 5.6.1E: Find the first four terms of each of the recursively defined sequen...
 5.6.2E: Find the first four terms of each of the recursively defined sequen...
 5.6.3E: Find the first four terms of each of the recursively defined sequen...
 5.6.4E: Find the first four terms of each of the recursively defined sequen...
 5.6.5E: Find the first four terms of each of the recursively defined sequen...
 5.6.6E: Find the first four terms of each of the recursively defined sequen...
 5.6.7E: Find the first four terms of each of the recursively defined sequen...
 5.6.8E: Find the first four terms of each of the recursively defined sequen...
 5.6.9E: Let a0, a1, a2, ... be defined by the formula an = 3n + 1, for all ...
 5.6.10E: Let b0, b1, b2, ... be defined by the formula bn = 4n, for all inte...
 5.6.11E: Let c0, c1, c2, ... be defined by the formula cn = 2n –1 for all in...
 5.6.12E: Let s0, s1, s2, ... be defined by the formula for all integers n ? ...
 5.6.13E: Let t0, t1, t2, ... be defined by the formula tn = 2 + n for all in...
 5.6.14E: Let d0, d1, d2, ... be defined by the formula dn = 3n – 2nfor all i...
 5.6.15E: For the sequence of Catalan numbers defined in Example prove that f...
 5.6.16E: Use the recurrence relation and values for the Tower of Hanoi seque...
 5.6.17E: Tower of Hanoi with Adjacency Requirement: Suppose that in addition...
 5.6.18E: Tower of Hanoi with Adjacency Requirement: Suppose the same situati...
 5.6.19E: FourPole Tower of Hanoi: Suppose that the Tower of Hanoi problem h...
 5.6.20E: Tower of Hanoi Poles in a Circle: Suppose that instead of being lin...
 5.6.21E: Double Tower of Hanoi: In this variation of the Tower of Hanoi ther...
 5.6.22E: Fibonacci Variation: A single pair of rabbits (male and female) is ...
 5.6.23E: Fibonacci Variation: A single pair of rabbits (male and female) is ...
 5.6.24E: Use the recurrence relation and values for F0, F1, F2, ... given in...
 5.6.25E: The Fibonacci sequence satisfies the recurrence relation Fk = Fk1 ...
 5.6.26E: Prove that Fk = 3Fk–3 + 2Fk–4 for all integers k ? 4.
 5.6.27E: Prove that for all integers k ? 1.
 5.6.28E: Prove that F F F = 2FFfor all integers k ?1.
 5.6.29E: Prove that for all integers k ? 1.
 5.6.30E: Use mathematical induction to prove that for all integers n ? 0, .
 5.6.31E: Use strong mathematical induction to prove that Fn<2n for all integ...
 5.6.32E: Let F0, F1, F2, … be the Fibonacci sequence defined in Section 5.5....
 5.6.33E: It turns out that the Fibonacci sequence satisfies the following ex...
 5.6.34E: (For students who have studied calculus) Find assuming that the lim...
 5.6.35E: (For students who have studied calculus) Prove that exists.
 5.6.36E: (For students who have studied calculus) Define x0, x1, x2, ... as ...
 5.6.37E: Compound Interest: Suppose a certain amount of money is deposited i...
 5.6.38E: Compound Interest: Suppose a certain amount of money is deposited i...
 5.6.39E: With each step you take when climbing a staircase, you can move up ...
 5.6.41E: Use the recursive definition of summation, together with mathematic...
 5.6.42E: Use the recursive definition of product, together with mathematical...
 5.6.43E: Use the recursive definition of product, together with mathematical...
 5.6.44E: The triangle inequality for absolute value states that for all real...
Solutions for Chapter 5.6: Discrete Mathematics with Applications 4th Edition
Full solutions for Discrete Mathematics with Applications  4th Edition
ISBN: 9780495391326
Solutions for Chapter 5.6
Get Full SolutionsSince 43 problems in chapter 5.6 have been answered, more than 24114 students have viewed full stepbystep solutions from this chapter. Discrete Mathematics with Applications was written by Sieva Kozinsky and is associated to the ISBN: 9780495391326. This expansive textbook survival guide covers the following chapters and their solutions. Chapter 5.6 includes 43 full stepbystep solutions. This textbook survival guide was created for the textbook: Discrete Mathematics with Applications , edition: 4th.

Augmented matrix [A b].
Ax = b is solvable when b is in the column space of A; then [A b] has the same rank as A. Elimination on [A b] keeps equations correct.

Back substitution.
Upper triangular systems are solved in reverse order Xn to Xl.

Basis for V.
Independent vectors VI, ... , v d whose linear combinations give each vector in V as v = CIVI + ... + CdVd. V has many bases, each basis gives unique c's. A vector space has many bases!

Cramer's Rule for Ax = b.
B j has b replacing column j of A; x j = det B j I det A

Determinant IAI = det(A).
Defined by det I = 1, sign reversal for row exchange, and linearity in each row. Then IAI = 0 when A is singular. Also IABI = IAIIBI and

Diagonalization
A = S1 AS. A = eigenvalue matrix and S = eigenvector matrix of A. A must have n independent eigenvectors to make S invertible. All Ak = SA k SI.

Dimension of vector space
dim(V) = number of vectors in any basis for V.

Fibonacci numbers
0,1,1,2,3,5, ... satisfy Fn = Fnl + Fn 2 = (A7 A~)I()q A2). Growth rate Al = (1 + .J5) 12 is the largest eigenvalue of the Fibonacci matrix [ } A].

Free variable Xi.
Column i has no pivot in elimination. We can give the n  r free variables any values, then Ax = b determines the r pivot variables (if solvable!).

Kirchhoff's Laws.
Current Law: net current (in minus out) is zero at each node. Voltage Law: Potential differences (voltage drops) add to zero around any closed loop.

Left nullspace N (AT).
Nullspace of AT = "left nullspace" of A because y T A = OT.

Normal equation AT Ax = ATb.
Gives the least squares solution to Ax = b if A has full rank n (independent columns). The equation says that (columns of A)·(b  Ax) = o.

Nullspace matrix N.
The columns of N are the n  r special solutions to As = O.

Partial pivoting.
In each column, choose the largest available pivot to control roundoff; all multipliers have leij I < 1. See condition number.

Pascal matrix
Ps = pascal(n) = the symmetric matrix with binomial entries (i1~;2). Ps = PL Pu all contain Pascal's triangle with det = 1 (see Pascal in the index).

Reflection matrix (Householder) Q = I 2uuT.
Unit vector u is reflected to Qu = u. All x intheplanemirroruTx = o have Qx = x. Notice QT = Q1 = Q.

Row picture of Ax = b.
Each equation gives a plane in Rn; the planes intersect at x.

Schur complement S, D  C A } B.
Appears in block elimination on [~ g ].

Spectrum of A = the set of eigenvalues {A I, ... , An}.
Spectral radius = max of IAi I.

Triangle inequality II u + v II < II u II + II v II.
For matrix norms II A + B II < II A II + II B II·