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Solutions for Chapter 90: Measuring Turns

Saxon Math, Course 1 | 1st Edition | ISBN: 9781591417835 | Authors: Stephan Hake

Full solutions for Saxon Math, Course 1 | 1st Edition

ISBN: 9781591417835

Saxon Math, Course 1 | 1st Edition | ISBN: 9781591417835 | Authors: Stephan Hake

Solutions for Chapter 90: Measuring Turns

Solutions for Chapter 90
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Textbook: Saxon Math, Course 1
Edition: 1
Author: Stephan Hake
ISBN: 9781591417835

Saxon Math, Course 1 was written by and is associated to the ISBN: 9781591417835. Chapter 90: Measuring Turns includes 30 full step-by-step solutions. Since 30 problems in chapter 90: Measuring Turns have been answered, more than 33787 students have viewed full step-by-step solutions from this chapter. This textbook survival guide was created for the textbook: Saxon Math, Course 1, edition: 1. This expansive textbook survival guide covers the following chapters and their solutions.

Key Math Terms and definitions covered in this textbook
  • 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 •

  • Conjugate Gradient Method.

    A sequence of steps (end of Chapter 9) to solve positive definite Ax = b by minimizing !x T Ax - x Tb over growing Krylov subspaces.

  • Dimension of vector space

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

  • 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.

  • Fast Fourier Transform (FFT).

    A factorization of the Fourier matrix Fn into e = log2 n matrices Si times a permutation. Each Si needs only nl2 multiplications, so Fnx and Fn-1c can be computed with ne/2 multiplications. Revolutionary.

  • 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.

  • Gauss-Jordan method.

    Invert A by row operations on [A I] to reach [I A-I].

  • 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.

  • Iterative method.

    A sequence of steps intended to approach the desired solution.

  • Kronecker product (tensor product) A ® B.

    Blocks aij B, eigenvalues Ap(A)Aq(B).

  • 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.

  • 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.

  • 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.

  • Nilpotent matrix N.

    Some power of N is the zero matrix, N k = o. The only eigenvalue is A = 0 (repeated n times). Examples: triangular matrices with zero diagonal.

  • 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.

  • 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.

  • Rotation matrix

    R = [~ CS ] rotates the plane by () and R- 1 = RT rotates back by -(). Eigenvalues are eiO and e-iO , eigenvectors are (1, ±i). c, s = cos (), sin ().

  • Row picture of Ax = b.

    Each equation gives a plane in Rn; the planes intersect at x.

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