Заглавная страница
Избранные статьи Случайная статья Познавательные статьи Новые добавления Обратная связь КАТЕГОРИИ: Археология Биология Генетика География Информатика История Логика Маркетинг Математика Менеджмент Механика Педагогика Религия Социология Технологии Физика Философия Финансы Химия Экология ТОП 10 на сайте Приготовление дезинфицирующих растворов различной концентрацииТехника нижней прямой подачи мяча. Франкопрусская война (причины и последствия) Организация работы процедурного кабинета Смысловое и механическое запоминание, их место и роль в усвоении знаний Коммуникативные барьеры и пути их преодоления Обработка изделий медицинского назначения многократного применения Образцы текста публицистического стиля Четыре типа изменения баланса Задачи с ответами для Всероссийской олимпиады по праву ЗНАЕТЕ ЛИ ВЫ?
Влияние общества на человека
Приготовление дезинфицирующих растворов различной концентрации Практические работы по географии для 6 класса Организация работы процедурного кабинета Изменения в неживой природе осенью Уборка процедурного кабинета Сольфеджио. Все правила по сольфеджио Балочные системы. Определение реакций опор и моментов защемления 
The real numbers. Supremum and Infimum of a set.Стр 1 из 3Следующая ⇒
Some set theory. Answer: Definition: Let S and T be set: a) S contains T, and we write S c T or T c S, if every number of T is also in S. in this case, T is a subset of S. b) S\T is the set of elements that are in S, but not in T. c) S equals T, and we write: S=T, is S contains T, and T contains S, thus, S=T if and only if S and T have the same members. d) The compecment of S denoted by S^{c} , is the set of elements in the universal set that are not in S. e) The union of S and T, denoted by S T, is the set of elements in at least one of S and T. f) The intersection of Sand T , denoted by S T is the set of element in both S and T. If S T 0(the empty set), then S and T are disjoint sets. Definition If x_{0} is real number and, ε>0, then the open interval (x_{0}ε,x_{0}+ε) in an neighborhood of x_{0}. If the set S contains an εneighborhood of x_{0} , then S is a neighborhood of x_{0}, and x_{0} is an interior point of S. Definition. Let S be subset of R. Then (a) x_{0} is a limitpoint of S, if every deleted neighborhood of x_{0} contains a point of S. (b) x_{0} is boundary point of S, if every neighborhood of x_{0} contains a least one point in S and one not in S. The set of boundary points of S denoted by S The closure of S, denoted by =S S (c)x_{0} is exterior to S, of x_{0} is in the interior of S^{c} Theorem. A set Sis closed, if and only if no point of S^{c}is a limit point of S. Corollary. A set is closed, if and only if it contains all its limit points. Definition . A collection H of open sets, is an open covering of a set S, if every point in S is contained in a set H belonging to H, that is, if Sc {HHcH}
Heineborel Theorem. BolzanoWeierstrass Theorem. Answer: If H is an open covering of a closed and bounded subset S of the real line, then S has an open covering H consisting of finitely many open sets belonging to H. Proof: Since S bounded, it has an infimumα and supremum β , and, since S is closed,α and β belong to S. Define S_{t}=S [α;t] for ≥α. and let F={tα<t<β and finitely many sets form H cover S_{t}} Since S_{β} =S_{1 }the theorem will be proved, if we can show that β F. To do this, we use the completeness of the reals. Since α S, S_{α}is the singleton set {α}, which is contained in some open set H_{α}form H, because H covers S; therefore, αcF. Since, F is nonempty and bounded above by β, it has supremumλ. First, we wish to show that λ=β by definition of F , it suffices to rull out the possibility that α<β. We consider two cases: Case 1. Suppose that α<β and λ S. Then, since S is closed λ is not a limitpointof S. Consequently, there is an ε>0 such that [λε, λ+ε] S= , so S_{λ}_{ε}=S_{λ+ε} However, the definition of λ implies that S_{λε} has a finite subcovering from H, while S_{λ+ε}does not. This is a contradiction. 3.2 Case 2.Suppose that λ<β and λ S. Then there is an open set H_{λ} in H that contains λ and along with λ, an interval [λε, λ+ε] for some positive ε. Since S_{λε} has a finite covering.{H_{1},…,H_{n}} of sets from H, it follows, that S_{λ+ε} has the finite covering{H_{1},….,H_{n},H_{λ}}. This contradicts the definition of λ. Now we know that α=β, which is in S. Therefore, there is an open set H_{β} in H that contains β and along with, an interval of the from [βε,β+ε], for some positive ε. Since S_{βε} is covered by a finite collection of set {H_{1},…,H_{k}}, S_{p} is covered by the finite collection {H_{1},…,H_{k},H_{β}}. Since S_{β}=S_{1} . Henceforth , we will say a closed and bounded set is compact. As an application of the HeineBorel theorem, we prove the following theorem of Bolzano andWeierstrass. BolzanoWeierstrass theorem. Every boumded infinite set of real numbers has at least one limit point. Proof: We will show that a bounded nonempty set without a limit point can contain only a finite number of points. If S has no limitpoints,then S is closed and every point x of S has an open neighborhood N_{x} that contains no point of other than x. The collection H={N_{x} x S} is an open covering for S. Since S is also bounded HeineBorel theorem implies that S can be covered by a finite collection of sets from H_{1},…. say N_{x1},…,N_{xn}. Since these set contain only x_{1},x_{2},…x_{n} from S, it follows that S={x_{1},…,x_{n}}
Taylor’s formula. Answer: A polynomial is of the form P(x)=Q_{0}+Q_{1}(xx_{0})+…+a_{n}(xx_{0})^{n} (1) where Q_{0},Q_{1},…Q_{n} and x_{0} are constants. In particular, a constant polynomial P(x)=Q_{0} is of degree zero, if Q_{0} 0 If f is differentiable all x_{0} , then f(x)=f(x_{0})+f’(x_{0})(xx_{0})+ (xx_{0}) Where =0 The polynomial T_{1}(x)=f(x_{0})+f’(x_{0})(xx_{0}) which is of degree 1 and satisfies T_{1}(x)=f(x_{0}) T^{’}_{1}(x_{0})=f’(x_{0}), approximates f so well near x_{0} that =0 (2) Now suppose that f has n derivatives at x_{0} and T_{n} is the polynomial of degree n such that T_{n}^{(r)}(x_{0})=f^{(r)}(x_{0}) 0 r n (3) T_{n} is polynomial of degree n; T_{n}(x)=a_{0}+a_{1}(xx_{0})+…+a_{n}(xx_{0})^{n} (4) Differentiating (4) yields T_{n}^{(r)}(x_{0})=r!a_{r} So (3) determines a_{r} uniquely as a_{r}=f^{(r)}(x_{0})/r! Therefore, T_{n}(x)=f(x_{0})+ (xx_{0}) + (xx_{0})^{2} + …+ (xx_{0})^{n} = (xx_{0})^{r}^{} We callT_{n} the nth Taylor polynomial of f about x_{0} (xxx дегенимиз x^{3}) 1. sinx=x – +– – +…. 2. cosx=1 – – + – – – + –  … 3. ln(1+x)=x – x^{2}/2 + x^{3}/3 – x^{4}/4 + x^{5}/5  … 4. e^{x}= 1+ x + x^{2}/2! +x^{3}/3! + x^{4}/4! + … 5. (1+x)^{a}=1 + ax + a(a1)x^{2}/2! + a(a1)(a2)x^{3}/3! + … + x^{a} The real numbers. Supremum and Infimum of a set. Answer: The real number system is first of all a set {a,b,c,…} on which the operation of addition and multiplication a defined that every pair of real numbers has a unique sum and product, both real numbers,with the following properties, A) a+b=b+a and ab=ba (commutative laws) B) (a+b)+c=a+(b+c) (ab)c=a(bc) (associative laws) C) a(b+c)=ab+ac (distributive law) D) There are distinct real numbers 0 and 1 such that a+0=a and a1=a for all a E) For each (a) there is a real number –a such that a+(a)=0 and if a 0 there is a real number such that a =1 A set on which two operations are difined so as to have properties (A)(E) called a field. The simplest possible field consist of two elements which we denote by 0 and 1, with addition defined by 0+0=1+(1)=0, 1+0=0+1=1 (1) and multiplication defined by 0×0=0×1=1×0=0, 1×1=1 (2) The order Relation The real number system is ordered by relation <, which has the following properties: F) For each pair of real numbers (a) and (b) exactly one of the following is true: a=b ,a<b or a>b G) If a<b and b<c, then a<c (The relation < is transitive) H) If a<b,thena+c<b+c for any c and if c>0, then a×c<b×c. A field with an order relation satisfying (F)(H) in an ordered field. Theorem 1 (The Triangle Inequality). If aandb are any two real numbers then: a+b≤a+b (3) Corollary 2. If a and b are any two real numbers then ab≥ab (4) and a+b≥a+b (5) supremum of a set: A set S of real numbers is bounded above if there is a real number b such that x≤b whenever x S. In this case, b is an upper bound of S. If b is an upper bound of S, then so is any large number, because of property (6) If is an upper bound of S, but no number less than β is, then β is a supremum of S and we write β=supS Example: If S is the set of negative numbers, then any nonnegative number is an upper bound of S,and sup=0 1.2 If S_{1} is the set of negative integers,then any number a such that a≥1 is an upper bound of S_{1} , sup=1 This example shows that a supremumof a set may or may not be in the set, since S_{1} , contains its supremum, but S does not. A nonempty set is a set that has at least one number. The empty set, denoted by , is the set that has no numbers. The Completeness Axiom. (I) If a nonempty set of reak numbers is bounded above, then it has a supremum. Property (I) is called conpleteness and we say that the real number system is a complete ordered field. Thorem 3. If a nonempty set S of real numbers is bounded above, then sups is the unique real number β such that (a) x≤β for all x in S. (b) if ε>0 there is an x_{0} in S such that x_{0}>βε Proof: we first show that β=supS has properties (a) and (b). Since β is an upper bound of it must satisfy (a). Since any real number a less than β can be written as βε with ε=βa>0, (b) is just another way of saying that no number less than β is an upper bound of S, β=supS (satisfies (a) and (b)) Now we show that, there cannot be more than one real number with properties (a) and (b). Suppose thatβ_{1}<β_{2} and β_{2} has property (b) thus, if ε>0, there is an x_{0} in S such that x_{0}> _{2}ε. Then by taking ε=β_{2}β_{1} , we see that there is an x_{0} in S such that x_{0}>β_{2}(β_{2}β_{1})=β_{1}, So β_{1} cannot have property (a). Therefore, there cannot be more than one real number that satisfies both (a) and (b). Some Natation: “x is a member of S”=>x S “x is not a member of S” => x S. Theorem 4. (The ArchemedeanProperty ). If p and ε are positive, then nε>p for some integer n. Proof: The proof is by contradiction. If the statement is false, p is an upper bound of the set S={ x  x = nε, n is a integer}. Therefore, S has a supremum β, by property (I). Therefore, nε≤β (b) 1.3For all integer whenever n is , (b) implies that (n+1)ε≤β and therefore nε≤βε for all integers n. Hence, βε is an upper bound of S. Since, βε<β, this contradicts the definition of β. Infimum of a set. A set S of real numbers is bounded below if there is a real number a such that x≥a, whenever x S. In this case, a is a lower bounded of S. but no number greater than α is, then α is an infimum of S, and we write: α=infS. Theorem 8. If nonempty set S of real numbers is bounded below, then infS is the unique real number α such that: (a) x≥α for all x in S. (b) If ε>0, there is an x_{0} in such that x_{0}< α+ε Proof: A set S in bounded, of there are numbersa and b such that a≤x≤b for all x in S. A bounded nonempty set has a unique supremum and a unique infimum, and: infS≤supS. A nonempty set S of real numbers is unbounded above if it no upper bound, or unbounded below, if it has no lower bound  < x <+ (7) We call points of infinity. If (S)is a nonempty set of real numbers, we write: supS= (8) to indicate that S is unbounded above, and infS= (9) to indicate that S is unbounded below. (a) If a is any real number, then: a+ = +a= , a = +a=  , = =0 (b) If a>0, then: a = a= , a( )=( )a= (c) If a<0 , then: a = a= , a( )=( )a= We also define: + = =( )( )= and   = ( )= Finally, we define  = = .
Some set theory. Answer: Definition: Let S and T be set: a) S contains T, and we write S c T or T c S, if every number of T is also in S. in this case, T is a subset of S. b) S\T is the set of elements that are in S, but not in T. c) S equals T, and we write: S=T, is S contains T, and T contains S, thus, S=T if and only if S and T have the same members. d) The compecment of S denoted by S^{c} , is the set of elements in the universal set that are not in S. e) The union of S and T, denoted by S T, is the set of elements in at least one of S and T. f) The intersection of Sand T , denoted by S T is the set of element in both S and T. If S T 0(the empty set), then S and T are disjoint sets. Definition If x_{0} is real number and, ε>0, then the open interval (x_{0}ε,x_{0}+ε) in an neighborhood of x_{0}. If the set S contains an εneighborhood of x_{0} , then S is a neighborhood of x_{0}, and x_{0} is an interior point of S. Definition. Let S be subset of R. Then (a) x_{0} is a limitpoint of S, if every deleted neighborhood of x_{0} contains a point of S. (b) x_{0} is boundary point of S, if every neighborhood of x_{0} contains a least one point in S and one not in S. The set of boundary points of S denoted by S The closure of S, denoted by =S S (c)x_{0} is exterior to S, of x_{0} is in the interior of S^{c} Theorem. A set Sis closed, if and only if no point of S^{c}is a limit point of S. Corollary. A set is closed, if and only if it contains all its limit points. Definition . A collection H of open sets, is an open covering of a set S, if every point in S is contained in a set H belonging to H, that is, if Sc {HHcH}


Последнее изменение этой страницы: 20160826; Нарушение авторского права страницы infopedia.su Все материалы представленные на сайте исключительно с целью ознакомления читателями и не преследуют коммерческих целей или нарушение авторских прав. Обратная связь  54.208.73.179 (0.017 с.) 