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Solutions for Chapter 11.2: Arithmetic Sequences

Full solutions for Intermediate Algebra for College Students | 6th Edition

ISBN: 9780321758934

Solutions for Chapter 11.2: Arithmetic Sequences

Solutions for Chapter 11.2
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Textbook: Intermediate Algebra for College Students
Edition: 6
Author: Robert F. Blitzer
ISBN: 9780321758934

Since 95 problems in chapter 11.2: Arithmetic Sequences have been answered, more than 29916 students have viewed full step-by-step solutions from this chapter. This textbook survival guide was created for the textbook: Intermediate Algebra for College Students, edition: 6. Chapter 11.2: Arithmetic Sequences includes 95 full step-by-step solutions. This expansive textbook survival guide covers the following chapters and their solutions. Intermediate Algebra for College Students was written by and is associated to the ISBN: 9780321758934.

Key Math Terms and definitions covered in this textbook
  • 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!

  • Cayley-Hamilton 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.)

  • Characteristic equation det(A - AI) = O.

    The n roots are the eigenvalues of A.

  • Diagonal matrix D.

    dij = 0 if i #- j. Block-diagonal: zero outside square blocks Du.

  • Distributive Law

    A(B + C) = AB + AC. Add then multiply, or mUltiply then add.

  • Fourier matrix F.

    Entries Fjk = e21Cijk/n give orthogonal columns FT F = nI. Then y = Fe is the (inverse) Discrete Fourier Transform Y j = L cke21Cijk/n.

  • Graph G.

    Set of n nodes connected pairwise by m edges. A complete graph has all n(n - 1)/2 edges between nodes. A tree has only n - 1 edges and no closed loops.

  • Incidence matrix of a directed graph.

    The m by n edge-node incidence matrix has a row for each edge (node i to node j), with entries -1 and 1 in columns i and j .

  • Left inverse A+.

    If A has full column rank n, then A+ = (AT A)-I AT has A+ A = In.

  • Linearly dependent VI, ... , Vn.

    A combination other than all Ci = 0 gives L Ci Vi = O.

  • 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.

  • 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 •

  • Pseudoinverse A+ (Moore-Penrose inverse).

    The n by m matrix that "inverts" A from column space back to row space, with N(A+) = N(AT). A+ A and AA+ are the projection matrices onto the row space and column space. Rank(A +) = rank(A).

  • Reduced row echelon form R = rref(A).

    Pivots = 1; zeros above and below pivots; the r nonzero rows of R give a basis for the row space of A.

  • Reflection matrix (Householder) Q = I -2uuT.

    Unit vector u is reflected to Qu = -u. All x intheplanemirroruTx = o have Qx = x. Notice QT = Q-1 = Q.

  • Special solutions to As = O.

    One free variable is Si = 1, other free variables = o.

  • 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.

  • Symmetric matrix A.

    The transpose is AT = A, and aU = a ji. A-I is also symmetric.

  • Wavelets Wjk(t).

    Stretch and shift the time axis to create Wjk(t) = woo(2j t - k).

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