 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: 4th. 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. Since 40 problems in chapter 5.3 have been answered, more than 24022 students have viewed full stepbystep solutions from this chapter. Chapter 5.3 includes 40 full stepbystep solutions.

Block matrix.
A matrix can be partitioned into matrix blocks, by cuts between rows and/or between columns. Block multiplication ofAB is allowed if the block shapes permit.

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 •

Commuting matrices AB = BA.
If diagonalizable, they share n eigenvectors.

Covariance matrix:E.
When random variables Xi have mean = average value = 0, their covariances "'£ ij are the averages of XiX j. With means Xi, the matrix :E = mean of (x  x) (x  x) T is positive (semi)definite; :E is diagonal if the Xi are independent.

Eigenvalue A and eigenvector x.
Ax = AX with x#O so det(A  AI) = o.

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!).

GramSchmidt orthogonalization A = QR.
Independent columns in A, orthonormal columns in Q. Each column q j of Q is a combination of the first j columns of A (and conversely, so R is upper triangular). Convention: diag(R) > o.

Hermitian matrix A H = AT = A.
Complex analog a j i = aU of a symmetric matrix.

Indefinite matrix.
A symmetric matrix with eigenvalues of both signs (+ and  ).

Multiplication Ax
= Xl (column 1) + ... + xn(column n) = combination of columns.

Plane (or hyperplane) in Rn.
Vectors x with aT x = O. Plane is perpendicular to a =1= O.

Projection p = a(aTblaTa) onto the line through a.
P = aaT laTa has rank l.

Rank r (A)
= number of pivots = dimension of column space = dimension of row space.

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

Singular Value Decomposition
(SVD) A = U:E VT = (orthogonal) ( diag)( orthogonal) First r columns of U and V are orthonormal bases of C (A) and C (AT), AVi = O'iUi with singular value O'i > O. Last columns are orthonormal bases of nullspaces.

Solvable system Ax = b.
The right side b is in the column space of A.

Subspace S of V.
Any vector space inside V, including V and Z = {zero vector only}.

Vector addition.
v + w = (VI + WI, ... , Vn + Wn ) = diagonal of parallelogram.

Volume of box.
The rows (or the columns) of A generate a box with volume I det(A) I.
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