node [above] {f} (m-1-3) (m-1-1) edge [bend right] node [left] {p\;} (m-2-2) (m-1-3) edge node [right] {\;q} (m-2-2) (m-2-2) edge node [right] {\epsilon} (m-1-1); \path[-stealth] ([yshift=5pt]m-1-1) edge node [right] {g} ([yshift=5pt]m-1-3); \end{tikzpicture}$
\end{document}

I dare you to compile it. Then you will get a disgusting output. What I want is to get the two arrow on top one over the other (just like corolary 1.3.1.5, page 70, in the notes on http://math.umn.edu/~kwlan/articles/cpt-PEL-type-thesis-revision.pdf).

Can anyone help? I would love if the solution didn't modify too much the way I coded the diagram since it is the way I always do it (with the matrix) but beggers can't be choosers, so.

-

Here is an attempt to fix it by using the .east and .west anchors:

## Notes:

• Not sure why you were using math mode -- I have removed that in the below MWE.

## Code:

\documentclass{article}

\usepackage{tikz}
\usetikzlibrary{matrix,arrows}

\begin{document}
\begin{tikzpicture}
\matrix (m) [matrix of math nodes,row sep=3em,column sep=1em,minimum width=2em]{
A &   & G \\
& S &   \\
};
\path[-stealth]
(m-1-1.east) edge node [above,yshift=1.0ex] {$f$} (m-1-3.west)
(m-1-1) edge [bend right] node [left] {$p\;$} (m-2-2)
(m-1-3) edge node [right] {$\;q$} (m-2-2)
(m-2-2) edge node [right] {$\epsilon$} (m-1-1);
\path[-stealth]
([yshift=5pt]m-1-1.east) edge node [below,,yshift=-1.0ex] {$g$} ([yshift=5pt]m-1-3.west);
\end{tikzpicture}
\end{document}

-
Amazing, thanks ! – ortholle Feb 16 '14 at 18:28

It's quite easy with tikz-cd:

\documentclass{article}
\usepackage{tikz-cd}
\begin{document}
$\begin{tikzcd}[column sep=1.5em] A \arrow[yshift=.85ex]{rr}{f} \arrow[yshift=-.45ex,swap]{rr}{g} \arrow[yshift=-.4ex,bend right,swap]{dr}{p} && G \arrow{dl}{q} \\ & S \arrow[swap]{ul}{\epsilon} \end{tikzcd}$
\end{document}

-

Here a variant of Peter's code but without matrix and some adjustment with the position of some arrows.

\documentclass{article}
\usepackage{tikz}
\usetikzlibrary{arrows}

\begin{document}
\begin{tikzpicture}
\path          node (A) {$A$}
(0:4cm)    node (G) {$G$}
(-60:4cm)  node (S) {$S$};
\path[-stealth]
(A) edge [bend right] node [left]  {$p\;$}      (S)
(G) edge              node [right] {$\;q$}      (S)
(S) edge              node [right] {$\epsilon$} (A)
([yshift=-2.5pt]A.east) edge node [above,yshift= 1.0ex]  {$f$} ([yshift=-2.5pt]G.west)
([yshift= 2.5pt]G.west) edge node [below,yshift=-1.0ex]  {$g$} ([yshift= 2.5pt]A.east);
\end{tikzpicture}
\end{document}

-


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# Volcano Watch — The ghosts of holiday eruptions past

Release Date:

Every year about this time, the HVO staff asks the question: Will the volcano behave over the holidays? Given the amount of volcanic and seismic activity on this island, our holidays are bound to be disrupted now and then.

The busiest holiday season in recent memory occurred during the current eruption of Kīlauea, when lavaflows cut a swath through the old Hawaiian village of Kapaahu on Thanksgiving Day 1986, then shifted 3 km (2 mi) to the east and destroyed 15 houses in Kalapana just before Christmas.

How many readers remember when the quickest way to get from Volcano to the black sand beach at Kaimu was to drive down Chain of Craters Road through the national park? Once you reached the coast, you sped past the beautiful picnic area at Kamoamoa and the visitor center at Wahaula, carelessly assuming that you had all the time in the world to go back and explore these places thoroughly. Then you were out of the park and passing the turnoff to Royal Gardens subdivision on the left and the dirt road to Punaluu (Queens Bath) on the right. A cluster of houses appeared on the inland side of the highway at Kapaahu, but you hurried on to the beach. All that changed in late 1986.

Kapaahu had forewarning of what was to come. Aa flows from episodic eruptions of Puu Oo spilled over the pali into nearby Royal Gardens several times in 1983-84, destroying 16 houses. In April 1984, an aa flow took a more easterly course and destroyed three houses on the northwestern edge of Kapaahu. None of these flows reached the highway, and the damage they did was limited, because most eruptive episodes lasted less than a day.

In July 1986, the eruption shifted to Kupaianaha, a new vent 3 km (2 mi) northeast of Puu Oo. The activity changed from episodic fountaining to continuous, quiet effusion, which promoted the development of lava tubes and allowed lava to move far from the vent before cooling and stagnating. By November 1986, tube-fed flows were slowly advancing down Pulama pali toward Kapaahu.

The day before Thanksgiving was sunny and clear, but a noxious cloud of black smoke settled over Kapaahu as lava crossed Highway 130, setting fire to the asphalt. The crackle of burning vegetation was punctuated by low booms from methane explosions near the edge of the flow. One of these explosions lobbed a football-sized chunk of rock onto the highway, startling the head of civil defense (now our mayor), who sprang into action and hustled onlookers away.

By nightfall, lava had overrun five houses above the highway and had flowed through the wrought-iron gates of Pacific Paradise Ocean Front Estates across the road. This subdivision had only two houses; both were destroyed by early Thanksgiving Day.

A historic structure within the subdivision lingered much longer. The ruins of Father Damien's Church, built in about 1870, escaped the first round of lava flows through Kapaahu. The high walls of the church, hidden beneath a tangle of monstera vines and shaded by huge plumeria trees, became a popular oasis for those of us working on the sweltering flows nearby. Unlike most structures destroyed by this eruption, the concrete walls of the old church were gradually buried, not burned, and finally disappeared beneath the rising tide of pahoehoe in early 1988.

The flows skirted Punaluu, leaving this lovely spot untouched for a few more months, but filled in the shallow tidal pools at the shoreline beyond. Lava entered the ocean the day after Thanksgiving for the first time since 1973 during the Mauna Ulu eruption.

A disruption in the tube system near the vent granted Kapaahu a temporary reprieve in early December but diverted lava toward Kalapana, where a week before Christmas, 14 homes burned in a single day.

Currently, the county's lava-viewing road ends at the site of Kapaahu, and lava is entering the ocean in the same place it first did in 1986, though the shoreline has extended seaward 280 m (900 ft) in the intervening years. For now, we can give thanks that no inhabited houses have been destroyed in the last decade, but we should not become complacent about the future.

### Volcano Activity Update

Eruptive activity of Kīlauea Volcano continued unabated at the Puu Oo vent during the past week. Lavamoves away from the vent toward the ocean in a network of tubes and descends Pulama pali in several separate tubes. Surface flows are observed above the pali and at various places along the trace of the Kamoamoa tube system. Lava continues to enter the ocean in two locations: Kamoamoa and the area east of Kupapau. The entry at Kupapa`u stopped some time last week, and the flow approaching the coast in the same area has stalled.

The public is reminded that the benches of the ocean entries are very hazardous, with possible collapses of the unstable new land. The steam clouds are extremely hot, highly acidic, and laced with glass particles. Swimming at the black sand beaches of the benches can be a blistering or even deadly venture.

There were no earthquakes reported felt during the week ending on November 21.


text:
Geometric meaning of L-genus - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-22T10:18:42Z http://mathoverflow.net/feeds/question/89981 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/89981/geometric-meaning-of-l-genus Geometric meaning of L-genus Mauricio 2012-03-01T18:13:43Z 2012-05-13T10:11:11Z
Is there any reasonable geometric meaning of the L-genus for smooth manifolds? or perhaps easier, for complex algebraic surfaces?

The question came up after a friend and I realized that we don't understand why one would expect to have such a formula for the signature in terms of Pontrjagin classes (i.e. the signature theorem). Any insights about this will be appreciated.
 http://mathoverflow.net/questions/89981/geometric-meaning-of-l-genus/89986#89986 Answer by Csar Lozano Huerta for Geometric meaning of L-genus Csar Lozano Huerta 2012-03-01T19:14:00Z 2012-03-01T19:14:00Z
Let $S$ be a smooth algebraic complex surface. Then, there is the following relation: $$p_1=c_1(S)^2-2c_2(S)=K_S^2-2\chi_{top}(S)=3L$$ where $p_1$ is the first Pontryagin class and $L$ the L-genus.

On the other hand, cobordism theory says $p_1[S]=3\tau$ where $\tau$ is the signature of $S$.

Now (by Hodge theory)

$\tau=4\chi(\mathcal{O_S})-\chi_{top}(S)$ therefore, (pairing off with the fundamental class) the relation among $L$ and $Td$ looks like

$$K^2+\chi_{top}(S)=3\tau+3\chi_{top}(S)= 12Td(S)$$

where the second Todd class satisfies $Td(S)=1/12 (K^2+\chi_{top}(S))$
 http://mathoverflow.net/questions/89981/geometric-meaning-of-l-genus/90001#90001 Answer by Greg Friedman for Geometric meaning of L-genus Greg Friedman 2012-03-02T00:31:03Z 2012-03-02T00:31:03Z
Hirzebruch himself has a very nice paper explaining (if I remember correctly) how he came up with the signature theorem and why the formulas arise in a fairly reasonable way. Here's the reference: MR0368023 (51 #4265) Hirzebruch, F. The signature theorem: reminiscences and recreation. Prospects in mathematics (Proc. Sympos., Princeton Univ., Princeton, N.J., 1970), pp. 3–31. Ann. of Math. Studies, No. 70, Princeton Univ. Press, Princeton, N.J. 1971.
 http://mathoverflow.net/questions/89981/geometric-meaning-of-l-genus/90030#90030 Answer by Liviu Nicolaescu for Geometric meaning of L-genus Liviu Nicolaescu 2012-03-02T11:56:09Z 2012-05-13T10:11:11Z
I take that you ask how in the world did Hirzebruch come up with the complicated expression of the L-genus?

The key fact behind this is that the signature is a genus, i.e. a ring morphism $\gamma:\Omega^\bullet_+\to\mathbb{R}$ from the oriented cobordism ring to the (ring of) reals. By Thom's work we deduce that a genus is determined by its values on $\mathbb{CP}^{2n}$ and thus by the generating series

$$r^\gamma(t)=1+\sum_{n\geq 1}\gamma(\mathbb{CP}^{2n})t^{2n}.$$

In the case of signature we have

$$r^\gamma(t)=\frac{1}{1-t^2}.$$

How does one go from this series to the the function $\frac{\xi}{\tanh \xi}$ that enters into the defintion of $L$? As a teaser, let me point out that

$$\xi =\log \left( \frac{1}{1-t^2}\right) \Leftrightarrow t=\tanh \xi.$$

The signature theorem follows from the above observations using a bit of algebraic combinatorics. For details see Chap. 7 of these lecture notes for a graduate course on this topic that I taught in 2008.


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# Normal and shear components of the stress vector

1. Aug 25, 2013

### roldy

1. The problem statement, all variables and given/known data
Find the normal and shear components ($\vec{\sigma}_{n},\vec{\tau}$) of the stress vector on the surface ABCD.

Given:
State of stress at a point
$\sigma_{xx}=3$
$\sigma_{yy}=0$
$\sigma_{zz}=-1$
$\sigma_{xy}=-\sqrt{2}$
$\sigma_{xz}=1000$

Unit normal:
$\widehat{n}=\frac{1}{\sqrt{2}}\widehat{i}+\frac{1}{\sqrt{2}}\widehat{j}+0\widehat{k}$

2. Relevant equations
$\vec{\sigma}_{n}=\vec{t}\cdot\widehat{n}$
$\vec{t}=\vec{\sigma}_{n}+\vec{\tau}$

3. The attempt at a solution

First I take the stress matrix and multiply it by the unit normal to find the traction components

$\sigma = \left[\begin{matrix} 3 & -\sqrt{2} & 1000 \\ -\sqrt{2} & 0 & -\sqrt{2} \\ 1000 & -\sqrt{2} & -1 \\ \end{matrix} \right]$

$\left[\begin{matrix} 3 & -\sqrt{2} & 1000 \\ -\sqrt{2} & 0 & -\sqrt{2} \\ 1000 & -\sqrt{2} & -1 \\ \end{matrix} \right] \left[\begin{matrix} \frac{1}{-\sqrt{2}} \\ \frac{1}{-\sqrt{2}} \\ 0 \\ \end{matrix} \right] = \left[\begin{matrix} t_{x} \\ t_{y} \\ t_{z} \\ \end{matrix} \right]$

This results in
$\vec{t}= \left[\begin{matrix} \frac{3}{\sqrt{2}}-1 \\ -1 \\ \frac{1000}{\sqrt{2}}-1 \\ \end{matrix} \right]$

Then I find the normal component

$\vec{\sigma}_{n}=\left[ \frac{3}{\sqrt{2}}-1,-1,\frac{1000}{\sqrt{2}}-1\right] \cdot \left[\begin{matrix} \frac{1}{-\sqrt{2}} \\ \frac{1}{-\sqrt{2}} \\ 0 \\ \end{matrix} \right]$

$\vec{\sigma}_{n} = \frac{3}{2}-\sqrt{2}$

Using the hint that was provided, I solve for $\vec{\tau}$
$\vec{\tau}=\vec{t}-\vec{\sigma}_n$

Is this procedure correct thus far? I've seen elsewhere that
$\tau=\vec{t}-(\vec{t}\cdot \vec{n})\vec{n}$

I'm a bit confused with the notation of the supplied equation for $\sigma$ because if you take the dot product of a vector with a unit normal you should only get a scalar. The notation shows that it is a vector.

2. Aug 25, 2013

### Staff: Mentor

In your notation, $\vec{\sigma}$ is the stress tensor. The stress vector on a surface perpendicular to the unit normal $\vec{n}$ is given by:
$$\vec{t}=\vec{\sigma}\centerdot\vec{n}$$
Note that $\vec{\sigma}$ is a second order tensor, and, when you dot it with the unit normal, you produce a vector (first order tensor) $\vec{t}$. The vector $\vec{t}$ is not pointing perpendicular to the surface. If you want the component perpendicular to the surface, you need to dot it with the unit normal.
Scalar component of stress vector perpendicular to the surface = $\vec{n}\centerdot\vec{t}$
The normal vector component of the stress vector on the surface is equal to the scalar component times the unit normal $\vec{\sigma_n}=(\vec{n}\centerdot\vec{t})\vec{n}$
The tangential vector component of the stress vector on the surface is equal to the stress vector minus the normal vector component:$\vec{\tau}=\vec{t}-(\vec{n}\centerdot\vec{t})\vec{n}$

Hope this helps.

Chet