 5.3.1: Based on the discussion of the product 1 1 2 1 1 3 1 1 4 1 1 n at t...
 5.3.2: Experiment with computing values of the product 1 + 1 1 1 + 1 2 1 +...
 5.3.3: Observe that 1 13 = 1 3 1 13 + 1 35 = 2 5 1 13 + 1 35 + 1 57 = 3 7 ...
 5.3.4: Observe that 1 = 1, 1 4 = (1 + 2), 1 4 + 9 = 1 + 2 + 3, 1 4 + 9 16 ...
 5.3.5: Evaluate the sum n k=1 k (k + 1)! for n = 1, 2, 3, 4, and 5. Make a...
 5.3.6: For each positive integer n, let P(n) be the property 5n 1 is divis...
 5.3.7: For each positive integer n, let P(n) be the property 2n < (n + 1)!...
 5.3.8: Prove each statement in 823 by mathematical induction.
 5.3.9: Prove each statement in 823 by mathematical induction.
 5.3.10: Prove each statement in 823 by mathematical induction.
 5.3.11: Prove each statement in 823 by mathematical induction.
 5.3.12: Prove each statement in 823 by mathematical induction.
 5.3.13: Prove each statement in 823 by mathematical induction.
 5.3.14: Prove each statement in 823 by mathematical induction.
 5.3.15: Prove each statement in 823 by mathematical induction.
 5.3.16: Prove each statement in 823 by mathematical induction.
 5.3.17: Prove each statement in 823 by mathematical induction.
 5.3.18: Prove each statement in 823 by mathematical induction.
 5.3.19: Prove each statement in 823 by mathematical induction.
 5.3.20: Prove each statement in 823 by mathematical induction.
 5.3.21: Prove each statement in 823 by mathematical induction.
 5.3.22: Prove each statement in 823 by mathematical induction.
 5.3.23: Prove each statement in 823 by mathematical induction.
 5.3.24: A sequence a1, a2, a3,... is defined by letting a1 = 3 and ak = 7ak...
 5.3.25: A sequence b0, b1, b2,... is defined by letting b0 = 5 and bk = 4 +...
 5.3.26: A sequence b0, b1, b2,... is defined by letting b0 = 5 and bk = 4 +...
 5.3.27: A sequence d1, d2, d3,... is defined by letting d1 = 2 and dk = dk1...
 5.3.28: Prove that for all integers n 1, 1 3 = 1 + 3 5 + 7 = 1 + 3 + 5 7 + ...
 5.3.29: Prove that for all integers n 1, 1 3 = 1 + 3 5 + 7 = 1 + 3 + 5 7 + ...
 5.3.30: Theorem: For any integer n 1, all the numbers in a set of n numbers...
 5.3.31: For all integers n 1, 3n 2 is even. Proof (by mathematical inductio...
 5.3.32: Some 5 5 checkerboards with one square removed can be completely co...
 5.3.33: Consider a 4 6 checkerboard. Draw a covering of the board by Lshap...
 5.3.34: a. Use mathematical induction to prove that any checkerboard with d...
 5.3.35: Let m and n be any integers that are greater than or equal to 1. a....
 5.3.36: In a roundrobin tournament each team plays every other team exactl...
 5.3.37: On the outside rim of a circular disk the integers from 1 through 3...
 5.3.38: Suppose that n as and n bs are distributed around the outside of a ...
 5.3.39: For a polygon to be convex means that all of its interior angles ar...
 5.3.40: a. Prove that in an 8 8 checkerboard with alternating black and whi...
Solutions for Chapter 5.3: Mathematical Induction II
Full solutions for Discrete Mathematics with Applications  4th Edition
ISBN: 9780495391326
Solutions for Chapter 5.3: Mathematical Induction II
Get Full SolutionsSince 40 problems in chapter 5.3: Mathematical Induction II have been answered, more than 23979 students have viewed full stepbystep solutions from this chapter. This textbook survival guide was created for the textbook: Discrete Mathematics with Applications , edition: 4th. Chapter 5.3: Mathematical Induction II includes 40 full stepbystep solutions. This expansive textbook survival guide covers the following chapters and their solutions. Discrete Mathematics with Applications was written by Sieva Kozinsky and is associated to the ISBN: 9780495391326.

Characteristic equation det(A  AI) = O.
The n roots are the eigenvalues of A.

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

Complete solution x = x p + Xn to Ax = b.
(Particular x p) + (x n in nullspace).

Condition number
cond(A) = c(A) = IIAIlIIAIII = amaxlamin. In Ax = b, the relative change Ilox III Ilx II is less than cond(A) times the relative change Ilob III lib II· Condition numbers measure the sensitivity of the output to change in the input.

Diagonal matrix D.
dij = 0 if i # j. Blockdiagonal: zero outside square blocks Du.

Elimination matrix = Elementary matrix Eij.
The identity matrix with an extra eij in the i, j entry (i # j). Then Eij A subtracts eij times row j of A from row i.

Full column rank r = n.
Independent columns, N(A) = {O}, no free variables.

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.

Hilbert matrix hilb(n).
Entries HU = 1/(i + j 1) = Jd X i 1 xj1dx. Positive definite but extremely small Amin and large condition number: H is illconditioned.

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

Lucas numbers
Ln = 2,J, 3, 4, ... satisfy Ln = L n l +Ln 2 = A1 +A~, with AI, A2 = (1 ± /5)/2 from the Fibonacci matrix U~]' Compare Lo = 2 with Fo = O.

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

Orthonormal vectors q 1 , ... , q n·
Dot products are q T q j = 0 if i =1= j and q T q i = 1. The matrix Q with these orthonormal columns has Q T Q = I. If m = n then Q T = Q 1 and q 1 ' ... , q n is an orthonormal basis for Rn : every v = L (v T q j )q j •

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

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

Right inverse A+.
If A has full row rank m, then A+ = AT(AAT)l has AA+ = 1m.

Semidefinite matrix A.
(Positive) semidefinite: all x T Ax > 0, all A > 0; A = any RT R.

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!

Spanning set.
Combinations of VI, ... ,Vm fill the space. The columns of A span C (A)!

Sum V + W of subs paces.
Space of all (v in V) + (w in W). Direct sum: V n W = to}.