 5.3.7E: For each positive integer n, let P(n) be the property 2n < (n + 1)!...
 5.3.1E: Based on the discussion of the product at the beginning of this sec...
 5.3.2E: Experiment with computing values of the product for small values of...
 5.3.3E: Observe that Guess a general formula and prove it by mathematical i...
 5.3.4E: Observe that Guess a general formula and prove it by mathematical i...
 5.3.5E: Evaluate the sum for n = 1, 2, 3, 4, and 5. Make a conjecture about...
 5.3.6E: For each positive integer n, let P(n) be the property5n –1 is divis...
 5.3.8E: Prove the statement by mathematical induction.Exercise5n –1 is divi...
 5.3.9E: Prove the statement by mathematical induction.Exercise7n –1 is divi...
 5.3.10E: Prove the statement by mathematical induction.Exercisen3 –7n + 3 is...
 5.3.11E: Prove the statement by mathematical induction.Exercise32n – 1 is di...
 5.3.12E: Prove the statement by mathematical induction.ExerciseFor any integ...
 5.3.13E: Prove the statement by mathematical induction.ExerciseFor any integ...
 5.3.14E: Prove the statement by mathematical induction.Exercisen3 – n is div...
 5.3.15E: Prove the statement by mathematical induction.Exercisen(n2 + 5) is ...
 5.3.16E: Prove the statement by mathematical induction.Exercise2n<(n + 1)!, ...
 5.3.17E: Prove the statement by mathematical induction.Exercise1 + 3n ? 4n, ...
 5.3.18E: Prove the statement by mathematical induction.Exercise5n + 9 ? 6n, ...
 5.3.19E: Prove the statement by mathematical induction.Exercisen2<2n, for al...
 5.3.20E: Prove the statement by mathematical induction.Exercise2n ? (n + 2)!...
 5.3.21E: Prove the statement by mathematical induction.Exercise for all inte...
 5.3.22E: Prove the statement by mathematical induction.Exercise1+ nx ? (1 + ...
 5.3.23E: Prove the statement by mathematical induction.Exercisea. n3 > 2n + ...
 5.3.24E: A sequence a1, a2, a3, ... is defined by letting a1 = 3 and ak = 7a...
 5.3.25E: A sequence b0, b1, b2, ... is defined by letting b0 = 5 and bk = 4 ...
 5.3.26E: A sequence c0, c1, c2, ... is defined by letting c0 = 3 and ck = (c...
 5.3.27E: A sequence d2, d3, ... is defined by letting d1 = 2 and for all int...
 5.3.28E: Prove that for all integers n ? 1,
 5.3.29E: As each of a group of businesspeople arrives at a meeting, each sha...
 5.3.30E: In order for a proof by mathematical induction to be valid, the bas...
 5.3.31E: In order for a proof by mathematical induction to be valid, the bas...
 5.3.32E: Some 5 × 5 checkerboards with one square removed can be completely ...
 5.3.33E: Consider a 4 × 6 checkerboard. Draw a covering of the board by Lsh...
 5.3.34E: a. Use mathematical induction to prove that any checkerboard with d...
 5.3.35E: Let m and n be any integers that are greater than or equal to 1.a. ...
 5.3.36E: In a roundrobin tournament each team plays every other team exactl...
 5.3.37E: On the outside rim of a circular disk the integers from 1 through 3...
 5.3.38E: Suppose that n a’s and n b’s are distributed around the outside of ...
 5.3.39E: For a polygon to be convex means that given any two points on or in...
 5.3.40E: a. Prove that in an 8 ×8 checkerboard with alternating black and wh...
Solutions for Chapter 5.3: Discrete Mathematics with Applications 4th Edition
Full solutions for Discrete Mathematics with Applications  4th Edition
ISBN: 9780495391326
Solutions for Chapter 5.3
Get Full SolutionsThis textbook survival guide was created for the textbook: Discrete Mathematics with Applications , edition: 4. Discrete Mathematics with Applications was written by and is associated to the ISBN: 9780495391326. This expansive textbook survival guide covers the following chapters and their solutions. Since 40 problems in chapter 5.3 have been answered, more than 91182 students have viewed full stepbystep solutions from this chapter. Chapter 5.3 includes 40 full stepbystep solutions.

Affine transformation
Tv = Av + Vo = linear transformation plus shift.

CayleyHamilton Theorem.
peA) = det(A  AI) has peA) = zero matrix.

Change of basis matrix M.
The old basis vectors v j are combinations L mij Wi of the new basis vectors. The coordinates of CI VI + ... + cnvn = dl wI + ... + dn Wn are related by d = M c. (For n = 2 set VI = mll WI +m21 W2, V2 = m12WI +m22w2.)

Cofactor Cij.
Remove row i and column j; multiply the determinant by (I)i + j •

Column picture of Ax = b.
The vector b becomes a combination of the columns of A. The system is solvable only when b is in the column space C (A).

Factorization
A = L U. If elimination takes A to U without row exchanges, then the lower triangular L with multipliers eij (and eii = 1) brings U back to 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!).

Fundamental Theorem.
The nullspace N (A) and row space C (AT) are orthogonal complements in Rn(perpendicular from Ax = 0 with dimensions rand n  r). Applied to AT, the column space C(A) is the orthogonal complement of N(AT) in Rm.

Least squares solution X.
The vector x that minimizes the error lie 112 solves AT Ax = ATb. Then e = b  Ax is orthogonal to all columns of A.

Linear combination cv + d w or L C jV j.
Vector addition and scalar multiplication.

Markov matrix M.
All mij > 0 and each column sum is 1. Largest eigenvalue A = 1. If mij > 0, the columns of Mk approach the steady state eigenvector M s = s > O.

Outer product uv T
= column times row = rank one matrix.

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).

Pivot.
The diagonal entry (first nonzero) at the time when a row is used in elimination.

Random matrix rand(n) or randn(n).
MATLAB creates a matrix with random entries, uniformly distributed on [0 1] for rand and standard normal distribution for randn.

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 space C (AT) = all combinations of rows of A.
Column vectors by convention.

Simplex method for linear programming.
The minimum cost vector x * is found by moving from comer to lower cost comer along the edges of the feasible set (where the constraints Ax = b and x > 0 are satisfied). Minimum cost at a comer!

Stiffness matrix
If x gives the movements of the nodes, K x gives the internal forces. K = ATe A where C has spring constants from Hooke's Law and Ax = stretching.

Tridiagonal matrix T: tij = 0 if Ii  j I > 1.
T 1 has rank 1 above and below diagonal.