Week 9 Difference Equations Discrete Math April 23, 2020 Marie - PowerPoint PPT Presentation

Cyclic groups Difference Equations, Recursive Equations Week 9 Difference Equations Discrete Math April 23, 2020 Marie Demlova: Discrete Math

Cyclic groups Subgroups of a Finite Cyclic Group Difference Equations, Recursive Equations Cyclic groups A cyclic group. Given a group G = ( G , ◦ , e ) . If there exists an element a ∈ G for which � a � = G we say that the group is cyclic and that a is a generating element of ( G , ◦ , e ) . Examples. ◮ ( Z n , + , 0 ) (for any natural number n > 1) is a cyclic group with its generating element 1. ◮ For every prime number p the group ( Z ⋆ p , · , 1 ) is a cyclic group. It is not straightforward to show it. Moreover, to find a generating element is a difficult task for some primes p . ◮ The group ( Z ⋆ 8 , · , 1 ) is not cyclic. We have Z ⋆ 8 = { 1 , 3 , 5 , 7 } and there is no element with order 4. Marie Demlova: Discrete Math

Cyclic groups Subgroups of a Finite Cyclic Group Difference Equations, Recursive Equations Cyclic groups Proposition. Given a finite cyclic group G = ( G , ◦ , e ) with n elements. Then for every natural number d which divides n there exists a subgroup of G with d elements. Remark. A finite cyclic group has only subgroups that itself are cyclic. Marie Demlova: Discrete Math

Cyclic groups Subgroups of a Finite Cyclic Group Difference Equations, Recursive Equations Cyclic groups Exercise 1. Given a group ( Z ⋆ 17 , · , 1 ) . Find all its subgroups. Exercise 2. Given a group ( Z ⋆ 17 , · , 1 ) . Find all its generating elements. Exercise 3. Given a group ( Z ⋆ 14 , · , 1 ) . a) Write down all its elements. b) Find orders r ( a ) for all its elements. c) Is the group a cyclic group? d) Find all its subgroups. Marie Demlova: Discrete Math

Cyclic groups Linear Difference Equations with Constant Coefficients Difference Equations, Recursive Equations Difference Equations, Recursive Equations Sequences. A sequence is a mapping from the set of all integers greater or equal to an integer n 0 into the set of all real numbers. Hence { a n 0 , a n 0 + 1 , a n 0 + 2 , . . . } where a i ∈ R . Linear Difference Equations. Let c i ( n ) , i ∈ { 0 , . . . , k − 1 } , be functions Z → R , c 0 ( n ) not identically zero, and let { b n } ∞ n = n 0 be a sequence. Then the equation a n + k + c k − 1 ( n ) a n + k − 1 + . . . + c 1 ( n ) a n + 1 + c 0 ( n ) a n = b n , n ≥ n 0 is a linear difference equation of order k (also a linear recursive equation of order k ). Marie Demlova: Discrete Math

Cyclic groups Linear Difference Equations with Constant Coefficients Difference Equations, Recursive Equations Difference Equations, Recursive Equations Functions c i ( n ) are coefficients of the equation, the sequence { b n } ∞ n = n 0 the right-hand side of the equation. If { b n } ∞ n = n 0 is the zero sequence then we speak about homogeneous equation, otherwise the equation is non-homogeneous. We write a linear difference equation also k − 1 � a n + k + c i ( n ) a n + i = b n , n ≥ n 0 . i = 0 Marie Demlova: Discrete Math

Cyclic groups Linear Difference Equations with Constant Coefficients Difference Equations, Recursive Equations Difference Equations, Recursive Equations Solutions of Linear Difference Equations. A solution of a linear difference equation is any sequence { u n } ∞ n = n 0 such that if we substitute u n for a n in it we obtain a statement that is valid. Initial Conditions. Given a linear difference equation of order k a n + k + c k − 1 ( n ) a n + k − 1 + . . . + c 1 ( n ) a n + 1 + c 0 ( n ) a n = b n , n ≥ n 0 By initial conditions we mean the following system a n 0 = A 0 , a n 0 + 1 = A 1 , . . . , a n 0 + k − 1 = A k − 1 , where A i are real numbers. Marie Demlova: Discrete Math

Cyclic groups Linear Difference Equations with Constant Coefficients Difference Equations, Recursive Equations Difference Equations, Recursive Equations The Associated Homogeneous Equation. Given a linear difference equation a n + k + c k − 1 ( n ) a n + k − 1 + . . . + c 1 ( n ) a n + 1 + c 0 ( n ) a n = b n , n ≥ n 0 . Then the equation a n + k + c k − 1 ( n ) a n + k − 1 + . . . + c 1 ( n ) a n + 1 + c 0 ( n ) a n = 0 , n ≥ n 0 . is the associated homogeneous equation to the equation above. Marie Demlova: Discrete Math

Cyclic groups Linear Difference Equations with Constant Coefficients Difference Equations, Recursive Equations Difference Equations, Recursive Equations Proposition. Given a linear difference equation. Then the following holds: 1. If { u n } ∞ n = n 0 and { v n } ∞ n = n 0 are two solutions of non homogeneous equation then { u n } ∞ n = n 0 − { v n } ∞ n = n 0 is a solution of the associated homogeneous equation. 2. If { u n } ∞ n = n 0 is a solution of non homogeneous equation and { w n } ∞ n = n 0 is a solution of the associated homogeneous equation, then { u n } ∞ n = n 0 + { w n } ∞ n = n 0 is a solution of non homogeneous equation. 3. Let { ˆ u n } ∞ n = n 0 be a fixed solution of the non homogeneous equation. Then for every solution { v n } ∞ n = n 0 of it there exists a solution solution { w n } ∞ n = n 0 of the associated homogeneous equation for which { v n } ∞ n = n 0 = { ˆ u n } ∞ n = n 0 + { w n } ∞ n = n 0 . Marie Demlova: Discrete Math

Cyclic groups Linear Difference Equations with Constant Coefficients Difference Equations, Recursive Equations Difference Equations, Recursive Equations Theorem. Given a homogeneous linear difference equation. Then for the set S of all solutions the following holds: 1. If { u n } ∞ n = n 0 and { v n } ∞ n = n 0 belong to S then so does { u n } ∞ n = n 0 + { v n } ∞ n = n 0 . 2. If { u n } ∞ n = n 0 belongs to S and α is any real number, then { k u n } ∞ n = n 0 belongs to S as well. Marie Demlova: Discrete Math

Cyclic groups Linear Difference Equations with Constant Coefficients Difference Equations, Recursive Equations Difference Equations, Recursive Equations The difference equation a n + k + c k − 1 a n + k − 1 + . . . + c 1 a n + 1 + c 0 a n = b n , n ≥ n 0 , c i ∈ R , i.e. coefficients c i ( n ) are constant functions, is difference equation with constance coefficients. Characteristic equation of the equation above is λ k + c k − 1 λ k − 1 + . . . + c 1 λ + c 0 = 0 . Any λ satisfying characteristic equation leads to one solution a n = { λ n } ∞ n = n 0 . Marie Demlova: Discrete Math

Cyclic groups Linear Difference Equations with Constant Coefficients Difference Equations, Recursive Equations Linear Difference Equations with Constant Coefficients Real roots of characteristic equation. If λ is a root of the characteristic equation of multiplicity t then the following are linearly independent solutions of its homogeneous equation n = 0 , { n 2 λ n } ∞ n = 0 , . . . , { n t − 1 λ n } ∞ { λ n } ∞ n = 0 , { n λ n } ∞ n = 0 . Marie Demlova: Discrete Math

Cyclic groups Linear Difference Equations with Constant Coefficients Difference Equations, Recursive Equations Linear Difference Equations with Constant Coefficients Complex roots of characteristic equation. If λ = a + ı b is a complex root of the characteristic equation of multiplicity t then the following are linearly independent complex solutions of its homogeneous equation { ( a + ı b ) n } ∞ n = 0 and { ( a − ı b ) n } ∞ n = 0 and the following real solutions { r n cos n ϕ } ∞ n = 0 and { r n sin n ϕ } ∞ n = 0 Marie Demlova: Discrete Math