- 4.4.1: Use the Composite Trapezoidal rule with the indicated values of n t...
- 4.4.2: Use the Composite Trapezoidal rule with the indicated values of n t...
- 4.4.3: Use the Composite Simpson's rule to approximate the integrals in Ex...
- 4.4.4: Use the Composite Simpson's rule to approximate the integrals in Ex...
- 4.4.5: Use the Composite Midpoint rule with n + 2 subintervals to approxim...
- 4.4.6: Use the Composite Midpoint rule with n + 2 subintervals to approxim...
- 4.4.7: Approximate /,( x 2 ln(x2 + 1) t/x using h 0.25. Use a. Composite T...
- 4.4.8: Approximate J0 x e x dx using h = 0.25. Use a. Composite Trapezoida...
- 4.4.9: Suppose that /(0) = 1, /(0.5) = 2.5, /(I) = 2, and /(0.25) = /(0.75...
- 4.4.10: The Midpoint rule for approximating /(x) dx gives the value 12, the...
- 4.4.11: Determine the values of n and h required to approximate I e 2xsin3x...
- 4.4.12: Repeat Exercise 11 for the integral JJ x 2 cosx dx
- 4.4.13: Determine the values of n and h required to approximate 'o x + A to...
- 4.4.14: Repeat Exercise 13 for the integral x \nx dx.
- 4.4.15: Let / be defined by 'x3 +1, 0 < A" < 0.1, 1.001 + 0.03(a - 0.1) + 0...
- 4.4.16: In multivariable calculus and in statistics courses, itis shown tha...
- 4.4.17: Determine to within 10 6 the length ofthe graph ofthe ellipse with ...
- 4.4.18: A car laps a race track in 84 seconds. The speed of the car at each...
- 4.4.19: A particle of mass tn moving through a fluid is subjected to a visc...
- 4.4.20: To simulate the thermal characteristics of disk brakes (see the fol...
- 4.4.21: Find an approximation to within 10 4 ofthe value ofthe integral con...
- 4.4.22: The equation / 48 / \/1 + (cosx)2 dx. Jo f -^=e-'2/2 dt - 0.45 can ...
- 4.4.23: Show that the error E(f) for the Composite Simpson's rule can be ap...
- 4.4.24: a. Derive an estimate for (/) in the Composite Trapezoidal rule usi...
- 4.4.25: Use the error estimates of Exercises 23 and 24 to estimate the erro...
- 4.4.26: Use the error estimates of Exercises 23 and 24 to estimate the erro...
Solutions for Chapter 4.4: Composite Numerical Integration
Full solutions for Numerical Analysis | 10th Edition
Adjacency matrix of a graph.
Square matrix with aij = 1 when there is an edge from node i to node j; otherwise aij = O. A = AT when edges go both ways (undirected). Adjacency matrix of a graph. Square matrix with aij = 1 when there is an edge from node i to node j; otherwise aij = O. A = AT when edges go both ways (undirected).
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.)
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).
Column space C (A) =
space of all combinations of the columns of A.
cond(A) = c(A) = IIAIlIIA-III = 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.
S. Permutation with S21 = 1, S32 = 1, ... , finally SIn = 1. Its eigenvalues are the nth roots e2lrik/n of 1; eigenvectors are columns of the Fourier matrix F.
Determinant IAI = det(A).
Defined by det I = 1, sign reversal for row exchange, and linearity in each row. Then IAI = 0 when A is singular. Also IABI = IAIIBI and
Dot product = Inner product x T y = XI Y 1 + ... + Xn Yn.
Complex dot product is x T Y . Perpendicular vectors have x T y = O. (AB)ij = (row i of A)T(column j of B).
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.
Hankel matrix H.
Constant along each antidiagonal; hij depends on i + j.
Hessenberg matrix H.
Triangular matrix with one extra nonzero adjacent diagonal.
Identity matrix I (or In).
Diagonal entries = 1, off-diagonal entries = 0.
Inverse matrix A-I.
Square matrix with A-I A = I and AA-l = I. No inverse if det A = 0 and rank(A) < n and Ax = 0 for a nonzero vector x. The inverses of AB and AT are B-1 A-I and (A-I)T. Cofactor formula (A-l)ij = Cji! detA.
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.
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.
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.
Solvable system Ax = b.
The right side b is in the column space of A.
Transpose matrix AT.
Entries AL = Ajj. AT is n by In, AT A is square, symmetric, positive semidefinite. The transposes of AB and A-I are BT AT and (AT)-I.
Unitary matrix UH = U T = U-I.
Orthonormal columns (complex analog of Q).