text: An injection may also be called a one-to-one (or 1–1) function; some people consider this less formal than "injection''. We An injective function is also called an injection. Taking the contrapositive, $f$ I know that there does not exist a continuos function from [0,1] onto (0,1) because the image of a compact set for a continous function f must be compact, but isn't it also the case that the inverse image of a compact set Cost function in linear regression is also called squared error function.True Statement 3 M. Hauskrecht Surjective function Definition: A function f from A to B is called onto, or surjective, if and only if for every b B there is an element a A such that f(a) = b. I was doing a math problem this morning and realized that the solution lied in the fact that if a function of A -> A is one to one then it is onto. A function is an onto function if its range is equal to its co-domain. How many injective functions are there from An "onto" function, also called a "surjection" (which is French for "throwing onto") moves the domain A ONTO B; that is, it completely covers B, so that all of B is the range of the function. In other words, the function F … that $g(b)=c$. If f: A → B and g: B → C are onto functions show that gof is an onto function. For example, the rule f(x) = x2 de nes a mapping from R to R which is NOT injective since it sometimes maps two inputs to the same output (e.g., both 2 and 2 get mapped onto 4). In an onto function, every possible value of the range is paired with an element in the domain. always positive, $f$ is not surjective (any $b\le 0$ has no preimages). For example, the rule f(x) = x2 de nes a mapping from R to R which is NOT injective since it sometimes maps two inputs to the same output (e.g., both 2 and 2 get mapped onto 4). For example, in mathematics, there is a sin function. If f and g both are onto function, then fog is also onto. EASY Answer since g: B → C is onto suppose z ∈ C,there exists a pre-image in B Let the pre-image be … I was doing a math problem this morning and realized that the solution lied in the fact that if a function of A -> A is one to one then it is onto. Onto Functions When each element of the f(2)=t&g(2)=t\\ Section 7.2: One-to-One, Onto and Inverse Functions In this section we shall developed the elementary notions of one-to-one, onto and inverse functions, similar to that developed in a basic algebra course. Surjective (Also Called "Onto") A function f (from set A to B ) is surjective if and only if for every y in B , there is at least one x in A such that f ( x ) = y , in other words f is surjective if and only if f(A) = B . On the other hand, $g$ fails to be injective, We are given domain and co-domain of 'f' as a set of real numbers. • A function f is a one-to-one correspondence, or a bijection, or reversible, or invertible, iff it is both one-to- one and onto. It is so obvious that I have been taking it for granted for so long time. [2] Alternative: all co-domain elements are covered A f: A B B Example 19 Show that if f : A → B and g : B → C are onto, then gof : A → C is also onto. $f\colon A\to B$ and an injection $g\,\colon B\to C$ such that $g\circ f$ \begin{array}{} factorizations.). If x = -1 then y is also 1. \begin{array}{} is neither injective nor surjective. b) Find an example of a surjection 7.2 One-to-one and onto Functions_0d7c552f25def335a170bcdbd6bcbafd.pdf - 7.2 One-to-One and Onto Function One-to-One A function \u2192 is called one-to-one • A function f is a one-to-one correspondence, or a bijection, or reversible, or invertible, iff it is both one-to- one and onto. In this section, we define these concepts In other words, the function F maps X onto … Example 4.3.8 a) Find a function $f\colon \N\to \N$ one-to-one and onto Function • Functions can be both one-to-one and onto. 3 M. Hauskrecht Surjective function Definition: A function f from A to B is called onto, or surjective, if and only if for every b B there is an element a A such that f(a) = b. Suppose $A$ and $B$ are non-empty sets with $m$ and $n$ elements That is, in B all the elements will be involved in mapping. An onto function is sometimes called a surjection or a surjective function. Proof. A function f: A -> B is called an onto function if the range of f is B. $f\colon A\to B$ and a surjection $g\,\colon B\to C$ such that $g\circ f$ 2. is onto (surjective)if every element of is mapped to by some element of . In other words, nothing is left out. An onto function is sometimes called a surjection or a surjective function. f(4)=t&g(4)=t\\ One-one and onto mapping are called bijection. Such functions are usually divided into two important classes: the real analytic functions and the complex analytic functions, which are commonly called holomorphic functions. called the projection onto $B$. $f\colon A\to B$ is injective if each $b\in "officially'' in terms of preimages, and explore some easy examples Here are the definitions: 1. is one-to-one (injective) if maps every element of to a unique element in . When working in the coordinate plane, the sets A and B may both become the Real numbers, stated as f : R→R In other words, every element of the function's codomain is the image of at most one element of its domain. For one-one function: 1 Surjective, It merely means that every value in the output set is connected to the input; no output values remain unconnected. the number of elements in$A$and$B$? f (a) = b, then f is an on-to function. If$f\colon A\to B$is a function,$A=X\cup Y$and Hence the given function is not one to one. If f: A → B and g: B → C are onto functions show that gof is an onto function. ), and ƒ (x) = x². Such functions are referred to as onto functions or surjections. h4��"����jY �Q � ѷ���N߸rirЗ�(�-���gLA� u�/��PR�����*�dY=�a_�ϯ3q�K�$�/1��,6�B"jX�^���G2��F��^8[qN�R�&.^�'�2�����N��3��c�����4��9�jN�D�ϼǦݐ�� 4. An injection may also be called a one-to-one (or 1–1) function; some people consider this less formal than "injection''. In other words, nothing is left out. 4. then the function is onto or surjective. exceptionally useful. f(2)=r&g(2)=r\\ If x = -1 then y is also 1. a) Suppose $A$ and $B$ are finite sets and An onto function is also called a surjective function. A function ƒ: A → B is onto if and only if ƒ (A) = B; that is, if the range of ƒ is B. B$has at most one preimage in$A$, that is, there is at most one To say that the elements of the codomain have at most Theorem 4.3.11$g(x)=2^x$. Onto function or Surjective function : Function f from set A to set B is onto function if each element of set B is connected with set of A elements. An injective function is called an injection. I'll first clear up some terms we will use during the explanation. The rule fthat assigns the square of an integer to this integer is a function. If f and fog are onto, then it is not necessary that g is also onto. Transcript Ex 1.2, 5 Show that the Signum Function f: R → R, given by f(x) = { (1 for >0@ 0 for =0@−1 for <0) is neither one-one nor onto. parameters) are the data items that are explicitly given tothe function for processing. Since$g$is injective, If f and fog both are one to one function, then g is also one to one. Two simple properties that functions may have turn out to be Or we could have said, that f is invertible, if and only if, f is onto and one surjective functions. Let A = {a 1 , a 2 , a 3 } and B = {b 1 , b 2 } then f : A -> B. The figure given below represents a onto function. the other hand, for any$b\in \R$the equation$b=g(x)$has a solution that is injective, but one-to-one and onto Function • Functions can be both one-to-one and onto. Example: The function f(x) = 2x from the set of natural numbers N to the set of non-negative even numbers E is one-to-one and onto. In computer science, a call stack is a stack data structure that stores information about the active subroutines of a computer program. f(1)=s&g(1)=t\\ Therefore$g$is The function f is an onto function if and only if fory Since$f$is injective,$a=a'$. onto function; some people consider this less formal than 1 $$. In other Note that the common English word "onto" has a technical mathematical meaning. A function f from the set of natural numbers to the set of integers defined by f ( n ) = ⎩ ⎪ ⎪ ⎨ ⎪ ⎪ ⎧ 2 n − 1 , when n is odd − 2 n , when n is even View solution We can say that a function that is a mapping from the domain x to the co-domain y is invertible, if and only if -- I'll write it out -- f is both surjective and injective. A function f(1)=s&g(1)=r\\ Ex 4.3.6 what conclusion is possible? One-one and onto mapping are called bijection. A, a\ne a' implies f(a)\ne f(a'). 1 Definition. So then when I try to render my grid it can't find the proper div to point to and doesn't ever render. If f and fog are onto, then it is not necessary that g is also onto. are injections, surjections, or both. 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. In other words, if each b ∈ B there exists at least one a ∈ A such that. doing proofs. Since 3^x is 233 Example 97. surjective. Ex 4.3.7 relationship from elements of one set X to elements of another set Y (X and Y are non-empty sets Can we construct a function stream We can say that a function that is a mapping from the domain x to the co-domain y is invertible, if and only if -- I'll write it out -- f is both surjective and injective. than "injection''. A surjective function is called a surjection. Onto functions are alternatively called surjective functions. But sometimes my createGrid() function gets called before my divIder is actually loaded onto the page. 3. is one-to-one onto (bijective) if it is both one-to-one and onto. Thus it is a . Illustration Check whether y = f(x) = x 3; f : R → R is one-one/many-one/into/onto function. Definition 4.3.6 2.1. . (fog)-1 = g-1 o f-1 Some Important Points: Example 5.4.1 The graph of the piecewise-defined functions h: [1, 3] → [2, 5] defined by h(x) = … An onto function is also called a surjection, and we say it is surjective. Since f is surjective, there is an a\in A, such that Ex 4.3.8 Here f is injective since r,s,t have one preimage and Definition. What conclusion is possible regarding If f and fog both are one to one function, then g is also one to one. An injective function is called an injection. In other words, ƒ is onto if and only if there for every b ∈ B exists a ∈ A such that ƒ (a) = b. Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. Onto Functions When each element of the \end{array} g\circ f\colon A \to C is surjective also. f(5)=r&g(5)=t\\ Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. Example 4.3.7 Suppose A=\{1,2,3,4,5\}, B=\{r,s,t\}, and,$$ b) Find a function$g\,\colon \N\to \N$that is surjective, but Transcript Ex 1.2, 5 Show that the Signum Function f: R → R, given by f(x) = { (1 for >0@ 0 for =0@−1 for <0) is neither one-one nor onto. An onto function is also called a surjection, and we say it is surjective. In this case the map is also called a one-to-one correspondence. A function is an onto function if its range is equal to its co-domain. Theorem 4.3.5 If$f\colon A\to B$and$g\,\colon B\to Cf\vert_X$and$f\vert_Y$are both injective, can we conclude that$f$Ex 4.3.1 This means that ƒ (A) = {1, 4, 9, 16, 25} ≠ N = B. Ex 4.3.4 We can flip it upside down by multiplying the whole function by −1: g(x) = −(x 2) This is also called reflection about the x-axis (the axis where y=0) We can combine a negative value with a scaling: one-to-one (or 1–1) function; some people consider this less formal Definition 7 A function f : X → Y is said to be one-one and onto (or bijective), if f is both one-one and onto. I know that there does not exist a continuos function from [0,1] onto (0,1) because the image of a compact set for a continous function f must be compact, but isn't it also the case that the inverse image of a compact set Let be a function whose domain is a set X. Indeed, every integer has an image: its square. Note: for the examples listed below, the cartesian products are assumed to be taken from all real numbers. Example 19 Show that if f : A → B and g : B → C are onto, then gof : A → C is also onto. Definition: A function f: A → B is onto B iff Rng(f) = B. A surjective function, also called a surjection or an onto function, is a function where every point in the range is mapped to from a point in the domain. �>�t�L��T�����Ù�7���Bd��Ya|��x�h'�W�G84 \end{array} 8. Ifyou were to ask a computer to find the sin(2), sin would be the functio… and consequences. x��i��U��X�_�|�I�N���B"��Rȇe�m�X��>���������;�!Eb�[ǫw_U_���w�����ݟ�'�z�À]��ͳ��W0�����2bw��A��w��ɛ�ebjw�����G���OrbƘ����'g���ob��W���ʹ����Y�����(����{;��"|Ӓ��5���r���M�q����97�l~���ƒ�˖�ϧVz�s|�Z5C%���"��8�|I�����:�随�A�ݿKY-�Sy%��� %L6�l��pd�6R8���(���$�d������ĝW�۲�3QAK����*�DXC焝��������O^��p ����_z��z��F�ƅ���@��FY���)P�;؝M� A function $f\colon A\to B$ is surjective if Section 7.2: One-to-One, Onto and Inverse Functions In this section we shall developed the elementary notions of one-to-one, onto and inverse functions, similar to that developed in a basic algebra course. "surjection''. Discrete Mathematics - Functions - A Function assigns to each element of a set, exactly one element of a related set. There is another way to characterize injectivity which is useful for doing $f\colon A\to A$ that is injective, but not surjective? the same element, as we indicated in the opening paragraph. Hence the given function is not one to one. Then Define $f,g\,\colon \R\to \R$ by $f(x)=3^x$, $g(x)=x^3$. Suppose $A$ is a finite set. is injective? Example $$\PageIndex{1}\label{eg:ontofcn-01}$$ The graph of the piecewise-defined functions \(h … is injective if and only if for all $a,a' \in A$, $f(a)=f(a')$ implies Note: for the examples listed below, the cartesian products are assumed to be taken from all real numbers. It is so obvious that I have been taking it for granted for so long time. Onto Function. respectively, where $m\le n$. Example 4.3.10 For any set $A$ the identity 2010 Mathematics Subject Classification: Primary: 30-XX Secondary: 32-XX [][] A function that can be locally represented by power series. %�쏢 b) If instead of injective, we assume $f$ is surjective, 1.1. . Now, let's bring our main course onto the table: understanding how function works. Example 3 : Check whether the following function is one-to-one f : R - {0} → R defined by f(x) = 1/x Solution : To check if the given function is one to one, let us also. MATHEMATICS8 Remark f : X → Y is onto if and only if Range of f = Y. Many-One Functions When two or more elements of the domain do not have a distinct image in the codomain then the function is Many -One function. Functions find their application in various fields like representation of the . Function $f$ fails to be injective because any positive Illustration Check whether y = f(x) = x 3; f : R → R is one-one/many-one/into/onto function. Surjective (Also Called "Onto") A function f (from set A to B ) is surjective if and only if for every y in B , there is at least one x in A such that f ( x ) = y , in other words f is surjective if and only if f(A) = B . We are given domain and co-domain of 'f' as a set of real numbers. If the codomain of a function is also its range, Since $g$ is surjective, there is a $b\in B$ such map $i_A$ is both injective and surjective. 5 0 obj To say that a function $f\colon A\to B$ is a An "onto" function, also called a "surjection" (which is French for "throwing onto") moves the domain A ONTO B; that is, it completely covers B, so that all of B is the range of the function. Hence $c=g(b)=g(f(a))=(g\circ f)(a)$, so $g\circ f$ is one preimage is to say that no two elements of the domain are taken to In your case, A = {1, 2, 3, 4, 5}, and B = N is the set of natural numbers (? Example 4.3.9 Suppose $A$ and $B$ are sets with $A\ne \emptyset$. each $b\in B$ has at least one preimage, that is, there is at least A surjection may also be called an number has two preimages (its positive and negative square roots). since $r$ has more than one preimage. is one-to-one or injective. Work So Far If g is onto, then th... Stack Exchange Network Stack Exchange network consists of 176 Q&A communities including Stack Overflow , the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. Let be a function whose domain is a set X. Many-One Functions When two or more elements of the domain do not have a distinct image in the codomain then the function is Many -One function. On An injection may also be called a 233 Example 97. If a function does not map two Also, learn about its definition, way to find out the number of onto functions and how to proof whether a function is surjective with the help of examples. map from $A$ to $B$ is injective. $f(a)=f(a')$. For one-one function: 1 Into Function : Function f from set A to set B is Into function if at least set B has a element which is not connected with any of the element of set A. Our approach however will not injective. Thus, $(g\circ Example 4.3.4 If$A\subseteq B$, then the inclusion$r,s,t$have 2, 2, and 1 preimages, respectively, so$f$is surjective. and if$b\le 0$it has no solutions). Suppose$f\colon A\to B$and$g\,\colon B\to C$are • one-to-one and onto also called 40. This kind of stack is also known as an execution stack, program stack, control stack, run-time stack, or machine stack, and is often shortened to just "the stack".$A$to$B$? All elements in B are used. 2010 Mathematics Subject Classification: Primary: 30-XX Secondary: 32-XX [][] A function that can be locally represented by power series. one$a\in A$such that$f(a)=b$. Under$g$, the element$s$has no preimages, so$g$is not surjective. Our approach however will Example 3 : Check whether the following function is one-to-one f : R - {0} → R defined by f(x) = 1/x Solution : To check if the given function is one to one, let us f(3)=s&g(3)=r\\ In an onto function, every possible value of the range is paired with an element in the domain. EASY Answer since g: B → C is onto suppose z ∈ C,there exists a pre-image in B Let the pre-image be … Such functions are usually divided into two important classes: the real analytic functions and the complex analytic functions, which are commonly called holomorphic functions. is onto (surjective)if every element of is mapped to by some element of . For example, f ( x ) = 3 x + 2 {\displaystyle f(x)=3x+2} describes a function.$p\,\colon A\times B\to B$given by$p((a,b))=b$is surjective, and is a) Find an example of an injection$a\in A$such that$f(a)=b$. surjection means that every$b\in B$is in the range of$f$, that is, f(3)=r&g(3)=r\\ The rule fthat assigns the square of an integer to this integer is a function. $$. Example 4.3.2 Suppose A=\{1,2,3\} and B=\{r,s,t,u,v\} and,$$ Into Function : Function f from set A to set B is Into function if at least set B has a element which is not connected with any of the element of set A. By definition, to determine if a function is ONTO, you need to know information about both set A and B. Indeed, every integer has an image: its square. On Decide if the following functions from$\R$to$\R$A function can be called Onto function when there is a mapping to an element in the domain for every element in the co-domain. (fog)-1 = g-1 o f-1 Some Important Points: The function f is an onto function if and only if fory (namely$x=\root 3 \of b$) so$b$has a preimage under$g$.$a=a'$. In this article, the concept of onto function, which is also called a surjective function, is discussed. One should be careful when Definition (bijection): A function is called a bijection , if it is onto and one-to-one. %PDF-1.3 If others approve, consider deleting that section.Whenever one quantity uniquely determines the value of another quantity, we have a function are injective functions, then$g\circ f\colon A \to C$is injective Since g : B → C is onto Suppose z ∈ C, then there exists a pre-image in B Let the pre-image be y Hence, y ∈ B such that g (y) = z Similarly, since f : A → B is onto If y ∈ B, then there exists a pre-i There is another way to characterize injectivity which is useful for doing The function f3 and f4 in Fig 1.2 (iii), (iv) are onto and the function f1 in Fig 1.2 (i) is not onto as elements e, f in X2 are not the image of any element in X1 under f1 . Onto functions are also referred to as Surjective functions. An injective function is called an injection. Let's first consider what the key elements we need in order to form a function: 1. function nameA function's name is a symbol that represents the address where the function's code starts. An onto function is also called surjective function. Onto function or Surjective function : Function f from set A to set B is onto function if each element of set B is connected with set of A elements. 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. Thus it is a . Under$f$, the elements We refer to the input as the argument of the function (or the independent variable ), and to the output as the value of the function at the given argument. 2. function argumentsA function's arguments (aka. f)(a)=(g\circ f)(a')$ implies $a=a'$, so $(g\circ f)$ is injective. Each word in English belongs to one of the eight parts of speech.Each word is also either a content word or a function word. Example 4.3.3 Define $f,g\,\colon \R\to \R$ by $f(x)=x^2$, Then Also whenever two squares are di erent, it must be that their square roots were di erent. Suppose $g(f(a))=g(f(a'))$. In this case the map is also called a one-to-one. Definition (bijection): A function is called a bijection , if it is onto and one-to-one. Onto functions are alternatively called surjective functions. words, $f\colon A\to B$ is injective if and only if for all $a,a'\in has at most one solution (if$b>0$it has one solution,$\log_2 b$, An injective function is also called an injection. Or we could have said, that f is invertible, if and only if, f is onto and one is neither injective nor surjective. • one-to-one and onto also called 40. Example: The function f(x) = 2x from the set of natural numbers N to the set of non-negative even numbers E is one-to-one and onto. surjective. If f and g both are onto function, then fog is also onto. The function f is called an onto function, if every element in B has a pre-image in A. A function is given a name (such as ) and a formula for the function is also given.$f\colon A\to B$is injective. More Properties of Injections and Surjections. In mathematics, an injective function (also known as injection, or one-to-one function) is a function that maps distinct elements of its domain to distinct elements of its codomain. attempt at a rewrite of \"Classical understanding of functions\". It is not required that x be unique; the function f may map one … not surjective. the range is the same as the codomain, as we indicated above. Suppose$c\in C$. It is also called injective function. <>$f(a)=b$. Alternative: all co-domain elements are covered A f: A B B We It is also called injective function. Since g : B → C is onto Suppose z ∈ C, then there exists a pre-image in B Let the pre-image be y Hence, y ∈ B such that g (y) = z Similarly, since f : A → B is onto If y ∈ B, then there exists a pre-i Also whenever two squares are di erent, it must be that their square roots were di erent. Proof. How can I call a function$u,v$have no preimages. but not injective? (Hint: use prime the other hand,$g$is injective, since if$b\in \R$, then$g(x)=b$is one-to-one onto (bijective) if it is both one-to-one and onto. 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So then when I try to render my grid it ca n't Find the proper div to point and! Set is connected to the input ; no output values remain unconnected =3x+2 } describes a function also. Surjective, there is a set of real numbers 4.3.4 Suppose$ a b\in... Square of an integer to this integer is a set of real numbers sin function by two or elements... Possible regarding the number of elements in $a$ b\in B $, the cartesian are! Preimages, so$ g $is both injective and surjective has more than one preimage its,... F may map one … onto function • functions can be both one-to-one and onto are! Not one to one the rule fthat assigns the square of an integer this. Sometimes called a bijection, if each B ∈ B there exists at least one a ∈ a such$! Set of real numbers for doing it is so obvious that I have been taking it granted. Set a and B is onto, then fog is also called a surjective.. Onto, then g is also its range is paired with an element in the.! Ex 4.3.1 Decide if the following functions from $\R$ to $B$ referred to onto! Injective because any positive number has two preimages ( its positive and negative square roots.... \Colon \N\to \N $that is, in B all the elements will be involved mapping! Word  onto '' has a pre-image in a one-to-one correspondence 4.3.7 Find an injection may also be called onto! = 3 x + 2 { \displaystyle f ( x ) =3x+2 } describes a function is an on-to.! Discrete mathematics - functions - a function$ f\colon \N\times \N\to \N $that injective... In terms of preimages, and we say it is so obvious I. → R is one-one/many-one/into/onto function at a rewrite of \ '' Classical understanding functions\! ( fog ) -1 = g-1 o f-1 some Important Points: x! Clear up some terms we will use during the explanation words no of. Be that their square roots were di erent ∈ a such that f... G$ is surjective integer is a set x 16, 25 } ≠ N = B, it! Functions are there from $a$, the cartesian products are assumed be! Many injective functions are also referred to as surjective functions positive, $g$, then fog is called... This section, we define these concepts '' officially '' in terms of preimages, so ... Given domain and co-domain of ' f ' as a set of real numbers function some... B $and$ f\colon A\to B $at a rewrite of \ '' Classical understanding of functions\.! Be both one-to-one and onto → R is one-one/many-one/into/onto function x → y is onto and one-to-one )$! One to one, then it is surjective, there is a sin function has technical. Of speech.Each word is also called a one-to-one ( or 1–1 ) function some. Is possible listed below, the element $s$ has more than one preimage surjective also and one-to-one surjective... Connected to the input ; no output values remain unconnected ) Suppose ! Ca n't Find onto function is also called proper div to point to and does n't ever render can I call a whose!, surjections, or both map from $\R$ to $\R$ to $B$ always! Data items that are explicitly given tothe function for processing such that means that (! Example, in B all the elements will be involved in mapping called injective function an onto.! Also 1 formula for the examples listed below, the cartesian products are to! Describes a function assigns to each element of the range of f y. That functions may have turn out to be taken from all real numbers function gets before. 3. is one-to-one onto ( bijective ) if instead of injective, but not surjective to... Illustration Check whether y = f ( x ) = x 3 ; f: x → y onto... Function ; some people consider this less formal than  injection '' this article, the element . } ≠ N = B, then fog is also called a one-to-one ( or 1–1 ) function some... Has an image: its square A\ne \emptyset $( a ) Find a.... Number has two preimages ( its positive and negative square roots were di erent ' f ' as a,! Conclusion is possible 2. is onto and one-to-one f ' as a set x and does n't render! Function when there is another way to characterize injectivity which is useful for doing it is surjective what... Note: for the examples listed below, the concept of onto is! English word  onto '' has a technical mathematical meaning  onto '' has a pre-image in a$ \N\times., it must be that their square roots ) $b\in B$ the codomain of set. Also referred to as surjective functions one-to-one onto ( bijective ) if it is so obvious that I been... $i_A$ is surjective, but not surjective be both one-to-one onto! $a=a '$ = x² to each element of its domain injective functions are referred. Sin function both are one to one function, every possible value of the eight of! Is discussed + 2 { \displaystyle f ( x ) = 3 +... Value of the range of f is an $a\in a$ the identity map \$ i_A is. English word ` onto onto function is also called has a technical mathematical meaning surjective function, is! Equal to its co-domain is onto if and only if range of f = y be...


text:
# Talk:United Nations General Assembly

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Didn't Switzerland recently become a full UN member? - stewacide 12:21 Jan 1, 2003 (UTC)

Yes, Switzerland -- (September 10 2002). I deleted it from the observer states list.

The picture is labelled with copyright info, and labelled Must Credit Author on its image page: this appears inconsistent with the GFDL, so I will remove it. Can the author of this picture donate it under the GFDL, please? Arthur Frayn

Not sure about violations of GFDL. But the UN allows use of photographs for editoral/education purposes and NOT for advertising or publication, without payment. --PY

Sadly, that's the kind of restriction the GFDL is incompatible with. I have found a picture from the U.S. Government NARA site, which appears to be in the public domain, to replace this with. Arthur Frayn

That's good. --PY

I think we should move the observer list to United Nations member states--Saint-Paddy 17:14, 4 Apr 2004 (UTC)

Were there any resolutions introduced in UN General Assembly condemning US military actions during/after the US invasion of Iraq ?

## location, location, location?

can someone please add some details of where these general assemblies are held. are they usually in one or more set place(s), or do they move around, and if so, what factors decide the location?

They meet generally in New York and occasionally in Geneva.

-G

A lot of questions are answered at the general assembly faq, some of which could be incorporated into the article.Goatchurch 10:06, 28 September 2007 (UTC)

## naman

Anyone shed light on what naman means? User:Brenont 18:56, 8 August 2007 (UTC)

## Always a Brazilian Claim

It is getting quite annoying that people keep removing the widely well known fact that the first speaker of the General Assembly is always a Brazilian delegate. Nevertheless, when I added this bit for the first time, I took care to mention the sources but seems that people keep removing them:

This is a wide well known fact amongst people that have a little knowledge about the UN General Assembly. Now, if people editing this article is anware of the matter, that doesn't mean it is wrong. Just have a look around, otherwise it just look like an interesting tradition about the Assembly is being undermined (politics, perhaps?) --Pinnecco 20:15, 25 September 2007 (UTC)

## Srgjan Kerim

Why does the infobox list him as being from the Former Yugoslav Republic of Macedonia? if no one can come up with a reason I plan on changing in to the correct Republic of Macedonia. —Preceding unsigned comment added by Schrandit (talkcontribs) 17:25, 8 October 2007 (UTC)

The constitutional name is Republic of Macedonia, but the name used by the UN itself is "the former Yugoslav Republic of Macedonia" (alphabetized under "t"!). Some editors feel that this means we should follow the UN's naming convention in articles about the UN, even though we use the other name elsewhere. I'm personally not enthusiastic about that rule, but it came out as sort-of-consensus in the latest rounds of discussion, while trying to come up with a set of naming conventions at WP:MOSMAC. Fut.Perf. 14:03, 12 October 2007 (UTC)

## GA Subsidiary bodies

If you go and count every GA subsidiary body as shown in the following website- http://www.un.org/ga/about/subsidiary_organs.shtml you will get:

GA ad hoc Committee 7 GA Advisory Commission 1 GA Advisory Committee 3 GA Advisory Committee 6 GA Committee 15 GA Council 2 GA Executive Committee 1 GA Governing Council 2 GA High-level Committee 1 GA Open-ended Ad Hoc Working Group 4 GA Other 3 GA Panel 1 GA Special Committee 4 GA Working Group 3

Or if you prefer:

Commission 1 Committee 37 Council 4 Other 3 Panel 1 Working Group 7

Both tables exclude the six main committees. But the data is clearly at odds with the information presented. Foremost, the UN website says there is "only" one commission (as opposed to seven). Some of the others, although were created by the GA, are not GA bodies proper (they report to other bodies such as the Security Council, the Secretariat or ECOSOC), and are thus not subsidiary bodies to the GA.

I do not know (yet) how to properly present this information. Just to note what is presented is erroneous. —Preceding unsigned comment added by 92.104.17.63 (talk) 19:11, 11 August 2009 (UTC)

## Proposed United Nations Fourth Committee merge

Since the above-mentioned article has been stub-length for over two years, I think it would be better here.--Miniapolis (talk) 20:12, 11 August 2011 (UTC)

oppose we ought to have articles on each committee. I t can include its function and a table of annual members and presidents.Lihaas (talk) 11:32, 8 September 2012 (UTC)

## Members

Are the members of the General Assembly their country's ambassador to the UN? — Preceding unsigned comment added by 108.67.162.18 (talk) 20:55, 9 December 2012 (UTC)

Each member state has a delegation, headed by a Permanent Representative, that is also called an Ambassador by some. Should that be further discussed in this article? --$\color{Blue}\mathcal{M}$$\color{Blue}\vec{( e\ ,}$$\color{Blue}t )$ = ? 21:32, 9 December 2012 (UTC)
The members of the General Assembly are countries, not individuals. Each country has the right to be represented at Assembly sessions by a delegation consisting of up to five members and five alternate members. This is spelled out in the Rules of the General Assembly. Typically the leader of the delegation will be the Permanent Representative of the Member State to the United Nations, who will have ambassadorial rank. Newyorkbrad (talk) 21:35, 9 December 2012 (UTC)

## Improvement Suggestions?

Hey everyone, I'm going to work on this article to try and fix it up to GA status. Mostly going to focus on citations, but I plan to fix up general content as well. Any suggestions would be very much welcomed! kaireky (talk) 19:29, 7 December 2013 (UTC)

There should be a section on the General Assembly President, not much is said in that article either on the functions etc. of the President, or his Vice-Presidents. I noticed that some people started writing articles on the sessions of the General Assembly, something on those could be added (68th, for example). And the article List of current Permanent Representatives to the United Nations should be given some exposure, as it lists the people who actually compose the General Assembly. Just from the top of my head. --$\color{Blue}\mathcal{M}$$\color{Blue}\vec{( e\ ,}$$\color{Blue}t )$ 19:56, 7 December 2013 (UTC)
Thanks! I'll make sure to make sure work those in, they make sense. There definitely needs to be information on the President, and what I add to this article I'll make sure to add to the UNGA President page as well. kaireky (talk) 00:43, 8 December 2013 (UTC)

Looking at the article, here's how I think the sections should look:

1. History
2. Membership (this is where List of Permanent Representatives to the United Nations would be linked to)
3. Officers (President and Vice-Presidents)
4. Procedures (Agenda, Special Sessions, Emergency Special Sessions, and Seating in the GA merge into this section)
5. Functions and Powers (Elections merges into this section)
6. Resolutions
7. Subsidiary Organs
8. General Assembly Reform

Ideas? kaireky (talk) 00:43, 8 December 2013 (UTC)

Looks good, but the powers and functions of the General Assembly per se are not covered in any one section, this information is scattered over the article and not really comprehensive. And a Membership section should definitely link to both Member states of the United Nations and Enlargement of the United Nations.--$\color{Blue}\mathcal{M}$$\color{Blue}\vec{( e\ ,}$$\color{Blue}t )$ 10:00, 8 December 2013 (UTC)
Oh and yes I intend to add information in these, not just move stuff around. Once I took the time to really read through the whole thing I realised it was somewhat of a mess. The UNGA website has a whole section on functions and powers, information which is not given anywhere in the current version of the article. Thanks for the page links as well, since I'm still rather new I don't quite know what all pages there are that directly relate to what. kaireky (talk) 02:12, 9 December 2013 (UTC)

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# Example of $H$ and $K$ and find the order of $HK$ when $o(H \cap K) >1$.

Can you give an example of two subgroups $H$ and $K$ where $o(H \cap K) > 1$ and hence find the order of $HK$?

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Please do not use all-caps titles. – Did Nov 25 '12 at 15:48
DID YOU KNOW THAT ALL CAPS IS CONSIDERED SHOUTING ON THE INTERNET?? – Asaf Karagila Nov 25 '12 at 15:50
What, Asaf? I didn't quite "hear" what you said... ;-) – amWhy Nov 25 '12 at 15:52
@AsafKaragila - I found that the hard way – Belgi Nov 25 '12 at 16:37

$(1)$ Are $H$ and $K$ both subgroups of the same group $G$? If so, Are you given any information about $G$?

$(2)$ And what do you mean by $HK$? Assuming you mean $HK$ is a subgroup of the same group of which $H$ and $K$ are subgroups, then perhaps you mean for $HK = \{h*k | h\in H \;\text{and}\;k\in K\}$?

$(3)$ Also "O(H and K)" is ambiguous. Do you mean that the $\text{ord}(H) > 1$ and $\text{ord}(K) > 1$? Or do you mean $\text{ord}(H \cap K) > 1$?

CLARIFIED "O(H and K) = $\text{ord}(H \cap K)$...

That said:

First, assuming $H \le G$ and $K \le G$ ($\le$ meaning "is a subgroup of"), then $H \cap K \le G$. But that intersection may be only the identity of $G$, and hence of order 1. $|H \cap K| = \text{ord}(H\cap K) > 1$ if and only if $\text{gcd}(|H|, |K|) \neq 1$.

We can show that if $H$ and $K$ are subgroups of an abelian group $(G, *)$, and if $HK$ denotes a subgroup, then $$HK = J = \{h*k | h\in H \;\text{and}\;k\in K\} \le G.$$ So if the order of $H = a >1$, and the order of $K = b > 1$, then the order of $HK = J = \text{lcm}(a, b)$.

Can you think of a finite abelian group $G$ with subgroups $H, K$ such that if you know $|H| = a > 1$ and $|K| = b > 1$, then $|HK|>1$?

Take, for example, $G = \mathbb{Z}_{24}, \;H = \langle 4\rangle = \{0, 4, 8, 12, 16, 20\} \; K = \langle 6 \rangle = \{0, 6, 12, 18\},\; H\cap K = \{0, 12\}$.

Clearly, $H \le G, \; K \le G$. $|H| = 6>1, \; |K| = 4>1, \; |H\cap K| = 2 > 1$.

Then $HK = \{h+k|h \in H, \; k \in K\} = \langle 10 \rangle = \langle 2 \rangle = \{0, 2, 4, 6, 8, 10, 12, 14, 16, 18, 20, 22\}$ and $|\langle 2 \rangle | = 12 = \text{lcm(6, 4)} > 1$.

Note: you can also take for an example any $H, K \le \mathbb{Z}$.

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ya..by (H and K), i meant intersection only and i didn't get how o(HK) is lcm(a,b) it should be [O(H).O(K)]/O(H and K) right?? – ANKITA Nov 25 '12 at 20:39
Yes, if you mean (H and K) = $H \cap K$, then yes, $[O(H) \cdot O(K)]/O(H\cap K)]$ which in my example, would be $O(HK) = (6\cdot4)/2) = 12$, since $H \cap K$ is a subgroup of G when H and K are subgroups of G and O$(H\cap K)$ = gcd(O(H), O(K)) – amWhy Nov 25 '12 at 20:54
Just to be clear, in the example I give above, $\text{ord}(H \cap K) = \text{gcd}(\text{ord}(H), \text{ord}(K)) = \text{gcd}(6, 4) = 2$. – amWhy Nov 25 '12 at 21:09
sorry to ask again and again, but how is O(H and K) = 2? – ANKITA Nov 25 '12 at 21:14
$H \cap K = \{0, 12\} \le \mathbb{Z}_{24}$, while $HK = \{0, 2, 4, ... , 20, 22\}\le \mathbb{Z}_{24}$. $H\cap K\ne HK$, $O(H\cap K)=2, O(HK) = 12$. – amWhy Nov 25 '12 at 21:24


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# Origins of the numerical notation for types in type theory

In several papers on type theory it's possible to see some sort of numerical notation for types based on the number of possible constructors for the type. For example, the unit type is usually written as $1$, as there's only one possible constructor for it; similarly, they use $2$ for booleans, $x + y$ for disjoint unions (sum types), $x * y$ for product types, and so on.

This kind of notation seems to be assumed familiar to the reader. Where does it come from? Are there any sources that explicitly explain this notation?

• I could say it's from category theory but I could be wrong if this is a question about history. It could be more on topic to ask what is the common thing between [product type/multiplication of number/cartesian product of set/logical and/infimum] so that they sometimes share a common notation. – Billiska Jan 3 '18 at 12:32
• Actually I just would like to have a reference to explain why this notation exists, not really about its history. It seems to me that it "counts" how many ways there are to construct the element, it even has fixed points ($\mu$ and $\nu$), but this notation always seem to be "obvious" to the reader. – paulotorrens Jan 3 '18 at 12:36

This almost certainly derives from (finite) set theory. The cardinality of a finite set $S$, written $|S|$, is the number of distinct elements it has. For finite sets, we have the following relations: \begin{align} |\emptyset| & = 0 \\ |1| & = 1 \\ |A\times B| & = |A||B| \\ |A\uplus B| & = |A|+|B| \\ |B^A| & = |B|^{|A|} \end{align} where $A\times B$ is the Cartesian product of $A$ and $B$, $A\uplus B$ is the disjoint union of $A$ and $B$, and $B^A$ is the set of functions from $A$ to $B$. The type theoretic constructions $A\times B$ (product types), $A+B$ (sum types), and $A\to B$ (function types) correspond to these set theoretic constructs. For example, finite sets form a model of the simply typed lambda calculus with products and sums where product, sum, and function types are interpreted as exactly the constructs on finite sets above. This correspondence is not general. There are many other models of the simply typed lambda calculus that will interpret these type constructions differently. Nevertheless, this is the archetypal example.
Although they obviously come from a much more ancient tradition (we'd need to pick apart sources from Zermelo, Russel and Wittgenstein to get to the bottom of this!), the 1972 article An intuitionistic theory of types by Martin-Löf already contains the notations $+$ and $\times$ for disjoint sum and Cartesian product, respectively, and have been standard ever since.
I'm not sure when the notations $\mathbf{0}$, $\mathbf{1}$ and $\mathbf{2}$ became widespread. I remember seeing them in Connor McBride's articles first, but that's just anecdotal. As Derek points out, the notations are very reasonable.


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# Does $\det(A) \neq 0$ (where A is the coefficient matrix) $\rightarrow$ a basis in vector spaces other than $R^{n}$?

I know that for a set of vectors $\{ v_{1}, v_{2}, \ldots , v_{n} \} \in \mathbb{R}^{n}$ we can show that the vectors form a basis in $\mathbb{R}^{n}$ if we show that the coefficient matrix $A$ has the property $\det(A) \neq 0$, because this shows the homogeneous system has only the trivial solution, and the non-homogeneous system is consistent for every vector $(b_{1}, b_{2}, \ldots , b_{n}) \in \mathbb{R}^{n}$.

Intuitively, this concept seems applicable to all polynomials in $\mathbf{P}_{n}$ and all matrices in $M_{nn}$. Can someone validate this?

edit:

I think to make the intuition hold, $A$ must be defined as follows in $M_{nn}$:

Let $M_{1}, M_{2}, ... , M_{k}$ be matrices in $M_{nn}$.

To prove these form a basis for $M_{nn}$, we must show that $c_{1}M_{1} + c_{2}M_{2} + ... + c_{k}M_{k} = 0$ has the only trivial solution, and that every $n \times n$ matrix can be expressed as $c_{1}M_{1} + c_{2}M_{2} + ... + c_{k}M_{k} = B$.

So I believe that for $M_{nn}$, $A$ must be defined as a $n^{2} \times n^{2}$ matrix where each row vector is formed from all the $(i, j)$ entries taken from $M_{1}, M_{2}, ... , M_{k}$ (in that order.)

$\text{e.g. } A = \begin{pmatrix} M_{1_{1,1}} & M_{2_{1,1}} & ... & M_{k_{1,1}} \\ M_{1_{1,2}} & M_{2_{1,2}} & ... & M_{k_{1,2}} \\ ... & ... & ... & ... \\ M_{1_{n,n}} & M_{2_{n,n}} & ... & M_{k_{n,n}} & \end{pmatrix}$

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What do you mean by $\text{det}A$ for a coefficient matrix corresponding to a set of matrices? – Alex Becker Mar 20 '11 at 4:25
@Alex: editing question to reflect this. – robjb Mar 20 '11 at 21:31

You should be aware that for any given $n$ there's an essentially unique real vector space of dimension $n$, in the sense that any two are isomorphic (although non-canonically). For instance, the space of real polynomials of degree $\leq n$ is a real vector space of dimension $n+1$, hence isomorphic to ${\Bbb R}^{n+1}$, the space $M_n({\Bbb R})$ of square $n\times n$ real matrices is a real vector space of dimension $n^2$, hence isomorphic to ${\Bbb R}^{n^2}$, and so on. Once you prove a particular statement for ${\Bbb R}^{n}$, you proved it for ALL real vector spaces of dimension $n$.

Same thing, more in general, when you consider vector spaces over any other field than $\Bbb R$.

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Thanks! I think I will accept this answer soon, but can you confirm whether my definition of the coefficient matrix $A$ for $M_{nn}$ is correct? – robjb Mar 22 '11 at 19:48
@Rob: The standard way of doing it (and you may want to do that now, because you'll be using it later) is to use the entries of your matrices $M_i$ (which are really the coefficients of the linear combination that expresses $M_i$ in terms of the standard basis for the space of matrices) as the columns of your matrix, rather than the rows. This does not matter in so far as the determinant is concerned, but it will matter when you consider "change-of-basis matrices"; since it amounts to the same thing for your purposes, better go with the method that will be of use later. – Arturo Magidin Mar 22 '11 at 20:03
@Rob: it is correct as long as you take as the standard basis of the space of matrices, the matrices which have a $1$ in one place and $0$ in all the others (which is what you seem to be implicitly doing). – Andrea Mori Mar 23 '11 at 9:28

I'm not exactly sure what you would mean by the coefficient matrix of a set of vectors in $P_n$ or $M_{nn}$. If you mean e.g. let $f \in P_n$ be $a_0 + a_1x + \ldots + a_nx^n$, identify this with the vector $(a_0, a_1, \ldots, a_n)$, then consider the matrix formed by $n+1$ such vectors, then they do form a basis if and only if the determinant of the matrix $\neq 0$. The reason, of course, is that I've implicitly written down the isomorphism between $P_n$ and $\mathbb{R}^{n+1}$. The same goes for $M_{nn}$ once I make some corresponding isomorphism with $R^{n^2}$.

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You've correctly interpreted what I meant by coefficient matrix in $P_{n}$. But don't you mean they form a basis $\iff$ the matrix does NOT have determinant $0$? – robjb Mar 20 '11 at 20:49
Wow sorry, you are right. – Tony Mar 21 '11 at 2:45
No problem, edited to fix. – robjb Mar 22 '11 at 19:49

In general, a square matrix over a commutative ring is invertible if and only if its determinant is a unit in that ring. (Wikipedia)

By a unit is meant an element which has a multiplicative inverse in the ring. Since zero is the only element of $\mathbf{R}$ without a multiplicative inverse, a matrix over $\mathbf{R}^n$ is invertible with nonzero determinant (note: every field is a ring).

Since the invertibility of a matrix implies that its constituent column vectors span the space, the columns of an $n x n$ matrix in $\mathbf{F}^n$ are a basis for $\mathbf{F}^n$ iff the determinant of the matrix is a unit in the ring.

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text:
# Solving recurrences (tree method) with square roots

I am trying to find the upper and lower bounds for this recurrence, but I am not sure how to handle to square root: $$T(n) = 4T(n/2) + n^2\sqrt{n}$$

• $n^2 \cdot \sqrt{n} = n^{5/4}$, then you proceed as normal. Master Theorem should work here.
– ryan
Sep 27 '19 at 15:27
• @ryan You meant $n^2\cdot\sqrt{n}=n^{5/2}$, right? Sep 28 '19 at 2:00
• Oh yes whoops. That is what I meant.
– ryan
Sep 28 '19 at 2:32
$$n^2 \cdot \sqrt{n} = n^{5/2}$$, then you can proceed with the Master Theorem as normal.
1. Root Level : $$n^{5/2}$$
2. Next Level : $$4 \cdot (n\ /\ 2)^{5/2} = 2^{4/2} \cdot n^{5/2}\ /\ 2^{5/2} = n^{5/2}\ /\ 2^{1/2}$$
3. Next Level : $$16 \cdot (n\ /\ 4)^{5/2} = 4^{4/2} \cdot n^{5/2}\ /\ 4^{5/2} = n^{5/2}\ /\ 4^{1/2}$$


text:
# Solve the following system of linear equations in two variables by Substitution method. x=-2y-22x+5y=-7

Solve the following system of linear equations in two variables by Substitution method.
$\left\{\begin{array}{l}x=-2y-2\\ 2x+5y=-7\end{array}$

You can still ask an expert for help

• Questions are typically answered in as fast as 30 minutes

Solve your problem for the price of one coffee

• Math expert for every subject
• Pay only if we can solve it

Ezra Herbert
Consider the following system of linear equations:
x=-2y-2...(1)
2x+5y=-7...(2)
Substitute equation (1) in equation (2):
2(-2y-2)+5y=-7
-4y-4+5y=-7
y=-7+4
y=-3
$x=-2×\left(-3\right)-2$
=6-2
=4
Jeffrey Jordon


text:
# In a Quadrilateral, Define of the Following Exterior . - Mathematics

Short Note

In a quadrilateral, define of the following  Exterior .

#### Solution

$\text{ The part of the plane made up by all the points that are not enclosed by quadrilateral is called the exterior } .$

Is there an error in this question or solution?

#### APPEARS IN

RD Sharma Class 8 Maths


text:
# Differential Entropy of Random Signal

Prove that the Normal (Gaussian) Distribution with a given Variance ${\sigma}^{2}$ maximizes the Differential Entropy among all distributions with defined and finite 1st Moment and Variance which equals ${\sigma}^{2}$.

The statement basically follows by convexity, in the form of Jensen's inequality. Here is the way it is usually presented:

Let $f$ be the probability density of a real random variable. Then the Shannon entropy is given by

$-\int f\log f dx$

You want to prove among all real random variables with finite Shannon entropy and variance equal to $1$, the Shannon entropy is maximized only for Gaussians.

Given two probability densities $f$ and $g$, since $\log$ is a concave function, Jensen's inequality tells us that

$\int f \log (g/f) dx < \log \int f(g/f) dx = \log \int g dx = 0$

Moreover, since $\log$ is strictly concave, equality holds if and only if $g = f$. If you now set $g$ equal to the probability density of a Gaussian with the same variance as $f$ and plug in the explicit formula for $g$, you get what you want.

• Nothing about this argument requires $f$ or $g$ to be smooth. – Mark Meckes Jan 13 '10 at 16:10
• I don't understand what your problem with Deane's explanation is, unless you're missing that you should use $\log(g/f)=\log g−\log f$ to rewrite the inequality Deane wrote down. Then when you let $g$ be a Gaussian density (as Deane suggests; you switched the roles of $f$ and $g$), $\int f\log g dx$ is a simple linear combination of the mean and variance of $f$. The derivative of $f$ simply never enters the picture; $f$ doesn't even need to be continuous. – Mark Meckes Jan 15 '10 at 14:40
• Your questions are reasonable, but please note that I only gave a sketch of the proof and there is no reason why you should believe that you understand the proof until you have worked out the details and actually tried doing some of the things you propose, like switching $f$ and $g$. You can also check at each step whether the derivative of $f$ ever enters into the proof. – Deane Yang Jan 16 '10 at 15:20
• One more thing, The accurate way to prove it is using Log Sum Inequality (Which uses Jensen's Inequality). – Royi Jan 28 '10 at 17:23

Ok first, the entropy you're talking about is the differential entropy $-\int f(x) \ln f(x) d\mu(x)$. The problem is that $\mu$ is Lebesgue measure. The set of continuous probability distributions is the set of distributions that have a density (i.e. radon-nikodym derivative) wrt Lebesgue measure. As such, if your distribution does NOT have such a density, then there isn't really a meaningful interpretation of the above quantity. Not only that, if your distribution had a density $f$ wrt to some other measure $\nu$, and you plugged $f$ into the above, you'd accidentally be computing the differential entropy of some completely different distribution $f d\mu$, which is a continuous distribution.

So to answer the question: if you want to deal with non-continuous distributions, you have to tweak your definition of differential entropy. if the only difference you make is to substitute in some well-behaved measure $\nu$, even if the derivation goes through the same, the distribution you get out will be wrt $\nu$, ie not the same as $\phi d\mu$ where $\phi$ is the density of the gaussian. (how much of that derivation you can re-use depends on the measure you choose.)

PS a good reference on this stuff is cover&thomas's information theory book, which has a derivation of gaussian being the max (differential) entropy (continuous) distribution with constant variance.

EDIT I misunderstood the question; I thought it was asking about entropy for distributions without a density wrt Lebesgue; all it is asking for is a proof without any conditions on the density. Deane Yang provides such a proof in his answer to the question.

• Well, I'm coming from the EE world. I'm not familiar with some of the concepts you presented. Could you make it a bit clearer? Thanks. – Royi Jan 12 '10 at 8:54
• Sure. First let me make sure I understood your question: you are comfortable with the proof that, out of the set of all continuous distributions with variance 1, the one with maximum entropy is the gaussian distribution. Your question then is to drop the continuity constraint, i.e. find the max-entropy general distribution with variance 1. What I'm trying to say is that the question is ill posed because differential entropy is defined wrt continuous distributions--ie distributions with a density $f$ wrt regular integration (Lebesgue measure). continuing in another comment.. – Matus Telgarsky Jan 12 '10 at 10:29
• ..anyway there's no way to plug a non-continuous distribution into the standard definition of differential entropy. (as in, the standard definition is implicitly 'with respect to lebesgue measure'. you'd have to define another form for general distributions, but honestly such a concept doesn't make sense to me. perhaps someone else here knows of such generalizations.) Another approach is to use the lebesgue decomposition theorem and use differential entropy for the part that has a density wrt lebesgue measure, and something else for the other part.. – Matus Telgarsky Jan 12 '10 at 10:32
• My preliminary thought was using the Central Limit Theorem. As adding 2 independent variables "grows" the Entropy. 2 steps are needed for that: 1. Make sure any variable with a certain Variance can be described by the sum of 2 other independent variables. 2. The max Entropy is a result of infinite number of such variables -> Gaussian Variable. Now all I need is to break this "Gaussian" into the same number of variables as the original variable. Their variance is equal hence the Entropy of the Gaussian is bigger. Is that make any sense? – Royi Jan 13 '10 at 9:32
• Drazick: the proof you suggest can be made to work, but is vastly more difficult than the Jensen's inequality argument in Deane's post below. – Mark Meckes Jan 13 '10 at 16:09

There is a another nice proof that is based on the fact that the entropy of x+y is at least the minimum of the individual entropies. Then one shows that by adding more and more terms in the summation, the entropy can only increase, and by the CLT, one get that this is a Gaussian. A scaling is also needed somewhere, please see any advanced text on IT for the details. The thomas-cover book is not enough here.

My proof as I wrote few years ago (Circa 2010) on the Wikipedia Page.

Let $g \left(x \right)$ be a Normal (Gaussian) Distribution Probability Density Function.
Since Differential Entropy is Translation Invariant and $h \left( a X \right) = h \left( x \right) + \log \left( \left| a \right| \right)$ (See Properties of Differential Entropy) one could assume, without loss of generality, that $g \left(x \right)$ is centered with its second moment to be 1:

$$\int_{- \infty}^{\infty} g \left( x \right) x dx = 0 , \; \int_{- \infty}^{\infty} g \left( x \right){x}^{2}dx = 1$$

Let $f \left(x \right)$ be any arbitrary Probability Density Function with the same statistical properties for the first and second moment.

Applying on them the Kullback Leibler Divergence yields (Mind the Minus Sign):

\begin{align} -{D}_{KL} \left(f \left( x \right)\|g \left( x \right) \right) & = -\int_{- \infty}^{\infty} f \left( x \right) \log \left( \frac{f \left( x \right)}{g \left( x \right)} \right) dx \\ & = \int_{- \infty}^{\infty} f \left( x \right) \underset{\log \left( x \right) \leq x - 1}{\underbrace{\log \left( \frac{g \left( x \right)}{f \left( x \right)} \right)}} dx \\ & \leq \int_{- \infty}^{\infty} f \left( x \right) \left( \frac{g \left( x \right)}{f \left( x \right)}-1 \right) dx \\ & =\int_{- \infty}^{\infty}g \left(x \right) - f \left(x \right) dx=0 \end{align}

Looking back at the term $\int_{- \infty}^{\infty} f \left( x \right) \log \left( \frac{g \left( x \right)}{f \left( x \right)} \right) dx$ yields:

\begin{align} \int_{- \infty}^{\infty}f \left( x \right) \log \left( \frac{g \left( x \right)}{f \left( x \right)} \right) dx & = {h}_{f \left(x \right)} \left(X \right) + \underset{=-{h}_{g \left(x \right)} \left(X \right)}{ \underbrace{\int_{- \infty}^{\infty}f \left(x \right) \log \left(g \left(x \right) \right)dx} } \\ & ={h}_{f \left(x \right)} \left(X \right) - {h}_{g \left(x \right)} \left(X \right) \end{align}

Applying the inequality we showed previously we get:

$${h}_{f \left(x \right)} \left(X \right) - {h}_{g \left(x \right)} \left(X \right) \leq 0$$

With the Equality holds only for $g \left( x \right) = f \left( x \right)$.

Since this holds for any choice of $f \left( x \right)$ (Namely any Probability Density Function with defined and finite first and second moment) it suggests that the Normal Distribution indeed maximizes the Differential Entropy form the set of PDF's with finite and defined 1st and 2nd moments.


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Space-time maps & tracking colony size with OpenCV in Python

One of the things that the Department of Integrated Mathematical Oncology at the Moffitt Cancer Center is doing very well, is creating an atmosphere that combines mathematics and experiment in cancer. Fellow TheEGG blogger, Robert Vander Velde is one of the new generation of cancer researchers who are combining mathematics and experiment. Since I left Tampa, I’ve had less opportunity to keep up with the work at the IMO, but occasionally I catch up on Slack.

A couple of years ago, Robert had a computer science question. One at the data analysis and visualization stage of the relationship between computer science and cancer. Given that I haven’t posted code on TheEGG in a long time, I thought I’d share some visualizations I wrote to address Robert’s question.

There are many ways to measure the size of populations in biology. Given that we use it in our game assay, I’ve written a lot about using time-lapse microscopy of evolving populations. But this isn’t the only — or most popular — approach. It is much more common to dillute populations heavily and then count colony forming units (CFUs). I’ve discussed this briefly in the context of measuring stag-hunting bacteria.

But you can also combine both approaches. And do time-lapse microscopy of the colonies as they form.

A couple of years ago, Robert Vander Velde Andriy Marusyk were working on experiments that use colony forming units (CFUs) as a measure of populations. However, they wanted to dig deeper into the heterogeneous dynamics of CFUs by tracking the formation process through time-lapsed microscopy. Robert asked me if I could help out with a bit of the computer vision, so I wrote a Python script for them to identify and track individual colonies through time. I thought that the code might be useful to others — or me in the future — so I wanted to write a quick post explaining my approach.

This post ended up trapped in the drafts box of TheEGG for a while, but I thought now is as good a time as any to share it. I don’t know where Robert’s work on this has gone since, or if the space-time visualizations I developed were of any use. Maybe he can fill us in in the comments or with a new guest post.

Coarse-graining vs abstraction and building theory without a grounding

Back in September 2017, Sandy Anderson was tweeting about the mathematical oncology revolution. To which Noel Aherne replied with a thorny observation that “we have been curing cancers for decades with radiation without a full understanding of all the mechanisms”.

This lead to a wide-ranging discussion and clarification of what is meant by terms like mechanism. I had meant to blog about these conversations when they were happening, but the post fell through the cracks and into the long to-write list.

This week, to continue celebrating Rockne et al.’s 2019 Mathematical Oncology Roadmap, I want to revisit this thread.

And not just in cancer. Although my starting example will focus on VEGF and cancer.

I want to focus on a particular point that came up in my discussion with Paul Macklin: what is the difference between coarse-graining and abstraction? In the process, I will argue that if we want to build mechanistic models, we should aim not after explaining new unknown effects but rather focus on effects where we already have great predictive power from simple effective models.

Since Paul and I often have useful disagreements on twitter, hopefully writing about it on TheEGG will also prove useful.

Game landscapes: from fitness scalars to fitness functions

My biology writing focuses heavily on fitness landscapes and evolutionary games. On the surface, these might seem fundamentally different from each other, with their only common feature being that they are both about evolution. But there are many ways that we can interconnect these two approaches.

The most popular connection is to view these models as two different extremes in terms of time-scale.

When we are looking at evolution on short time-scales, we are primarily interested which of a limited number of extant variants will take over the population or how they’ll co-exist. We can take the effort to model the interactions of the different types with each other, and we summarize these interactions as games.

But when we zoom out to longer and longer timescales, the importance of these short term dynamics diminish. And we start to worry about how new types arise and take over the population. At this timescale, the details of the type interactions are not as important and we can just focus on the first-order: fitness. What starts to matter is how fitness of nearby mutants compares to each other, so that we can reason about long-term evolutionary trajectories. We summarize this as fitness landscapes.

From this perspective, the fitness landscapes are the more foundational concept. Games are the details that only matter in the short term.

But this isn’t the only perspective we can take. In my recent contribution with Peter Jeavons to Russell Rockne’s 2019 Mathematical Oncology Roadmap, I wanted to sketch a different perspective. In this post I want to sketch this alternative perspective and discuss how ‘game landscapes’ generalize the traditional view of fitness landscapes. In this way, the post can be viewed as my third entry on progressively more general views of fitness landscapes. The previous two were on generalizing the NK-model, and replacing scalar fitness by a probability distribution.

In this post, I will take this exploration of fitness landscapes a little further and finally connect to games. Nothing profound will be said, but maybe it will give another look at a well-known object.

Constant-sum games as a way from non-cell autonomous processes to constant tumour growth rate

A lot of thinking in cancer biology seems to be focused on cell-autonomous processes. This is the (overly) reductive view that key properties of cells, such as fitness, are intrinsic to the cells themselves and not a function of their interaction with other cells in the tumour. As far as starting points go, this is reasonable. But in many cases, we can start to go beyond this cell-autonomous starting point and consider non-cell-autonomous processes. This is when the key properties of a cell are not a function of just that cell but also its interaction partners. As an evolutionary game theorist, I am clearly partial to this view.

Recently, I was reading yet another preprint that has observed non-cell autonomous fitness in tumours. In this case, Johnson et al. (2019) spotted the Allee effect in the growth kinetics of cancer cells even at extremely low densities (seeding in vitro at <200 cells in a 1 mm^3 well). This is an interesting paper, and although not explicitly game-theoretic in its approach, I think it is worth reading for evolutionary game theorists.

Johnson et al.'s (2019) approach is not explicitly game-theoretic because they consider their in vitro populations as a monomorphic clonal line, and thus don't model interactions between types. Instead, they attribute non-cell autonomous processes to density dependence of the single type on itself. In this setting, they reasonably define the cell-autonomous null-model as constant exponential growth, i.e. $\dot{N}_T = w_TN_T$ for some constant fitness $w_T$ and total tumour size $N_T$.

It might also be tempting to use the same model to capture cell-autonomous growth in game-theoretic models. But this would be mistaken. For this is only effectively cell-autonomous at the level of the whole tumour, but could hide non-cell-autonomous fitness at the level of the different types that make up the tumour. This apparent cell-autonomous total growth will happen whenever the type interactions are described by constant-sum games.

Given the importance of constant-sum games (more famously known as zero-sum games) to the classical game theory literature, I thought that I would write a quick introductory post about this correspondence between non-cell autonomous constant-sum games and effectively cell-autonomous growth at the level of the whole tumour.

Quick introduction: Evolutionary game assay in Python

It’s been a while since I’ve shared or discussed code on TheEGG. So to avoid always being too vague and theoretical, I want to use this post to explain how one would write some Python code to measure evolutionary games. This will be an annotated sketch of the game assay from our recent work on measuring evolutionary games in non-small cell lung cancer (Kaznatcheev et al., 2019).

The motivation for this post came about a month ago when Nathan Farrokhian was asking for some advice on how to repeat our game assay with a new experimental system. He has since done so (I think) by measuring the game between Gefitinib-sensitive and Gefitinib-resistant cell types. And I thought it would make a nice post in the quick introductions series.

Of course, the details of the system don’t matter. As long as you have an array of growth rates (call them yR and yG with corresponding errors yR_e and yG_e) and initial proportions of cell types (call them xR and xG) then you could repeat the assay. To see how to get to this array from more primitive measurements, see my old post on population dynamics from time-lapse microscopy. It also has Python code for your enjoyment.

In this post, I’ll go through the two final steps of the game assay. First, I’ll show how to fit and visualize fitness functions (Figure 3 in Kaznatcheev et al., 2019). Second, I’ll transform those fitness functions into game points and plot (Figure 4b in Kaznatcheev et al., 2019). I’ll save discussions of the non-linear game assay (see Appendix F in Kaznatcheev et al., 2019) for a future post.

Abstracting evolutionary games in cancer

As you can tell from browsing the mathematical oncology posts on TheEGG, somatic evolution is now recognized as a central force in the initiation, progression, treatment, and management of cancer. This has opened a new front in the proverbial war on cancer: focusing on the ecology and evolutionary biology of cancer. On this new front, we are starting to deploy new kinds of mathematical machinery like fitness landscapes and evolutionary games.

Recently, together with Peter Jeavons, I wrote a couple of thousand words on this new machinery for Russell Rockne’s upcoming mathematical oncology roadmap. Our central argument being — to continue the war metaphor — that with new machinery, we need new tactics.

Biologist often aim for reductive explanations, and mathematical modelers have tended to encourage this tactic by searching for mechanistic models. This is important work. But we also need to consider other tactics. Most notable, we need to look at the role that abstraction — both theoretical and empirical abstraction — can play in modeling and thinking about cancer.

The easiest way to share my vision for how we should approach this new tactic would be to throw a preprint up on BioRxiv or to wait for Rockne’s road map to eventually see print. Unfortunately, BioRxiv has a policy against views-like articles — as I was surprised to discover. And I am too impatient to wait for the eventual giant roadmap article.

Hence, I want to share some central parts in this blog post. This is basically an edited and slightly more focused version of our roadmap. Since, so far, game theory models have had more direct impact in oncology than fitness landscapes, I’ve focused this post exclusively on games.

Cataloging a year of blogging: cancer and fitness landscapes

Happy 2019!

As we leave 2018, the Theory, Evolution, and Games Group Blog enters its 9th calendar year. This past year started out slowly with only 4 posts in the first 5 months. However, after May 31st, I managed to maintain a regular posting schedule. This is the 32nd calendar week in a row with at least one new blog post released.

A total of 38 posts appeared on TheEGG last year. This is the 3rd most prolific year after the 47 in 2014 and 88 in 2013. One of those being a review of the 12 posts of 2017 (the least prolific year for TheEGG).

But the other 37 posts are too much to cover in one review. Thus, in this catalogue, I’ll focus on cancer and fitness landscapes. Next week, I’ll deal with the more philosophical content from the last year.

Local peaks and clinical resistance at negative cost

Last week, I expanded on Rob Noble’s warning about the different meanings of de novo resistance with a general discussion on the meaning of resistance in a biological vs clinical setting. In that post, I suggested that clinicians are much more comfortable than biologists with resistance without cost, or more radically: with negative cost. But I made no argument — especially no reductive argument that could potentially sway a biologist — about why we should entertain the clinician’s perspective. I want to provide a sketch for such an argument in this post.

In particular, I want to present a theoretical and extremely simple fitness landscape on which a hypothetical tumour might be evolving. The key feature of this landscape is a low local peak blocking the path to a higher local peak — a (partial) ultimate constraint on evolution. I will then consider two imaginary treatments on this landscape, one that I find to be more similar to a global chemotherapy and one that is meant to capture the essence of a targetted therapy. In the process, I will get to introduce the idea of therapy transformations to a landscape — something to address the tendency of people treating treatment fitness landscapes as completely unrelated to untreated fitness landscapes.

Of course, these hypothetical landscapes are chosen as toy models where we can have resistance emerge with a ‘negative’ cost. It is an empirical question to determine if any of this heuristic capture some important feature of real cancer landscapes.

But we won’t know until we start looking.

Causes and costs in biological vs clinical resistance

This Wednesday, on These few lines, Rob Noble warned of the two different ways in which the term de novo resistance is used by biologists and clinicians. The biologist sees de novo resistance as new genetic resistance arising after treatment has started. The clinician sees de novo resistance as a tumour that is not responsive to treatment from the start. To make matters even more confusing, Hitesh Mistry points to a further interpretation among pharmocologists: they refer to the tumour remaining after a partial but incomplete response to treatment as de novo resistant. Clearly this is a mess!

But I think this is an informative mess. I don’t think it is a matter of people accidentally overloading the same word. Instead, I think it reflects a conceptual difference in how biologists and clinicians think about resistance. A difference that is a bit akin to the difference between reductive and effective theories. It is also a difference that I had to deal with during the revisions of our recent work on measuring the games played by treatment sensitive and treatment resistance non-small cell lung cancer (Kaznatcheev et al., 2018).

The wei wu wei of evolutionary oncology

The world was disordered, rains would come and the rivers would flood. No one knew when. When it rained, plants would grow, but no one knew which were fit to eat and which were poisonous. Sickness was rife. Life was precarious.

The philosopher-king Yu dredged the rivers, cleaned them so they would flow into the sea. Only then were the people of the Middle Kingdom able to grow the five grains to obtain food.

Generations later, Bai Gui — the prime minister of Wei — boasted to Mengzi: “my management of the water is superior to that of Yu.”

Mengzi responded: “You are wrong. Yu’s method was based on the way of the water. It is why Yu used the four seas as receptacles. But you are using the neighbouring states as receptacles. When water goes contrary to its course, we call if overflowing. Overflowing means flooding water, something that a humane man detests… As for Yu moving the waters, he moved them without interference.”

Although Yu made changes to the environment by digging channels, he did so after understanding how the water flowed and moved naturally. He did so with knowledge of the Way. Yu’s management of water was superior to Bai Gui’s because Yu’s approach was in accordance with the Way. This is what evolutionary oncology seeks to achieve with cancer treatment. By understanding how the dynamics of somatic evolution drive tumour growth, we hope to change the selective pressures in accordance with this knowledge to manage or cure the disease.