On density of old sets in Prikry type extensions.
|
|
- Εἰλείθυια Σπανός
- 5 χρόνια πριν
- Προβολές:
Transcript
1 On density of old sets in Prikry type extensions. Moti Gitik December 31, 2015 Abstract Every set of ordinals of cardinality κ in a Prikry extension with a measure over κ contains an old set of arbitrary large cardinality below κ, and, actually, it can be split into countably many old sets. What about sets bigger cardinalities? Clearly, any set of ordinals in a forcing extension of a regular cardinality above the cardinality of the forcing used, contains an old set of the same cardinality. Still cardinals in the interval (κ, 2 κ ] remain. Here we would like to address this type of questions. 1 A situation under 2 κ = κ +. Let us start with the following observation. Proposition 1.1 Suppose that 2 κ = κ +. Let U be a normal ultrafilter over κ and P U be the Prikry forcing with U. Then in V P U there is a subset of κ + without old subsets of the same size. Proof. Work in V. Pick a generating sequence A α α < κ + for U such that for every α β < κ +, A β A 1 α. It is possible since 2 κ = κ + and U is normal. Define an other generating sequence A α α < κ + as follows. Let C κ + be a club such that [α ν, α ν+1 ] = κ, for every ν < κ +, where {α ν ν < κ + } is an increasing enumeration of C. We set A α ν = A ν. Pick a surjective map h : κ [κ] <ω. Let, for every ν < κ +, g ν : (α ν, α ν+1 ) κ. Now if β (α ν, α ν+1 ), for some ν < κ +, then set A β = A α ν h(g ν (β)). Clearly, A α α < κ + is a generating sequence for U and for every α β < κ +, A β A α. Now let G P U be generic and {κ n n < ω} be the corresponding Prikry sequence. Set X = {α < κ + A α {κ n n < ω}}. 1 A B means that A \ B < κ 1
2 Lemma 1.2 X = κ +. Proof. Work in V. Let t, A P U and δ < κ +. Pick α ν C \ δ + 1 such that A A α ν. Find β (α ν, α ν+1 ) such that A β = A α ν t. Then So, we are done, using a density argument. of the lemma. t, A A α ν β X. Suppose now that there is X X, X = κ + and X V. Set A = α X A α. Then, clearly, A V and A {κ n n < ω}. But then A U. Hence, there is α < κ +, A α A. Now, for every β X, β α, we have A β A α A. But each such A β A. Hence, for every β X, β α, we have A β = A. But this is impossible. Thus just split A in V into two disjoint sets each of cardinality κ. One of them should be in U, and so must almost contain on of such A β s. Contradiction. So there are sets of ordinals of cardinality κ + without old subsets of cardinality κ +. But what about subsets of size κ? The next proposition provides an answer. Proposition 1.3 Every set of ordinals of size > κ contains an old subset of size κ. Proof. Suppose otherwise. Pick t, A G and a name a such that t, A ( a = κ + and a does not contain an old subset of size κ). Let a = {α ν ν < κ + }. For every ν < κ +, pick t ν, A ν G, t ν, A ν t, A such that t ν, A ν α ν. Find S κ +, S = κ + and t such that for every ν S we have t ν = t. Clearly, in order to derive a contradiction, it is enough to find A U, A A such that the condition 2
3 t, A decides simultaneously α ν s for κ many ν s. Work in V. Set Z = {ν < κ + B U(B A t, B αν)}. Then Z = κ +, since Z S. Let us chose for every ν Z a set B ν U, B ν A such that t, B α ν. Now let us use the following pretty observation of F. Galvin (see [1], but for a reader convenience let state the proof here): Proposition 1.4 (Galvin) Suppose that 2 <λ = λ, I is a normal ideal on λ and {B ν ν < λ + } I. Then there is X λ +, X = λ such that ν X B ν I. Proof. Set H αξ = {β < λ + B α ξ = B β ξ}, for every α < λ + and ξ < λ. Lemma 1.5 There is α < λ + such that for every ξ < λ we have H αξ = λ +. Proof. Suppose otherwise. Then for every α < λ + there is ξ α < λ such that H αξ λ. But for every ξ < λ we have 2 ξ λ, so there are at most λ possibilities for B β ξ s. Hence, H αξα λ. α<λ + But, clearly, α H αξα, for every α < λ +. Contradiction. of the lemma. Pick α such that for every ξ < λ, H αξ = λ +. Define a sequence η ξ ξ < λ by induction as follows. η ξ H αξ+1 \ {η ξ ξ < ξ}. Set Then B I, since B = ξ<λ B ηξ. B \ B α ξ<λ(b ηξ \ ξ + 1), and ξ<λ (B η ξ \ ξ + 1) I due to normality of I. 3
4 So, there will be X Z of cardinality κ and B U such that B B ν. Then t, B t, B ν, for every ν X, and so, t, B decides κ many of α ν s. ν X 2 A situation without 2 κ = κ +. Let us show that the assumption 2 κ = κ + cannot be dropped from 1.1. Start with the following general result. Proposition 2.1 Suppose that U is a normal ultrafilter over κ which has a generating family A ν ν < χ such that A β A α, for every α < β < χ. Let E be a set of ordinals in V P U. Then 1. if E = κ, then for every η < E, there is an old subset of E of cardinality η. 2. there is subset of κ of cardinality κ without an old subset of cardinality κ; 3. if ω < E < κ, then for every η E, there is an old subset of E of cardinality η; 4. there is a subset of κ of cardinality ω without infinite old subsets; 5. if E > cof(χ), then for every η E, there is an old subset of E of cardinality η; 6. there is a subset of χ of cardinality cof(χ) without old subsets of the same cardinality; 7. for every cardinal µ < χ with (cof(µ)) V = κ, there is a subset of µ without old subsets of the same cardinality; 8. if κ < E χ, then for every η E, such that E = cof(χ) or (cof( E )) V = κ imply η < E, there is an old subset of E of cardinality η. 2 Proof. The first four items are trivial. Item 5 is trivial as well, since P U will have a dense subset of cardinality cof(χ). Item 6 follows from the argument of 1.1 only replacing κ + by cof(χ). Let us deal with Item 7. 2 Remember that by König s theorem, cof(χ) κ +. 4
5 Suppose that κ < E χ. Let η E, η cof(χ) and η < E, if (cof( E )) V = κ. If η = κ, then the argument of 1.3 provides an old subset of E of cardinality κ. Assume that cof(η) > κ. Proceed as in 1.3. Let {α ξ ξ < η} be a set of η elements of E. Find Z η, Z = η, t, B ν U, for each ν Z such that t, B ν α ν. For every ν Z there is β ν such that A βν B ν. Then there will be β < χ and Z Z, Z = η, such that for every ν Z we have A β A βν B ν. Then for every ν Z there is τ ν < κ such that A β \ τ ν B ν. Pick Z Z, Z = η and τ < κ such that for every ν Z, τ ν = τ. It is possible since cof(η) > κ. Now we will have t, B ν \ τ t, A β \ τ, for every ν Z. Hence, the condition t, A β \ τ decides simultaneously α ν s for η many ν s. So we are done. The existence of such generating families with 2 κ > κ + follows from [4]: Theorem 2.2 Let κ be an almost huge cardinal with a measurable target λ. 3 Then for every cardinal χ, κ < χ < λ, cof(χ) > κ there is a cofinalities preserving extension with a normal ultrafilter over κ which has a generating family A ν ν < χ such that A β A α, for every α < β < χ. In particular, we can conclude the following: Corollary 2.3 It is consistent that in the Prikry forcing extension every set of ordinals of cardinality κ + contains an old subset of the same cardinality. Working a bit harder it is possible to show the following: 3 I.e. there is j : V M such that λ = j(κ) is a measurable cardinal and λ> M M. 5
6 Proposition 2.4 It is consistent that in the Prikry forcing extension every set of ordinals of cardinality κ + is a countable union of old sets. Proof. We start with the model with 2 κ = κ ++ and a normal ultrafilter U over κ with a generating family A ν ν < κ ++ such that A β A α, for every α < β < κ ++. Let G P U be generic and X = {α ν ν < κ + } be a set of ordinals of cardinality κ + in V [G]. For every ν < κ +, pick t ν, B ν G which decides αν. Let κ n n < ω be the Prikry sequence. For every n < ω, set X n = {α ν B U( κ 0,..., κ n, B G κ 0,..., κ n, B α ν)}. It is enough to show that each X n can be split into ω old sets. Fix n < ω. Set t = κ 0,..., κ n. So, for every α ν X n, we have t, B ν G which decides α ν. Find ξ < κ ++ large enough such that A ξ B ν, for every α ν X n, there is m, n m < ω, κ 0,..., κ m, A ξ \ κ m + 1 G. Now, for every k < ω, set X nk = {α ν A ξ \ κ k B ν }. Note that for every α ν X nk. Then and B ν (A ξ \ κ k ) {κ n+1,..., κ max(k,m) }, t, B ν t, (A ξ \ κ k ) {κ n+1,..., κ max(k,m) } t, (A ξ \ κ k ) {κ n+1,..., κ max(k,m) } G, for every α ν X nk. It follows now that X nk = {ρ ν < κ + ( t, (A ξ \ κ k ) {κ n+1,..., κ max(k,m) } α ν = ˇρ)}. 6
7 Clearly, the set on the right is in V, and, hence X nk V as well, for every k < ω. But also clear that X n = X nk. k<ω So we are done. Let us point out that 2 κ = κ ++ does not imply the conclusion of 2.3. Proposition 2.5 It is consistent with 2 κ = κ ++ that in the Prikry forcing extension there is a set of ordinals of cardinality κ + without old subsets of the same cardinality. Proof. Use the construction of [2]. We start with V = L[ E] model with o(κ) = κ ++. Let U 0 denotes the normal measure on κ which concentrates on nonmeasurable cardinals and which extends to normal measure U in the final model V of 2 κ = κ ++. The extension is of the form V [G <κ, G κ ], where V [G <κ ] is an extension of V by a forcing of size κ and G κ is the forcing over κ which consists of adding κ ++ Cohen subsets and may be some additional things which does not add new subsets to κ. Let U be a normal ultrafilter over κ in V = V [G <κ, G κ ]. Force with P U and let κ n n < ω be the Prikry sequence. Then κ n n < ω be a Prikry sequence also over V for U 0. Pick in V a generating family A α α < κ + for U 0. Proceed exactly as in 1.1 and define A β β < κ+ and X = {β < κ + A β {κ n n < ω}}. As in 1.1, then X = κ +. We claim that X does not contain old (i.e. those in V ) subsets of cardinality κ +. Suppose otherwise. Let X witnesses this. Consider A = A β. β X Now, this A need not be in V, since X was picked in V and, so may not be in V. However A {κ n n < ω}, and so A U. Consider j U : V M U (V ) κ /U. 7
8 Then j U V is an iterated ultrapower of V by its measures, but since U extends U 0, it follows that U 0 was applied first in this iteration process. In particular, it is not on the sequence of core model K U of M U and so not in M U. But A M U and so be can define U 0 in M U as follows: {Z P(κ) K U A Z}. Contradiction. 3 A remark on Extender based Prikry forcing. Let E be an extender over κ of the length at least κ +. Denote by P E the Extender based forcing as defined in [3]. 4 Let G P E be a generic subset. Proposition 3.1 In V [G] there is a subset of κ + without old subsets of cardinality κ. Proof. Let us denote by b α : ω κ the Prikry sequence of G for the α th measure E α of E. Then, for each α, there is the least n α < ω such that for every n, n α n < ω, π ακ (b α (n)) = b κ (n), where π ακ denotes the canonical projection of E α onto the normal measure E κ of the extender. There are A κ +, A = κ + and n < ω such that for every α A, n α = n. We claim that A does not contain old subsets of cardinality κ. Suppose otherwise. Let B be such a subset. Pick some p = p γ γ supp(p) p mc, T G forcing this and deciding B. We can assume that each α B belongs to supp(p), since otherwise we are completely free about b α and can easily to make π ακ (b α (n)) b κ (n), for some n n. Without loss of generality we can assume that for every α B, n m := p α and m = p mc (and then = p 0 ). Pick now some ν Suc T (p mc ). Let ν 0, as usual be π mcκ (ν). By the definition of the forcing, the set {α supp(p) ν is permitted for p α } 4 A very similar argument works for the Merimovich variations [5]. 8
9 has cardinality < κ (actually at most ν 0 ). Now, since B = κ, there is α B such that ν is not permitted for α. This means that in the extension p ν of p by ν, p α does not extend. But then, p ν π ακ (b α (m )) = b κ (m ). Contradiction. 9
10 References [1] J. Baumgartner, A. Hajnal and A. Mate, Weak saturation properties of ideals, Coll. Math. Soc. J. Bolyai 10, Infinite and Finite Sets, Keszthely, 1973, [2] M. Gitik, Not SCH from o(κ) = κ ++, [3] M. Gitik and M. Magidor, The SCH revisited, [4] M. Gitik and S. Shelah, On density of the box products, Topology and its Applications, Volume 88, Issue 3, 6 November 1998, [5] C. Merimovich, Prikry on Extenders, Revisited. Israel Journal of Mathematics, Volume 160, Issue 2, August 2007, pages
Every set of first-order formulas is equivalent to an independent set
Every set of first-order formulas is equivalent to an independent set May 6, 2008 Abstract A set of first-order formulas, whatever the cardinality of the set of symbols, is equivalent to an independent
Διαβάστε περισσότερα2 Composition. Invertible Mappings
Arkansas Tech University MATH 4033: Elementary Modern Algebra Dr. Marcel B. Finan Composition. Invertible Mappings In this section we discuss two procedures for creating new mappings from old ones, namely,
Διαβάστε περισσότεραST5224: Advanced Statistical Theory II
ST5224: Advanced Statistical Theory II 2014/2015: Semester II Tutorial 7 1. Let X be a sample from a population P and consider testing hypotheses H 0 : P = P 0 versus H 1 : P = P 1, where P j is a known
Διαβάστε περισσότεραOrdinal Arithmetic: Addition, Multiplication, Exponentiation and Limit
Ordinal Arithmetic: Addition, Multiplication, Exponentiation and Limit Ting Zhang Stanford May 11, 2001 Stanford, 5/11/2001 1 Outline Ordinal Classification Ordinal Addition Ordinal Multiplication Ordinal
Διαβάστε περισσότεραORDINAL ARITHMETIC JULIAN J. SCHLÖDER
ORDINAL ARITHMETIC JULIAN J. SCHLÖDER Abstract. We define ordinal arithmetic and show laws of Left- Monotonicity, Associativity, Distributivity, some minor related properties and the Cantor Normal Form.
Διαβάστε περισσότεραFractional Colorings and Zykov Products of graphs
Fractional Colorings and Zykov Products of graphs Who? Nichole Schimanski When? July 27, 2011 Graphs A graph, G, consists of a vertex set, V (G), and an edge set, E(G). V (G) is any finite set E(G) is
Διαβάστε περισσότεραOther Test Constructions: Likelihood Ratio & Bayes Tests
Other Test Constructions: Likelihood Ratio & Bayes Tests Side-Note: So far we have seen a few approaches for creating tests such as Neyman-Pearson Lemma ( most powerful tests of H 0 : θ = θ 0 vs H 1 :
Διαβάστε περισσότεραChapter 3: Ordinal Numbers
Chapter 3: Ordinal Numbers There are two kinds of number.. Ordinal numbers (0th), st, 2nd, 3rd, 4th, 5th,..., ω, ω +,... ω2, ω2+,... ω 2... answers to the question What position is... in a sequence? What
Διαβάστε περισσότεραExample Sheet 3 Solutions
Example Sheet 3 Solutions. i Regular Sturm-Liouville. ii Singular Sturm-Liouville mixed boundary conditions. iii Not Sturm-Liouville ODE is not in Sturm-Liouville form. iv Regular Sturm-Liouville note
Διαβάστε περισσότεραC.S. 430 Assignment 6, Sample Solutions
C.S. 430 Assignment 6, Sample Solutions Paul Liu November 15, 2007 Note that these are sample solutions only; in many cases there were many acceptable answers. 1 Reynolds Problem 10.1 1.1 Normal-order
Διαβάστε περισσότεραUniform Convergence of Fourier Series Michael Taylor
Uniform Convergence of Fourier Series Michael Taylor Given f L 1 T 1 ), we consider the partial sums of the Fourier series of f: N 1) S N fθ) = ˆfk)e ikθ. k= N A calculation gives the Dirichlet formula
Διαβάστε περισσότεραCardinals. Y = n Y n. Define h: X Y by f(x) if x X n X n+1 and n is even h(x) = g
Cardinals 1. Introduction to Cardinals We work in the base theory ZF. The definitions and many (but not all) of the basic theorems do not require AC; we will indicate explicitly when we are using choice.
Διαβάστε περισσότεραSection 8.3 Trigonometric Equations
99 Section 8. Trigonometric Equations Objective 1: Solve Equations Involving One Trigonometric Function. In this section and the next, we will exple how to solving equations involving trigonometric functions.
Διαβάστε περισσότεραEE512: Error Control Coding
EE512: Error Control Coding Solution for Assignment on Finite Fields February 16, 2007 1. (a) Addition and Multiplication tables for GF (5) and GF (7) are shown in Tables 1 and 2. + 0 1 2 3 4 0 0 1 2 3
Διαβάστε περισσότεραFourier Series. MATH 211, Calculus II. J. Robert Buchanan. Spring Department of Mathematics
Fourier Series MATH 211, Calculus II J. Robert Buchanan Department of Mathematics Spring 2018 Introduction Not all functions can be represented by Taylor series. f (k) (c) A Taylor series f (x) = (x c)
Διαβάστε περισσότεραFinite Field Problems: Solutions
Finite Field Problems: Solutions 1. Let f = x 2 +1 Z 11 [x] and let F = Z 11 [x]/(f), a field. Let Solution: F =11 2 = 121, so F = 121 1 = 120. The possible orders are the divisors of 120. Solution: The
Διαβάστε περισσότεραHomework 3 Solutions
Homework 3 Solutions Igor Yanovsky (Math 151A TA) Problem 1: Compute the absolute error and relative error in approximations of p by p. (Use calculator!) a) p π, p 22/7; b) p π, p 3.141. Solution: For
Διαβάστε περισσότεραA Note on Intuitionistic Fuzzy. Equivalence Relation
International Mathematical Forum, 5, 2010, no. 67, 3301-3307 A Note on Intuitionistic Fuzzy Equivalence Relation D. K. Basnet Dept. of Mathematics, Assam University Silchar-788011, Assam, India dkbasnet@rediffmail.com
Διαβάστε περισσότεραSINGULAR CARDINALS AND SQUARE PROPERTIES
SINGULAR CARDINALS AND SQUARE PROPERTIES MENACHEM MAGIDOR AND DIMA SINAPOVA Abstract. We analyze the effect of singularizing cardinals on square properties. By work of Džamonja-Shelah and of Gitik, if
Διαβάστε περισσότερα4.6 Autoregressive Moving Average Model ARMA(1,1)
84 CHAPTER 4. STATIONARY TS MODELS 4.6 Autoregressive Moving Average Model ARMA(,) This section is an introduction to a wide class of models ARMA(p,q) which we will consider in more detail later in this
Διαβάστε περισσότεραHomomorphism in Intuitionistic Fuzzy Automata
International Journal of Fuzzy Mathematics Systems. ISSN 2248-9940 Volume 3, Number 1 (2013), pp. 39-45 Research India Publications http://www.ripublication.com/ijfms.htm Homomorphism in Intuitionistic
Διαβάστε περισσότεραNowhere-zero flows Let be a digraph, Abelian group. A Γ-circulation in is a mapping : such that, where, and : tail in X, head in
Nowhere-zero flows Let be a digraph, Abelian group. A Γ-circulation in is a mapping : such that, where, and : tail in X, head in : tail in X, head in A nowhere-zero Γ-flow is a Γ-circulation such that
Διαβάστε περισσότερα2. Let H 1 and H 2 be Hilbert spaces and let T : H 1 H 2 be a bounded linear operator. Prove that [T (H 1 )] = N (T ). (6p)
Uppsala Universitet Matematiska Institutionen Andreas Strömbergsson Prov i matematik Funktionalanalys Kurs: F3B, F4Sy, NVP 2005-03-08 Skrivtid: 9 14 Tillåtna hjälpmedel: Manuella skrivdon, Kreyszigs bok
Διαβάστε περισσότεραLecture 2: Dirac notation and a review of linear algebra Read Sakurai chapter 1, Baym chatper 3
Lecture 2: Dirac notation and a review of linear algebra Read Sakurai chapter 1, Baym chatper 3 1 State vector space and the dual space Space of wavefunctions The space of wavefunctions is the set of all
Διαβάστε περισσότεραMath 446 Homework 3 Solutions. (1). (i): Reverse triangle inequality for metrics: Let (X, d) be a metric space and let x, y, z X.
Math 446 Homework 3 Solutions. (1). (i): Reverse triangle inequalit for metrics: Let (X, d) be a metric space and let x,, z X. Prove that d(x, z) d(, z) d(x, ). (ii): Reverse triangle inequalit for norms:
Διαβάστε περισσότεραThe Simply Typed Lambda Calculus
Type Inference Instead of writing type annotations, can we use an algorithm to infer what the type annotations should be? That depends on the type system. For simple type systems the answer is yes, and
Διαβάστε περισσότεραMINIMAL CLOSED SETS AND MAXIMAL CLOSED SETS
MINIMAL CLOSED SETS AND MAXIMAL CLOSED SETS FUMIE NAKAOKA AND NOBUYUKI ODA Received 20 December 2005; Revised 28 May 2006; Accepted 6 August 2006 Some properties of minimal closed sets and maximal closed
Διαβάστε περισσότεραBounding Nonsplitting Enumeration Degrees
Bounding Nonsplitting Enumeration Degrees Thomas F. Kent Andrea Sorbi Università degli Studi di Siena Italia July 18, 2007 Goal: Introduce a form of Σ 0 2-permitting for the enumeration degrees. Till now,
Διαβάστε περισσότεραStatistical Inference I Locally most powerful tests
Statistical Inference I Locally most powerful tests Shirsendu Mukherjee Department of Statistics, Asutosh College, Kolkata, India. shirsendu st@yahoo.co.in So far we have treated the testing of one-sided
Διαβάστε περισσότεραLecture 2. Soundness and completeness of propositional logic
Lecture 2 Soundness and completeness of propositional logic February 9, 2004 1 Overview Review of natural deduction. Soundness and completeness. Semantics of propositional formulas. Soundness proof. Completeness
Διαβάστε περισσότεραPhys460.nb Solution for the t-dependent Schrodinger s equation How did we find the solution? (not required)
Phys460.nb 81 ψ n (t) is still the (same) eigenstate of H But for tdependent H. The answer is NO. 5.5.5. Solution for the tdependent Schrodinger s equation If we assume that at time t 0, the electron starts
Διαβάστε περισσότεραConcrete Mathematics Exercises from 30 September 2016
Concrete Mathematics Exercises from 30 September 2016 Silvio Capobianco Exercise 1.7 Let H(n) = J(n + 1) J(n). Equation (1.8) tells us that H(2n) = 2, and H(2n+1) = J(2n+2) J(2n+1) = (2J(n+1) 1) (2J(n)+1)
Διαβάστε περισσότεραStrain gauge and rosettes
Strain gauge and rosettes Introduction A strain gauge is a device which is used to measure strain (deformation) on an object subjected to forces. Strain can be measured using various types of devices classified
Διαβάστε περισσότεραCongruence Classes of Invertible Matrices of Order 3 over F 2
International Journal of Algebra, Vol. 8, 24, no. 5, 239-246 HIKARI Ltd, www.m-hikari.com http://dx.doi.org/.2988/ija.24.422 Congruence Classes of Invertible Matrices of Order 3 over F 2 Ligong An and
Διαβάστε περισσότερα1. Introduction and Preliminaries.
Faculty of Sciences and Mathematics, University of Niš, Serbia Available at: http://www.pmf.ni.ac.yu/filomat Filomat 22:1 (2008), 97 106 ON δ SETS IN γ SPACES V. Renuka Devi and D. Sivaraj Abstract We
Διαβάστε περισσότεραderivation of the Laplacian from rectangular to spherical coordinates
derivation of the Laplacian from rectangular to spherical coordinates swapnizzle 03-03- :5:43 We begin by recognizing the familiar conversion from rectangular to spherical coordinates (note that φ is used
Διαβάστε περισσότεραChapter 6: Systems of Linear Differential. be continuous functions on the interval
Chapter 6: Systems of Linear Differential Equations Let a (t), a 2 (t),..., a nn (t), b (t), b 2 (t),..., b n (t) be continuous functions on the interval I. The system of n first-order differential equations
Διαβάστε περισσότερα3.4 SUM AND DIFFERENCE FORMULAS. NOTE: cos(α+β) cos α + cos β cos(α-β) cos α -cos β
3.4 SUM AND DIFFERENCE FORMULAS Page Theorem cos(αβ cos α cos β -sin α cos(α-β cos α cos β sin α NOTE: cos(αβ cos α cos β cos(α-β cos α -cos β Proof of cos(α-β cos α cos β sin α Let s use a unit circle
Διαβάστε περισσότεραHOMEWORK 4 = G. In order to plot the stress versus the stretch we define a normalized stretch:
HOMEWORK 4 Problem a For the fast loading case, we want to derive the relationship between P zz and λ z. We know that the nominal stress is expressed as: P zz = ψ λ z where λ z = λ λ z. Therefore, applying
Διαβάστε περισσότεραω ω ω ω ω ω+2 ω ω+2 + ω ω ω ω+2 + ω ω+1 ω ω+2 2 ω ω ω ω ω ω ω ω+1 ω ω2 ω ω2 + ω ω ω2 + ω ω ω ω2 + ω ω+1 ω ω2 + ω ω+1 + ω ω ω ω2 + ω
0 1 2 3 4 5 6 ω ω + 1 ω + 2 ω + 3 ω + 4 ω2 ω2 + 1 ω2 + 2 ω2 + 3 ω3 ω3 + 1 ω3 + 2 ω4 ω4 + 1 ω5 ω 2 ω 2 + 1 ω 2 + 2 ω 2 + ω ω 2 + ω + 1 ω 2 + ω2 ω 2 2 ω 2 2 + 1 ω 2 2 + ω ω 2 3 ω 3 ω 3 + 1 ω 3 + ω ω 3 +
Διαβάστε περισσότεραMatrices and Determinants
Matrices and Determinants SUBJECTIVE PROBLEMS: Q 1. For what value of k do the following system of equations possess a non-trivial (i.e., not all zero) solution over the set of rationals Q? x + ky + 3z
Διαβάστε περισσότεραSOME PROPERTIES OF FUZZY REAL NUMBERS
Sahand Communications in Mathematical Analysis (SCMA) Vol. 3 No. 1 (2016), 21-27 http://scma.maragheh.ac.ir SOME PROPERTIES OF FUZZY REAL NUMBERS BAYAZ DARABY 1 AND JAVAD JAFARI 2 Abstract. In the mathematical
Διαβάστε περισσότεραSome new generalized topologies via hereditary classes. Key Words:hereditary generalized topological space, A κ(h,µ)-sets, κµ -topology.
Bol. Soc. Paran. Mat. (3s.) v. 30 2 (2012): 71 77. c SPM ISSN-2175-1188 on line ISSN-00378712 in press SPM: www.spm.uem.br/bspm doi:10.5269/bspm.v30i2.13793 Some new generalized topologies via hereditary
Διαβάστε περισσότεραTHE GENERAL INDUCTIVE ARGUMENT FOR MEASURE ANALYSES WITH ADDITIVE ORDINAL ALGEBRAS
THE GENERAL INDUCTIVE ARGUMENT FOR MEASURE ANALYSES WITH ADDITIVE ORDINAL ALGEBRAS STEFAN BOLD, BENEDIKT LÖWE In [BoLö ] we gave a survey of measure analyses under AD, discussed the general theory of order
Διαβάστε περισσότεραLecture 15 - Root System Axiomatics
Lecture 15 - Root System Axiomatics Nov 1, 01 In this lecture we examine root systems from an axiomatic point of view. 1 Reflections If v R n, then it determines a hyperplane, denoted P v, through the
Διαβάστε περισσότεραOn a four-dimensional hyperbolic manifold with finite volume
BULETINUL ACADEMIEI DE ŞTIINŢE A REPUBLICII MOLDOVA. MATEMATICA Numbers 2(72) 3(73), 2013, Pages 80 89 ISSN 1024 7696 On a four-dimensional hyperbolic manifold with finite volume I.S.Gutsul Abstract. In
Διαβάστε περισσότεραReminders: linear functions
Reminders: linear functions Let U and V be vector spaces over the same field F. Definition A function f : U V is linear if for every u 1, u 2 U, f (u 1 + u 2 ) = f (u 1 ) + f (u 2 ), and for every u U
Διαβάστε περισσότεραF A S C I C U L I M A T H E M A T I C I
F A S C I C U L I M A T H E M A T I C I Nr 46 2011 C. Carpintero, N. Rajesh and E. Rosas ON A CLASS OF (γ, γ )-PREOPEN SETS IN A TOPOLOGICAL SPACE Abstract. In this paper we have introduced the concept
Διαβάστε περισσότεραΚΥΠΡΙΑΚΗ ΕΤΑΙΡΕΙΑ ΠΛΗΡΟΦΟΡΙΚΗΣ CYPRUS COMPUTER SOCIETY ΠΑΓΚΥΠΡΙΟΣ ΜΑΘΗΤΙΚΟΣ ΔΙΑΓΩΝΙΣΜΟΣ ΠΛΗΡΟΦΟΡΙΚΗΣ 19/5/2007
Οδηγίες: Να απαντηθούν όλες οι ερωτήσεις. Αν κάπου κάνετε κάποιες υποθέσεις να αναφερθούν στη σχετική ερώτηση. Όλα τα αρχεία που αναφέρονται στα προβλήματα βρίσκονται στον ίδιο φάκελο με το εκτελέσιμο
Διαβάστε περισσότεραEcon 2110: Fall 2008 Suggested Solutions to Problem Set 8 questions or comments to Dan Fetter 1
Eon : Fall 8 Suggested Solutions to Problem Set 8 Email questions or omments to Dan Fetter Problem. Let X be a salar with density f(x, θ) (θx + θ) [ x ] with θ. (a) Find the most powerful level α test
Διαβάστε περισσότεραProblem Set 3: Solutions
CMPSCI 69GG Applied Information Theory Fall 006 Problem Set 3: Solutions. [Cover and Thomas 7.] a Define the following notation, C I p xx; Y max X; Y C I p xx; Ỹ max I X; Ỹ We would like to show that C
Διαβάστε περισσότεραCHAPTER 25 SOLVING EQUATIONS BY ITERATIVE METHODS
CHAPTER 5 SOLVING EQUATIONS BY ITERATIVE METHODS EXERCISE 104 Page 8 1. Find the positive root of the equation x + 3x 5 = 0, correct to 3 significant figures, using the method of bisection. Let f(x) =
Διαβάστε περισσότεραSolution Series 9. i=1 x i and i=1 x i.
Lecturer: Prof. Dr. Mete SONER Coordinator: Yilin WANG Solution Series 9 Q1. Let α, β >, the p.d.f. of a beta distribution with parameters α and β is { Γ(α+β) Γ(α)Γ(β) f(x α, β) xα 1 (1 x) β 1 for < x
Διαβάστε περισσότεραLecture 34 Bootstrap confidence intervals
Lecture 34 Bootstrap confidence intervals Confidence Intervals θ: an unknown parameter of interest We want to find limits θ and θ such that Gt = P nˆθ θ t If G 1 1 α is known, then P θ θ = P θ θ = 1 α
Διαβάστε περισσότεραDIRECT PRODUCT AND WREATH PRODUCT OF TRANSFORMATION SEMIGROUPS
GANIT J. Bangladesh Math. oc. IN 606-694) 0) -7 DIRECT PRODUCT AND WREATH PRODUCT OF TRANFORMATION EMIGROUP ubrata Majumdar, * Kalyan Kumar Dey and Mohd. Altab Hossain Department of Mathematics University
Διαβάστε περισσότεραApproximation of distance between locations on earth given by latitude and longitude
Approximation of distance between locations on earth given by latitude and longitude Jan Behrens 2012-12-31 In this paper we shall provide a method to approximate distances between two points on earth
Διαβάστε περισσότεραSCHOOL OF MATHEMATICAL SCIENCES G11LMA Linear Mathematics Examination Solutions
SCHOOL OF MATHEMATICAL SCIENCES GLMA Linear Mathematics 00- Examination Solutions. (a) i. ( + 5i)( i) = (6 + 5) + (5 )i = + i. Real part is, imaginary part is. (b) ii. + 5i i ( + 5i)( + i) = ( i)( + i)
Διαβάστε περισσότερα12. Radon-Nikodym Theorem
Tutorial 12: Radon-Nikodym Theorem 1 12. Radon-Nikodym Theorem In the following, (Ω, F) is an arbitrary measurable space. Definition 96 Let μ and ν be two (possibly complex) measures on (Ω, F). We say
Διαβάστε περισσότεραIntuitionistic Fuzzy Ideals of Near Rings
International Mathematical Forum, Vol. 7, 202, no. 6, 769-776 Intuitionistic Fuzzy Ideals of Near Rings P. K. Sharma P.G. Department of Mathematics D.A.V. College Jalandhar city, Punjab, India pksharma@davjalandhar.com
Διαβάστε περισσότεραSplitting stationary sets from weak forms of Choice
Splitting stationary sets from weak forms of Choice Paul Larson and Saharon Shelah February 16, 2008 Abstract Working in the context of restricted forms of the Axiom of Choice, we consider the problem
Διαβάστε περισσότεραThe challenges of non-stable predicates
The challenges of non-stable predicates Consider a non-stable predicate Φ encoding, say, a safety property. We want to determine whether Φ holds for our program. The challenges of non-stable predicates
Διαβάστε περισσότεραSCITECH Volume 13, Issue 2 RESEARCH ORGANISATION Published online: March 29, 2018
Journal of rogressive Research in Mathematics(JRM) ISSN: 2395-028 SCITECH Volume 3, Issue 2 RESEARCH ORGANISATION ublished online: March 29, 208 Journal of rogressive Research in Mathematics www.scitecresearch.com/journals
Διαβάστε περισσότεραTHE SECOND ISOMORPHISM THEOREM ON ORDERED SET UNDER ANTIORDERS. Daniel A. Romano
235 Kragujevac J. Math. 30 (2007) 235 242. THE SECOND ISOMORPHISM THEOREM ON ORDERED SET UNDER ANTIORDERS Daniel A. Romano Department of Mathematics and Informatics, Banja Luka University, Mladena Stojanovića
Διαβάστε περισσότεραHomomorphism of Intuitionistic Fuzzy Groups
International Mathematical Forum, Vol. 6, 20, no. 64, 369-378 Homomorphism o Intuitionistic Fuzz Groups P. K. Sharma Department o Mathematics, D..V. College Jalandhar Cit, Punjab, India pksharma@davjalandhar.com
Διαβάστε περισσότεραMath221: HW# 1 solutions
Math: HW# solutions Andy Royston October, 5 7.5.7, 3 rd Ed. We have a n = b n = a = fxdx = xdx =, x cos nxdx = x sin nx n sin nxdx n = cos nx n = n n, x sin nxdx = x cos nx n + cos nxdx n cos n = + sin
Διαβάστε περισσότεραMath 6 SL Probability Distributions Practice Test Mark Scheme
Math 6 SL Probability Distributions Practice Test Mark Scheme. (a) Note: Award A for vertical line to right of mean, A for shading to right of their vertical line. AA N (b) evidence of recognizing symmetry
Διαβάστε περισσότεραLecture 21: Properties and robustness of LSE
Lecture 21: Properties and robustness of LSE BLUE: Robustness of LSE against normality We now study properties of l τ β and σ 2 under assumption A2, i.e., without the normality assumption on ε. From Theorem
Διαβάστε περισσότεραKnaster-Reichbach Theorem for 2 κ
Knaster-Reichbach Theorem for 2 κ Micha l Korch February 9, 2018 In the recent years the theory of the generalized Cantor and Baire spaces was extensively developed (see, e.g. [1], [2], [6], [4] and many
Διαβάστε περισσότεραAbstract Storage Devices
Abstract Storage Devices Robert König Ueli Maurer Stefano Tessaro SOFSEM 2009 January 27, 2009 Outline 1. Motivation: Storage Devices 2. Abstract Storage Devices (ASD s) 3. Reducibility 4. Factoring ASD
Διαβάστε περισσότεραSpace-Time Symmetries
Chapter Space-Time Symmetries In classical fiel theory any continuous symmetry of the action generates a conserve current by Noether's proceure. If the Lagrangian is not invariant but only shifts by a
Διαβάστε περισσότεραSequent Calculi for the Modal µ-calculus over S5. Luca Alberucci, University of Berne. Logic Colloquium Berne, July 4th 2008
Sequent Calculi for the Modal µ-calculus over S5 Luca Alberucci, University of Berne Logic Colloquium Berne, July 4th 2008 Introduction Koz: Axiomatisation for the modal µ-calculus over K Axioms: All classical
Διαβάστε περισσότεραCompleteness Theorem for System AS1
11 The Completeness Theorem for System AS1 1. Introduction...2 2. Previous Definitions...2 3. Deductive Consistency...2 4. Maximal Consistent Sets...5 5. Lindenbaum s Lemma...6 6. Every Maximal Consistent
Διαβάστε περισσότεραTridiagonal matrices. Gérard MEURANT. October, 2008
Tridiagonal matrices Gérard MEURANT October, 2008 1 Similarity 2 Cholesy factorizations 3 Eigenvalues 4 Inverse Similarity Let α 1 ω 1 β 1 α 2 ω 2 T =......... β 2 α 1 ω 1 β 1 α and β i ω i, i = 1,...,
Διαβάστε περισσότεραΚΥΠΡΙΑΚΗ ΕΤΑΙΡΕΙΑ ΠΛΗΡΟΦΟΡΙΚΗΣ CYPRUS COMPUTER SOCIETY ΠΑΓΚΥΠΡΙΟΣ ΜΑΘΗΤΙΚΟΣ ΔΙΑΓΩΝΙΣΜΟΣ ΠΛΗΡΟΦΟΡΙΚΗΣ 6/5/2006
Οδηγίες: Να απαντηθούν όλες οι ερωτήσεις. Ολοι οι αριθμοί που αναφέρονται σε όλα τα ερωτήματα είναι μικρότεροι το 1000 εκτός αν ορίζεται διαφορετικά στη διατύπωση του προβλήματος. Διάρκεια: 3,5 ώρες Καλή
Διαβάστε περισσότεραInverse trigonometric functions & General Solution of Trigonometric Equations. ------------------ ----------------------------- -----------------
Inverse trigonometric functions & General Solution of Trigonometric Equations. 1. Sin ( ) = a) b) c) d) Ans b. Solution : Method 1. Ans a: 17 > 1 a) is rejected. w.k.t Sin ( sin ) = d is rejected. If sin
Διαβάστε περισσότεραk A = [k, k]( )[a 1, a 2 ] = [ka 1,ka 2 ] 4For the division of two intervals of confidence in R +
Chapter 3. Fuzzy Arithmetic 3- Fuzzy arithmetic: ~Addition(+) and subtraction (-): Let A = [a and B = [b, b in R If x [a and y [b, b than x+y [a +b +b Symbolically,we write A(+)B = [a (+)[b, b = [a +b
Διαβάστε περισσότεραNOTES ON SQUARES, SCALES, AND REFLECTION
NOTES ON SQUARES, SCALES, AND REFLECTION MAXWELL LEVINE 1. Some Basic Facts Proposition 1. If κ
Διαβάστε περισσότερα5.4 The Poisson Distribution.
The worst thing you can do about a situation is nothing. Sr. O Shea Jackson 5.4 The Poisson Distribution. Description of the Poisson Distribution Discrete probability distribution. The random variable
Διαβάστε περισσότεραΠΤΥΧΙΑΚΗ ΕΡΓΑΣΙΑ ΒΑΛΕΝΤΙΝΑ ΠΑΠΑΔΟΠΟΥΛΟΥ Α.Μ.: 09/061. Υπεύθυνος Καθηγητής: Σάββας Μακρίδης
Α.Τ.Ε.Ι. ΙΟΝΙΩΝ ΝΗΣΩΝ ΠΑΡΑΡΤΗΜΑ ΑΡΓΟΣΤΟΛΙΟΥ ΤΜΗΜΑ ΔΗΜΟΣΙΩΝ ΣΧΕΣΕΩΝ ΚΑΙ ΕΠΙΚΟΙΝΩΝΙΑΣ ΠΤΥΧΙΑΚΗ ΕΡΓΑΣΙΑ «Η διαμόρφωση επικοινωνιακής στρατηγικής (και των τακτικών ενεργειών) για την ενδυνάμωση της εταιρικής
Διαβάστε περισσότεραANSWERSHEET (TOPIC = DIFFERENTIAL CALCULUS) COLLECTION #2. h 0 h h 0 h h 0 ( ) g k = g 0 + g 1 + g g 2009 =?
Teko Classes IITJEE/AIEEE Maths by SUHAAG SIR, Bhopal, Ph (0755) 3 00 000 www.tekoclasses.com ANSWERSHEET (TOPIC DIFFERENTIAL CALCULUS) COLLECTION # Question Type A.Single Correct Type Q. (A) Sol least
Διαβάστε περισσότεραIterated trilinear fourier integrals with arbitrary symbols
Cornell University ICM 04, Satellite Conference in Harmonic Analysis, Chosun University, Gwangju, Korea August 6, 04 Motivation the Coifman-Meyer theorem with classical paraproduct(979) B(f, f )(x) :=
Διαβάστε περισσότεραΑπόκριση σε Μοναδιαία Ωστική Δύναμη (Unit Impulse) Απόκριση σε Δυνάμεις Αυθαίρετα Μεταβαλλόμενες με το Χρόνο. Απόστολος Σ.
Απόκριση σε Δυνάμεις Αυθαίρετα Μεταβαλλόμενες με το Χρόνο The time integral of a force is referred to as impulse, is determined by and is obtained from: Newton s 2 nd Law of motion states that the action
Διαβάστε περισσότερα5. Choice under Uncertainty
5. Choice under Uncertainty Daisuke Oyama Microeconomics I May 23, 2018 Formulations von Neumann-Morgenstern (1944/1947) X: Set of prizes Π: Set of probability distributions on X : Preference relation
Διαβάστε περισσότερα( y) Partial Differential Equations
Partial Dierential Equations Linear P.D.Es. contains no owers roducts o the deendent variables / an o its derivatives can occasionall be solved. Consider eamle ( ) a (sometimes written as a ) we can integrate
Διαβάστε περισσότεραGenerating Set of the Complete Semigroups of Binary Relations
Applied Mathematics 06 7 98-07 Published Online January 06 in SciRes http://wwwscirporg/journal/am http://dxdoiorg/036/am067009 Generating Set of the Complete Semigroups of Binary Relations Yasha iasamidze
Διαβάστε περισσότεραNON-NORMALITY POINTS OF βx \ X
NON-NORMALITY POINTS OF βx \ X LYNNE YENGULALP 1. INTRODUCTION A point p in a space X is called a non-normality point if X \ {p} is not normal and is called a butterfly point if there are closed subsets
Διαβάστε περισσότεραTMA4115 Matematikk 3
TMA4115 Matematikk 3 Andrew Stacey Norges Teknisk-Naturvitenskapelige Universitet Trondheim Spring 2010 Lecture 12: Mathematics Marvellous Matrices Andrew Stacey Norges Teknisk-Naturvitenskapelige Universitet
Διαβάστε περισσότεραChapter 6: Systems of Linear Differential. be continuous functions on the interval
Chapter 6: Systems of Linear Differential Equations Let a (t), a 2 (t),..., a nn (t), b (t), b 2 (t),..., b n (t) be continuous functions on the interval I. The system of n first-order differential equations
Διαβάστε περισσότεραSection 9.2 Polar Equations and Graphs
180 Section 9. Polar Equations and Graphs In this section, we will be graphing polar equations on a polar grid. In the first few examples, we will write the polar equation in rectangular form to help identify
Διαβάστε περισσότεραSOLUTIONS TO MATH38181 EXTREME VALUES AND FINANCIAL RISK EXAM
SOLUTIONS TO MATH38181 EXTREME VALUES AND FINANCIAL RISK EXAM Solutions to Question 1 a) The cumulative distribution function of T conditional on N n is Pr T t N n) Pr max X 1,..., X N ) t N n) Pr max
Διαβάστε περισσότερα6.1. Dirac Equation. Hamiltonian. Dirac Eq.
6.1. Dirac Equation Ref: M.Kaku, Quantum Field Theory, Oxford Univ Press (1993) η μν = η μν = diag(1, -1, -1, -1) p 0 = p 0 p = p i = -p i p μ p μ = p 0 p 0 + p i p i = E c 2 - p 2 = (m c) 2 H = c p 2
Διαβάστε περισσότεραSection 7.6 Double and Half Angle Formulas
09 Section 7. Double and Half Angle Fmulas To derive the double-angles fmulas, we will use the sum of two angles fmulas that we developed in the last section. We will let α θ and β θ: cos(θ) cos(θ + θ)
Διαβάστε περισσότεραPractice Exam 2. Conceptual Questions. 1. State a Basic identity and then verify it. (a) Identity: Solution: One identity is csc(θ) = 1
Conceptual Questions. State a Basic identity and then verify it. a) Identity: Solution: One identity is cscθ) = sinθ) Practice Exam b) Verification: Solution: Given the point of intersection x, y) of the
Διαβάστε περισσότεραRight Rear Door. Let's now finish the door hinge saga with the right rear door
Right Rear Door Let's now finish the door hinge saga with the right rear door You may have been already guessed my steps, so there is not much to describe in detail. Old upper one file:///c /Documents
Διαβάστε περισσότεραDynamic types, Lambda calculus machines Section and Practice Problems Apr 21 22, 2016
Harvard School of Engineering and Applied Sciences CS 152: Programming Languages Dynamic types, Lambda calculus machines Apr 21 22, 2016 1 Dynamic types and contracts (a) To make sure you understand the
Διαβάστε περισσότεραChapter 2. Ordinals, well-founded relations.
Chapter 2. Ordinals, well-founded relations. 2.1. Well-founded Relations. We start with some definitions and rapidly reach the notion of a well-ordered set. Definition. For any X and any binary relation
Διαβάστε περισσότερα6.3 Forecasting ARMA processes
122 CHAPTER 6. ARMA MODELS 6.3 Forecasting ARMA processes The purpose of forecasting is to predict future values of a TS based on the data collected to the present. In this section we will discuss a linear
Διαβάστε περισσότεραHomework 8 Model Solution Section
MATH 004 Homework Solution Homework 8 Model Solution Section 14.5 14.6. 14.5. Use the Chain Rule to find dz where z cosx + 4y), x 5t 4, y 1 t. dz dx + dy y sinx + 4y)0t + 4) sinx + 4y) 1t ) 0t + 4t ) sinx
Διαβάστε περισσότεραSolutions to Exercise Sheet 5
Solutions to Eercise Sheet 5 jacques@ucsd.edu. Let X and Y be random variables with joint pdf f(, y) = 3y( + y) where and y. Determine each of the following probabilities. Solutions. a. P (X ). b. P (X
Διαβάστε περισσότεραOverview. Transition Semantics. Configurations and the transition relation. Executions and computation
Overview Transition Semantics Configurations and the transition relation Executions and computation Inference rules for small-step structural operational semantics for the simple imperative language Transition
Διαβάστε περισσότερα