CSL105: Discrete Mathematical Structures. Ragesh Jaiswal, CSE, IIT Delhi
|
|
- Madlyn Summers
- 5 years ago
- Views:
Transcription
1
2
3 Definition (Linear homogeneous recurrence) A linear homogeneous recurrence relation of degree k with constant coefficients is a recurrence relation of the form a n = c 1 a n 1 + c 2 a n c k a n k, where c 1, c 2,..., c k are real numbers, and c k 0. Linear means that that RHS is a sum of linear terms of the previous elements of the sequence. a n = a n 1 + a n 2 is a linear recurrence relation whereas a n = a n 1 + an 2 2 is not.
4 Definition (Linear homogeneous recurrence) A linear homogeneous recurrence relation of degree k with constant coefficients is a recurrence relation of the form a n = c 1 a n 1 + c 2 a n c k a n k, where c 1, c 2,..., c k are real numbers, and c k 0. Linear means that that RHS is a sum of linear terms of the previous elements of the sequence. Homogeneous means that there are no terms in the RHS that are not multiples of a j s. a n = a n 1 + a n 2 is homogeneous whereas a n = a n 1 + a n is not.
5 Definition (Linear homogeneous recurrence) A linear homogeneous recurrence relation of degree k with constant coefficients is a recurrence relation of the form a n = c 1 a n 1 + c 2 a n c k a n k, where c 1, c 2,..., c k are real numbers, and c k 0. Linear means that that RHS is a sum of linear terms of the previous elements of the sequence. Homogeneous means that there are no terms in the RHS that are not multiples of a j s. The coefficients of all the terms on the RHS are constants. The degree is k since a n is expressed as the previous k terms of the sequence.
6 Definition (Linear homogeneous recurrence) A linear homogeneous recurrence relation of degree k with constant coefficients is a recurrence relation of the form a n = c 1 a n 1 + c 2 a n c k a n k, where c 1, c 2,..., c k are real numbers, and c k 0. a n = r n is a solution of the recurrence if and only if r k c 1 r k 1... c k = 0. (1) (1) is called the characteristic equation of the recurrence relation. The solutions of the characteristic equation are called the characteristic roots of the recurrence relation.
7 Let c 1 and c 2 be real numbers. Suppose r 2 c 1 r c 2 = 0 has two distinct roots r 1 and r 2. Then the sequence {a n } is a solution of the linear homogeneous recurrence relation a n = c 1 a n 1 + c 2 a n 2 if and only if a n = α 1 r n 1 + α 2r n 2 for all n = 0, 1, 2,..., where α 1 and α 2 are constants.
8 Let c 1 and c 2 be real numbers. Suppose r 2 c 1 r c 2 = 0 has two distinct roots r 1 and r 2. Then the sequence {a n } is a solution of the linear homogeneous recurrence relation a n = c 1 a n 1 + c 2 a n 2 if and only if a n = α 1 r n 1 + α 2r n 2 for all n = 0, 1, 2,..., where α 1 and α 2 are constants. What is the solution of the recurrence relation a n = a n a n 2 with a 0 = 2 and a 1 = 7?
9 Let c 1 and c 2 be real numbers. Suppose r 2 c 1 r c 2 = 0 has two distinct roots r 1 and r 2. Then the sequence {a n } is a solution of the linear homogeneous recurrence relation a n = c 1 a n 1 + c 2 a n 2 if and only if a n = α 1 r n 1 + α 2r n 2 for all n = 0, 1, 2,..., where α 1 and α 2 are constants. Let c 1 and c 2 be real numbers with c 2 0. Suppose that r 2 c 1 r c 2 = 0 has only one root r 0. A sequence {a n } is a solution of the recurrence relation a n = c 1 a n 1 + c 2 a n 2 if and only if a n = α 1 r n 0 + α 2nr n 0, for n = 0, 1, 2,..., where α 1 and α 2 are constants. What is the solution of the recurrence relation a n = 6a n 1 9 a n 2 with a 0 = 1 and a 1 = 6?
10 Let c 1, c 2,..., c k be real numbers. Consider the linear homogeneous recurrence relation a n = c 1 a n 1 + c 2 a n c k a n k. Suppose the characteristic equation of the recurrence relation has k distinct characteristic roots r 1, r 2,..., r k. Then {a n } is a solution of the recurrence relation if and only if a n = α 1 r n 1 + α 2r n α kr n k for n = 0, 1, 2,..., where α 1, α 2,..., α k are constants. What is the solution of the recurrence relation a n = 6a n 1 11 a n 2 + 6a n 3 with a 0 = 2, a 1 = 5, and a 2 = 15?
11 Let c 1, c 2,..., c k be real numbers. Consider the linear homogeneous recurrence relation a n = c 1 a n 1 + c 2 a n c k a n k. Suppose the characteristic equation of the recurrence relation has t k distinct characteristic roots r 1, r 2,..., r t with multiplicities m 1, m 2,..., m t, respectively, so that m i 1 for i = 1, 2,..., t and m 1 + m m t = k. Then {a n } is a solution of the recurrence relation if and only if a n = (α 1,0 + α 1,1 n α 1,m1 1n m 1 1 )r n 1 +(α 2,0 + α 2,1 n α 2,m2 1n m 2 1 )r n (α t,0 + α t,1 n α t,mt 1n mt 1 )r n t for n = 0, 1, 2,..., where α i,j are constants for 1 i t and 0 j m i 1. What is the solution of the recurrence relation a n = 3a n 1 3 a n 2 a n 3 with a 0 = 1, a 1 = 2, and a 2 = 1?
12 A linear non-homogeneous recurrence relation with constant coefficients is a recurrence of the form: a n = c 1 a n 1 + c 2 a n c k a n k + F (n), where F (n) is a function not identically equal to zero and depending only on n. The recurrence relation a n = c 1 a n 1 + c 2 a n c k a n k is called the associated homogeneous recurrence relation. If {a n (p) } is a particular solution of the non-homogeneous linear recurrence relation with constant coefficients a n = c 1 a n 1 + c 2 a n c k a n k + F (n), then every solution is of the form {a n (p) + a n (h) }, where {a n (h) } is a solution of the associated homogeneous recurrence relation a n = c 1 a n 1 + c 2 a n c k a n k.
13 If {a n (p) } is a particular solution of the non-homogeneous linear recurrence relation with constant coefficients a n = c 1 a n 1 + c 2 a n c k a n k + F (n), then every solution is of the form {a n (p) + a n (h) }, where {a n (h) } is a solution of the associated homogeneous recurrence relation a n = c 1 a n 1 + c 2 a n c k a n k. Find all solutions of the recurrence relation a n = 3a n 1 + 2n. What is the solution with a 1 = 3?
14 If {a n (p) } is a particular solution of the non-homogeneous linear recurrence relation with constant coefficients a n = c 1 a n 1 + c 2 a n c k a n k + F (n), then every solution is of the form {a n (p) + a n (h) }, where {a n (h) } is a solution of the associated homogeneous recurrence relation a n = c 1 a n 1 + c 2 a n c k a n k. Find all solutions of the recurrence relation a n = 3a n 1 + 2n. What is the solution with a 1 = 3? Findall solutions if the recurrence relation a n = 5a n 1 6a n n.
15 Suppose {a n } satisfies the linear non-homogeneous recurrence relation a n = c 1 a n 1 + c 2 a n c k a n k + F (n), where c 1, c 2,..., c k are real numbers, and F (n) = (b t n t + b t 1 n t b 1 n + b 0 )s n, where b 0, b 1,..., b t are s real numbers. When s is not a root of the characteristic equation of the associated linear homogeneous recurrence relation, there is a particular solution of the form (p t n t + p t 1 n t p 1 n + p 0 )s n. When s is a root of this characteristic equation and its multiplicity is m, there is a particular solution of the form n m (p t n t + p t 1 n t p 1 n + p 0 )s n.
16 End
Consider an infinite row of dominoes, labeled by 1, 2, 3,, where each domino is standing up. What should one do to knock over all dominoes?
1 Section 4.1 Mathematical Induction Consider an infinite row of dominoes, labeled by 1,, 3,, where each domino is standing up. What should one do to knock over all dominoes? Principle of Mathematical
More informationSolving Recurrence Relations 1. Guess and Math Induction Example: Find the solution for a n = 2a n 1 + 1, a 0 = 0 We can try finding each a n : a 0 =
Solving Recurrence Relations 1. Guess and Math Induction Example: Find the solution for a n = 2a n 1 + 1, a 0 = 0 We can try finding each a n : a 0 = 0 a 1 = 2 0 + 1 = 1 a 2 = 2 1 + 1 = 3 a 3 = 2 3 + 1
More informationRecurrence Relations
Recurrence Relations Winter 2017 Recurrence Relations Recurrence Relations A recurrence relation for the sequence {a n } is an equation that expresses a n in terms of one or more of the previous terms
More informationRecursion. Slides by Christopher M. Bourke Instructor: Berthe Y. Choueiry. Fall 2007
Slides by Christopher M. Bourke Instructor: Berthe Y. Choueiry Fall 2007 1 / 47 Computer Science & Engineering 235 to Discrete Mathematics Sections 7.1-7.2 of Rosen Recursive Algorithms 2 / 47 A recursive
More informationTutorial 2 WANG PENG. Department of Systems Engineering and Engineering Management, The Chinese University of Hong Kong. September 28, 2017
Tutorial 2 WANG PENG Department of Systems Engineering and Engineering Management, The Chinese University of Hong Kong September 28, 2017 WANG PENG (ENGG 2440B) Tutorial 2 September 28, 2017 1 / 17 Outline
More informationSection 5.2 Solving Recurrence Relations
Section 5.2 Solving Recurrence Relations If a g(n) = f (a g(0),a g(1),..., a g(n 1) ) find a closed form or an expression for a g(n). Recall: nth degree polynomials have n roots: a n x n + a n 1 x n 1
More informationLogic and Discrete Mathematics. Section 6.7 Recurrence Relations and Their Solution
Logic and Discrete Mathematics Section 6.7 Recurrence Relations and Their Solution Slides version: January 2015 Definition A recurrence relation for a sequence a 0, a 1, a 2,... is a formula giving a n
More informationMarkov Chains. X(t) is a Markov Process if, for arbitrary times t 1 < t 2 <... < t k < t k+1. If X(t) is discrete-valued. If X(t) is continuous-valued
Markov Chains X(t) is a Markov Process if, for arbitrary times t 1 < t 2
More informationNotes for Recitation 14
6.04/18.06J Mathematics for Computer Science October 7, 006 Tom Leighton and Ronitt Rubinfeld Notes for Recitation 14 Guessing a Particular Solution A general linear recurrence has the form: fn) = b 1
More informationSection 7.2 Solving Linear Recurrence Relations
Section 7.2 Solving Linear Recurrence Relations If a g(n) = f (a g(0),a g(1),..., a g(n 1) ) find a closed form or an expression for a g(n). Recall: nth degree polynomials have n roots: a n x n + a n 1
More informationRecurrence Relations. fib(0) = 1 fib(1) = 1 q(0) = 1
Recurrence Relations Reading: Gossett Sections 7.1 and 7.. Definition: Aone-wayinfinite sequence is a function from the natural numbers to some other set. E.g. fib(0) = 1, fib(1) = 1, fib() =,fib(3) =
More informationDefn: A linear recurrence relation has constant coefficients if the a i s are constant. Ch 14: 1, 4, 5, 10, 13, 18, 22, 24, 25 and
7.4: linear homogeneous recurrence relation: Defn: A recurrence relation is linear if h n = a 1 (n)h n 1 + a (n)h n +... + a k (n)h n k + b(n) A recurrence relation has order k if a k 0 Ex: Derangement
More informationReview for Exam 2. Review for Exam 2.
Review for Exam 2. 5 or 6 problems. No multiple choice questions. No notes, no books, no calculators. Problems similar to homeworks. Exam covers: Regular-singular points (5.5). Euler differential equation
More informationLinear Recurrence Relations
Linear Recurrence Relations Linear Homogeneous Recurrence Relations The Towers of Hanoi According to legend, there is a temple in Hanoi with three posts and 64 gold disks of different sizes. Each disk
More informationSystem of Linear Equations
Chapter 7 - S&B Gaussian and Gauss-Jordan Elimination We will study systems of linear equations by describing techniques for solving such systems. The preferred solution technique- Gaussian elimination-
More informationOn second order non-homogeneous recurrence relation
Annales Mathematicae et Informaticae 41 (2013) pp. 20 210 Proceedings of the 1 th International Conference on Fibonacci Numbers and Their Applications Institute of Mathematics and Informatics, Eszterházy
More information1 Series Solutions Near Regular Singular Points
1 Series Solutions Near Regular Singular Points All of the work here will be directed toward finding series solutions of a second order linear homogeneous ordinary differential equation: P xy + Qxy + Rxy
More informationF. T. HOWARD AND CURTIS COOPER
SOME IDENTITIES FOR r-fibonacci NUMBERS F. T. HOWARD AND CURTIS COOPER Abstract. Let r 1 be an integer. The r-generalized Fibonacci sequence {G n} is defined as 0, if 0 n < r 1; G n = 1, if n = r 1; G
More informationMATH 243E Test #3 Solutions
MATH 4E Test # Solutions () Find a recurrence relation for the number of bit strings of length n that contain a pair of consecutive 0s. You do not need to solve this recurrence relation. (Hint: Consider
More informationJong C. Park Computer Science Division, KAIST
Jong C. Park Computer Science Division, KAIST Today s Topics Introduction Solving Recurrence Relations Discrete Mathematics, 2008 2 Computer Science Division, KAIST Definition A recurrence relation for
More informationCS 2210 Discrete Structures Advanced Counting. Fall 2017 Sukumar Ghosh
CS 2210 Discrete Structures Advanced Counting Fall 2017 Sukumar Ghosh Compound Interest A person deposits $10,000 in a savings account that yields 10% interest annually. How much will be there in the account
More informationBessel s and legendre s equations
Chapter 12 Bessel s and legendre s equations 12.1 Introduction Many linear differential equations having variable coefficients cannot be solved by usual methods and we need to employ series solution method
More informationPower Series Solutions We use power series to solve second order differential equations
Objectives Power Series Solutions We use power series to solve second order differential equations We use power series expansions to find solutions to second order, linear, variable coefficient equations
More informationAlgorithms Test 1. Question 1. (10 points) for (i = 1; i <= n; i++) { j = 1; while (j < n) {
Question 1. (10 points) for (i = 1; i
More informationTaylor Series Method for Second-Order Polynomial Differential Equations
Taylor Series Method for Second-Order Polynomial Differential Equations Viktor N. Latypov Sergey V. Sokolov Saint-Petersburg State University SCP-2015 Introduction New numerical algorithm to solve second-order
More informationContinuous-Time Markov Chain
Continuous-Time Markov Chain Consider the process {X(t),t 0} with state space {0, 1, 2,...}. The process {X(t),t 0} is a continuous-time Markov chain if for all s, t 0 and nonnegative integers i, j, x(u),
More informationSeries Solutions Near a Regular Singular Point
Series Solutions Near a Regular Singular Point MATH 365 Ordinary Differential Equations J. Robert Buchanan Department of Mathematics Fall 2018 Background We will find a power series solution to the equation:
More informationP i [B k ] = lim. n=1 p(n) ii <. n=1. V i :=
2.7. Recurrence and transience Consider a Markov chain {X n : n N 0 } on state space E with transition matrix P. Definition 2.7.1. A state i E is called recurrent if P i [X n = i for infinitely many n]
More informationLearning Objectives
Learning Objectives Learn about recurrence relations Learn the relationship between sequences and recurrence relations Explore how to solve recurrence relations by iteration Learn about linear homogeneous
More informationOutlines. Discrete Time Markov Chain (DTMC) Continuous Time Markov Chain (CTMC)
Markov Chains (2) Outlines Discrete Time Markov Chain (DTMC) Continuous Time Markov Chain (CTMC) 2 pj ( n) denotes the pmf of the random variable p ( n) P( X j) j We will only be concerned with homogenous
More informationMore about partitions
Partitions 2.4, 3.4, 4.4 02 More about partitions 3 + +, + 3 +, and + + 3 are all the same partition, so we will write the numbers in non-increasing order. We use greek letters to denote partitions, often
More informationTrigonometric Pseudo Fibonacci Sequence
Notes on Number Theory and Discrete Mathematics ISSN 30 532 Vol. 2, 205, No. 3, 70 76 Trigonometric Pseudo Fibonacci Sequence C. N. Phadte and S. P. Pethe 2 Department of Mathematics, Goa University Taleigao
More informationRecurrence Relations
Recurrence Relations Recurrence Relations Reading (Epp s textbook) 5.6 5.8 1 Recurrence Relations A recurrence relation for a sequence aa 0, aa 1, aa 2, ({a n }) is a formula that relates each term a k
More information8. Statistical Equilibrium and Classification of States: Discrete Time Markov Chains
8. Statistical Equilibrium and Classification of States: Discrete Time Markov Chains 8.1 Review 8.2 Statistical Equilibrium 8.3 Two-State Markov Chain 8.4 Existence of P ( ) 8.5 Classification of States
More informationAdvanced Counting Techniques
. All rights reserved. Authorized only for instructor use in the classroom. No reproduction or further distribution permitted without the prior written consent of McGraw-Hill Education. Advanced Counting
More informationMaximum exponent of boolean circulant matrices with constant number of nonzero entries in its generating vector
Maximum exponent of boolean circulant matrices with constant number of nonzero entries in its generating vector MI Bueno, Department of Mathematics and The College of Creative Studies University of California,
More information= main diagonal, in the order in which their corresponding eigenvectors appear as columns of E.
3.3 Diagonalization Let A = 4. Then and are eigenvectors of A, with corresponding eigenvalues 2 and 6 respectively (check). This means 4 = 2, 4 = 6. 2 2 2 2 Thus 4 = 2 2 6 2 = 2 6 4 2 We have 4 = 2 0 0
More informationRecurrences COMP 215
Recurrences COMP 215 Analysis of Iterative Algorithms //return the location of the item matching x, or 0 if //no such item is found. index SequentialSearch(keytype[] S, in, keytype x) { index location
More informationChapter 9. Perturbations in the Universe
Chapter 9 Perturbations in the Universe In this chapter the theory of linear perturbations in the universe are studied. 9.1 Differential Equations of Linear Perturbation in the Universe A covariant, linear,
More informationSTOCHASTIC PROCESSES Basic notions
J. Virtamo 38.3143 Queueing Theory / Stochastic processes 1 STOCHASTIC PROCESSES Basic notions Often the systems we consider evolve in time and we are interested in their dynamic behaviour, usually involving
More information17 Advancement Operator Equations
November 14, 2017 17 Advancement Operator Equations William T. Trotter trotter@math.gatech.edu Review of Recurrence Equations (1) Problem Let r(n) denote the number of regions determined by n lines that
More informationThe Combinatorialization of Linear Recurrences
The Combinatorialization of Linear Recurrences Arthur T Benjamin Harvey Mudd College Claremont, CA, USA benjamin@hmcedu Jennifer J Quinn University of Washington Tacoma Tacoma, WA USA jjquinn@uwedu Halcyon
More information12d. Regular Singular Points
October 22, 2012 12d-1 12d. Regular Singular Points We have studied solutions to the linear second order differential equations of the form P (x)y + Q(x)y + R(x)y = 0 (1) in the cases with P, Q, R real
More informationRecurrence Relations
Recurrence Relations Simplest Case of General Solutions Ioan Despi despi@turing.une.edu.au University of New England September 13, 2013 Outline Ioan Despi AMTH140 2 of 1 Simplest Case of General Solutions
More informationRecursion: Introduction and Correctness
Recursion: Introduction and Correctness CSE21 Winter 2017, Day 7 (B00), Day 4-5 (A00) January 25, 2017 http://vlsicad.ucsd.edu/courses/cse21-w17 Today s Plan From last time: intersecting sorted lists and
More information1. Introduction Definition 1.1. Let r 1 be an integer. The r-generalized Fibonacci sequence {G n } is defined as
SOME IDENTITIES FOR r-fibonacci NUMBERS F. T. HOWARD AND CURTIS COOPER Abstract. Let r 1 be an integer. The r-generalized Fibonacci sequence {G n} is defined as 8 >< 0, if 0 n < r 1; G n = 1, if n = r
More informationDiscrete Mathematics -- Chapter 10: Recurrence Relations
Discrete Mathematics -- Chapter 10: Recurrence Relations Hung-Yu Kao ( 高宏宇 ) Department of Computer Science and Information Engineering, National Cheng Kung University First glance at recurrence F n+2
More informationLECTURE 12: SOLUTIONS TO SIMULTANEOUS LINEAR EQUATIONS. Prof. N. Harnew University of Oxford MT 2012
LECTURE 12: SOLUTIONS TO SIMULTANEOUS LINEAR EQUATIONS Prof. N. Harnew University of Oxford MT 2012 1 Outline: 12. SOLUTIONS TO SIMULTANEOUS LINEAR EQUATIONS 12.1 Methods used to solve for unique solution
More informationUNIT 1 DETERMINANTS 1.0 INTRODUCTION 1.1 OBJECTIVES. Structure
UNIT 1 DETERMINANTS Determinants Structure 1.0 Introduction 1.1 Objectives 1.2 Determinants of Order 2 and 3 1.3 Determinants of Order 3 1.4 Properties of Determinants 1.5 Application of Determinants 1.6
More informationElementary Differential Equations, Section 2 Prof. Loftin: Practice Test Problems for Test Find the radius of convergence of the power series
Elementary Differential Equations, Section 2 Prof. Loftin: Practice Test Problems for Test 2 SOLUTIONS 1. Find the radius of convergence of the power series Show your work. x + x2 2 + x3 3 + x4 4 + + xn
More information+ + ( + ) = Linear recurrent networks. Simpler, much more amenable to analytic treatment E.g. by choosing
Linear recurrent networks Simpler, much more amenable to analytic treatment E.g. by choosing + ( + ) = Firing rates can be negative Approximates dynamics around fixed point Approximation often reasonable
More informationNote Set 5: Hidden Markov Models
Note Set 5: Hidden Markov Models Probabilistic Learning: Theory and Algorithms, CS 274A, Winter 2016 1 Hidden Markov Models (HMMs) 1.1 Introduction Consider observed data vectors x t that are d-dimensional
More informationTribhuvan University Institute of Science and Technology Micro Syllabus
Tribhuvan University Institute of Science and Technology Micro Syllabus Course Title: Discrete Structure Course no: CSC-152 Full Marks: 80+20 Credit hours: 3 Pass Marks: 32+8 Nature of course: Theory (3
More informationCalifornia State University Northridge MATH 280: Applied Differential Equations Midterm Exam 3
California State University Northridge MATH 280: Applied Differential Equations Midterm Exam 3 April 29, 2013. Duration: 75 Minutes. Instructor: Jing Li Student Name: Signature: Do not write your student
More informationAPPENDIX B GRAM-SCHMIDT PROCEDURE OF ORTHOGONALIZATION. Let V be a finite dimensional inner product space spanned by basis vector functions
301 APPENDIX B GRAM-SCHMIDT PROCEDURE OF ORTHOGONALIZATION Let V be a finite dimensional inner product space spanned by basis vector functions {w 1, w 2,, w n }. According to the Gram-Schmidt Process an
More informationAnalysis of Algorithms Generating Functions II. Holger Findling
Generating Functions II Holger Findling Lecture notes are obtained from Fundamentals of Algorithmics, Gilles Brassard and Paul Bratley. Inhomogeneous Recurrences. The solution of a linear recurrence with
More informationEnumerating Binary Strings
International Mathematical Forum, Vol. 7, 2012, no. 38, 1865-1876 Enumerating Binary Strings without r-runs of Ones M. A. Nyblom School of Mathematics and Geospatial Science RMIT University Melbourne,
More informationOn the asymptotic dynamics of 2D positive systems
On the asymptotic dynamics of 2D positive systems Ettore Fornasini and Maria Elena Valcher Dipartimento di Elettronica ed Informatica, Univ. di Padova via Gradenigo 6a, 353 Padova, ITALY e-mail: meme@paola.dei.unipd.it
More informationHandout 7: Recurrences (Cont d)
ENGG 2440B: Discrete Mathematics for Engineers Handout 7: Recurrences (Cont d) 2018 19 First Term Instructor: Anthony Man Cho So October 8, 2018 In the last handout, we studied techniques for solving linear
More informationAM 121: Intro to Optimization Models and Methods Fall 2018
AM 121: Intro to Optimization Models and Methods Fall 2018 Lecture 5: The Simplex Method Yiling Chen Harvard SEAS Lesson Plan This lecture: Moving towards an algorithm for solving LPs Tableau. Adjacent
More informationAnalysis of Markov Reward Models with Partial Reward Loss Based on a Time Reverse Approach
Analysis of Markov Reward Models with Partial Reward Loss Based on a Time Reverse Approach Gábor Horváth, Miklós Telek Technical University of Budapest, 1521 Budapest, Hungary {hgabor,telek}@webspn.hit.bme.hu
More informationThe Method of Frobenius
The Method of Frobenius Department of Mathematics IIT Guwahati If either p(x) or q(x) in y + p(x)y + q(x)y = 0 is not analytic near x 0, power series solutions valid near x 0 may or may not exist. If either
More informationBALANCING NUMBERS : SOME IDENTITIES KABERI PARIDA. Master of Science in Mathematics. Dr. GOPAL KRISHNA PANDA
BALANCING NUMBERS : SOME IDENTITIES A report submitted by KABERI PARIDA Roll No: 1MA073 for the partial fulfilment for the award of the degree of Master of Science in Mathematics under the supervision
More informationCounting Palindromic Binary Strings Without r-runs of Ones
1 3 47 6 3 11 Journal of Integer Sequences, Vol. 16 (013), Article 13.8.7 Counting Palindromic Binary Strings Without r-runs of Ones M. A. Nyblom School of Mathematics and Geospatial Science RMIT University
More informationEquations with regular-singular points (Sect. 5.5).
Equations with regular-singular points (Sect. 5.5). Equations with regular-singular points. s: Equations with regular-singular points. Method to find solutions. : Method to find solutions. Recall: The
More informationChapter 1 Numerical approximation of data : interpolation, least squares method
Chapter 1 Numerical approximation of data : interpolation, least squares method I. Motivation 1 Approximation of functions Evaluation of a function Which functions (f : R R) can be effectively evaluated
More informationSection Notes 9. Midterm 2 Review. Applied Math / Engineering Sciences 121. Week of December 3, 2018
Section Notes 9 Midterm 2 Review Applied Math / Engineering Sciences 121 Week of December 3, 2018 The following list of topics is an overview of the material that was covered in the lectures and sections
More informationME 3560 Fluid Mechanics
ME 3560 Fluid Mechanics 1 7.1 Dimensional Analysis Many problems of interest in fluid mechanics cannot be solved using the integral and/or differential equations. Examples of problems that are studied
More informationRemark By definition, an eigenvector must be a nonzero vector, but eigenvalue could be zero.
Sec 6 Eigenvalues and Eigenvectors Definition An eigenvector of an n n matrix A is a nonzero vector x such that A x λ x for some scalar λ A scalar λ is called an eigenvalue of A if there is a nontrivial
More informationTHE HALL EFFECT. Theory
THE HALL EFFECT Theory For a charge moving in a magnetic field, it experiences a force acting at right angles to both its direction and the direction of the magnetic field H. Hence in a semiconductor,
More informationThe Theory of Linear Homogeneous (Constant Coefficient) Recurrences
The Theory of Linear Homogeneous (Constant Coefficient) Recurrences by Evan Houston The paper below was prepared by Evan Houston for his Combinatorics classes several years ago. The mathematics is due
More informationPrepared by: M. S. KumarSwamy, TGT(Maths) Page
Prepared by: M. S. KumarSwamy, TGT(Maths) Page - 50 - CHAPTER 3: MATRICES QUICK REVISION (Important Concepts & Formulae) MARKS WEIGHTAGE 03 marks Matrix A matrix is an ordered rectangular array of numbers
More informationCOSMIC INFLATION AND THE REHEATING OF THE UNIVERSE
COSMIC INFLATION AND THE REHEATING OF THE UNIVERSE Francisco Torrentí - IFT/UAM Valencia Students Seminars - December 2014 Contents 1. The Friedmann equations 2. Inflation 2.1. The problems of hot Big
More informationMEETING 9 - INDUCTION AND RECURSION
MEETING 9 - INDUCTION AND RECURSION We do some initial Peer Instruction... Predicates Before we get into mathematical induction we will repeat the concept of a predicate. A predicate is a mathematical
More informationPower Series and Analytic Function
Dr Mansoor Alshehri King Saud University MATH204-Differential Equations Center of Excellence in Learning and Teaching 1 / 21 Some Reviews of Power Series Differentiation and Integration of a Power Series
More informationScaling Theory. Roger Herrigel Advisor: Helmut Katzgraber
Scaling Theory Roger Herrigel Advisor: Helmut Katzgraber 7.4.2007 Outline The scaling hypothesis Critical exponents The scaling hypothesis Derivation of the scaling relations Heuristic explanation Kadanoff
More informationConvergence rates in generalized Zeckendorf decomposition problems
Convergence rates in generalized Zeckendorf decomposition problems Ray Li (ryli@andrew.cmu.edu) 1 Steven J Miller (sjm1@williams.edu) 2 Zhao Pan (zhaop@andrew.cmu.edu) 3 Huanzhong Xu (huanzhox@andrew.cmu.edu)
More informationAlgorithms and Data Structures Strassen s Algorithm. ADS (2017/18) Lecture 4 slide 1
Algorithms and Data Structures Strassen s Algorithm ADS (2017/18) Lecture 4 slide 1 Tutorials Start in week (week 3) Tutorial allocations are linked from the course webpage http://www.inf.ed.ac.uk/teaching/courses/ads/
More informationFinite Math - J-term Section Systems of Linear Equations in Two Variables Example 1. Solve the system
Finite Math - J-term 07 Lecture Notes - //07 Homework Section 4. - 9, 0, 5, 6, 9, 0,, 4, 6, 0, 50, 5, 54, 55, 56, 6, 65 Section 4. - Systems of Linear Equations in Two Variables Example. Solve the system
More informationPAPER 71 COSMOLOGY. Attempt THREE questions There are seven questions in total The questions carry equal weight
MATHEMATICAL TRIPOS Part III Friday 31 May 00 9 to 1 PAPER 71 COSMOLOGY Attempt THREE questions There are seven questions in total The questions carry equal weight You may make free use of the information
More informationMATH 564/STAT 555 Applied Stochastic Processes Homework 2, September 18, 2015 Due September 30, 2015
ID NAME SCORE MATH 56/STAT 555 Applied Stochastic Processes Homework 2, September 8, 205 Due September 30, 205 The generating function of a sequence a n n 0 is defined as As : a ns n for all s 0 for which
More informationRank, Homogenous Systems, Linear Independence
Sept 11,1 Rank, Homogenous Systems, Linear Independence If (A B) is a linear system represented as an augmented matrix then A is called the COEFFICIENT MATRIX and B is (usually) called the RIGHT HAND SIDE
More information= P{X 0. = i} (1) If the MC has stationary transition probabilities then, = i} = P{X n+1
Properties of Markov Chains and Evaluation of Steady State Transition Matrix P ss V. Krishnan - 3/9/2 Property 1 Let X be a Markov Chain (MC) where X {X n : n, 1, }. The state space is E {i, j, k, }. The
More informationMomentum Equation-Derivation-1 Consider the 2D continuity equation and integrate over the height of the boundary layer
Module 2 : Convection Lecture 25 : Integral Boundary Layer Equations Objectives In this class: Integral equations for the momentum and thermal boundary layers are obtained Solution of these equations for
More informationORIE 6300 Mathematical Programming I August 25, Recitation 1
ORIE 6300 Mathematical Programming I August 25, 2016 Lecturer: Calvin Wylie Recitation 1 Scribe: Mateo Díaz 1 Linear Algebra Review 1 1.1 Independence, Spanning, and Dimension Definition 1 A (usually infinite)
More informationEXPLICIT EXPRESSIONS FOR MOMENTS OF χ 2 ORDER STATISTICS
Bulletin of the Institute of Mathematics Academia Sinica (New Series) Vol. 3 (28), No. 3, pp. 433-444 EXPLICIT EXPRESSIONS FOR MOMENTS OF χ 2 ORDER STATISTICS BY SARALEES NADARAJAH Abstract Explicit closed
More informationDEPARTMENT OF MATHEMATICS
DEPARTMENT OF MATHEMATICS Ma322 - Final Exam Spring 2011 May 3,4, 2011 DO NOT TURN THIS PAGE UNTIL YOU ARE INSTRUCTED TO DO SO. Be sure to show all work and justify your answers. There are 8 problems and
More informationStochastic Learning in a Neural Network with Adapting. Synapses. Istituto Nazionale di Fisica Nucleare, Sezione di Bari
Stochastic Learning in a Neural Network with Adapting Synapses. G. Lattanzi 1, G. Nardulli 1, G. Pasquariello and S. Stramaglia 1 Dipartimento di Fisica dell'universita di Bari and Istituto Nazionale di
More informationMSc MT15. Further Statistical Methods: MCMC. Lecture 5-6: Markov chains; Metropolis Hastings MCMC. Notes and Practicals available at
MSc MT15. Further Statistical Methods: MCMC Lecture 5-6: Markov chains; Metropolis Hastings MCMC Notes and Practicals available at www.stats.ox.ac.uk\ nicholls\mscmcmc15 Markov chain Monte Carlo Methods
More informationMaximum sum contiguous subsequence Longest common subsequence Matrix chain multiplication All pair shortest path Kna. Dynamic Programming
Dynamic Programming Arijit Bishnu arijit@isical.ac.in Indian Statistical Institute, India. August 31, 2015 Outline 1 Maximum sum contiguous subsequence 2 Longest common subsequence 3 Matrix chain multiplication
More informationCS 5321: Advanced Algorithms - Recurrence. Acknowledgement. Outline. Ali Ebnenasir Department of Computer Science Michigan Technological University
CS 5321: Advanced Algorithms - Recurrence Ali Ebnenasir Department of Computer Science Michigan Technological University Acknowledgement Eric Torng Moon Jung Chung Charles Ofria Outline Motivating example:
More information(please print) (1) (18) H IIA IIIA IVA VA VIA VIIA He (2) (13) (14) (15) (16) (17)
CHEM 10113, Quiz 3 September 28, 2011 Name (please print) All equations must be balanced and show phases for full credit. Significant figures count, show charges as appropriate, and please box your answers!
More informationReduced Synchronization Overhead on. December 3, Abstract. The standard formulation of the conjugate gradient algorithm involves
Lapack Working Note 56 Conjugate Gradient Algorithms with Reduced Synchronization Overhead on Distributed Memory Multiprocessors E. F. D'Azevedo y, V.L. Eijkhout z, C. H. Romine y December 3, 1999 Abstract
More informationMarkov Chains (Part 3)
Markov Chains (Part 3) State Classification Markov Chains - State Classification Accessibility State j is accessible from state i if p ij (n) > for some n>=, meaning that starting at state i, there is
More informationStatistics 992 Continuous-time Markov Chains Spring 2004
Summary Continuous-time finite-state-space Markov chains are stochastic processes that are widely used to model the process of nucleotide substitution. This chapter aims to present much of the mathematics
More informationCS 5321: Advanced Algorithms Analysis Using Recurrence. Acknowledgement. Outline
CS 5321: Advanced Algorithms Analysis Using Recurrence Ali Ebnenasir Department of Computer Science Michigan Technological University Acknowledgement Eric Torng Moon Jung Chung Charles Ofria Outline Motivating
More informationExam 2 Solutions. x 1 x. x 4 The generating function for the problem is the fourth power of this, (1 x). 4
Math 5366 Fall 015 Exam Solutions 1. (0 points) Find the appropriate generating function (in closed form) for each of the following problems. Do not find the coefficient of x n. (a) In how many ways can
More informationMarkov Chains Absorption Hamid R. Rabiee
Markov Chains Absorption Hamid R. Rabiee Absorbing Markov Chain An absorbing state is one in which the probability that the process remains in that state once it enters the state is (i.e., p ii = ). A
More informationWeighted Least Squares
Weighted Least Squares The standard linear model assumes that Var(ε i ) = σ 2 for i = 1,..., n. As we have seen, however, there are instances where Var(Y X = x i ) = Var(ε i ) = σ2 w i. Here w 1,..., w
More informationMonitoring Climate Change using Satellites: Lessons from MSU
Monitoring Climate Change using Satellites: Lessons from MSU Peter Thorne, Simon Tett Hadley Centre, Met Office, Exeter, UK UAH data from John Christy Residual uncertainty work in collaboration with John
More information