References Software see at the beginning of Chaps. 19 and 24. General References [GR I ] Abramowitz, M. and I. A. Stegun (eds.). Handbook of Mathematical Functions. 10th printing, with corrections. Washington. DC: National Bureau of Standards. 1972 (also New York: Dover, 1965). fGR2] Cajori, F., History of Mathematics. 5th ed. Reprinted. Providence. RI: American Mathematical Society. 2002. [GR3] Courant, R. and D. Hilbert, Methods of Mathematical Physics. 2 vols. Hoboken, NJ: Wiley, 2003. [GR4] Courant, R., Differential and Integral Calculus. 2 vols. Hoboken. NJ: Wiley, 2003. [GR5J Graham, R. L. el al., Concrete Mathematics. 2nd ed. Reading. MA: Addison-Wesley, 1994. |GR6] Ito. K. (cd.). Encyclopedic Dictionary of Mathematics. 4 vols. 2nd ed. Cambridge. MA: MIT Press. 1993. [GR7] Kreyszig, E., Introductory Functional Analysis with Applications. New York: Wiley, 1989. [GR81 Krcyszig, E., Differential Geometry. Mineola, NY: Dover. 1991. [GR9] Kreyszig, E. Introduction to Differential Geometry and Riemannian Geometry. Toronto: University of Toronto Press, 1975. IGR10] Szegö, G.. Orthogonal Polynomials. 4th ed. Reprinted. New York: American Mathematical Society, 2003. |GR1I] Thomas, G. et al., Thomas' Calculus, Early Transcendentals Update. 10th ed. Reading, MA: Addison-Wesley. 2003. Part A. Ordinary Differential Equations (ODEs) (Chaps. 1-6) See also Part E: Numeric Analysis [Al I Arnold, V. I., Ordinary Differential Equations. 3rd ed. New York: Springer, 1997. [A2] Bhatia, N. P. and G. P. Szego, Stability Theory of Dynamical Systems. New York: Springer, 2002. [A31 Birkhoff, G. and G.-C. Rota, Ordinary Differential Equations. 4th ed. New York: Wiley. 1989. [A4] Brauer, F. and J. A. Nohel, Qualitative Theory of Ordinary Differential Equations. Mineola. NY: Dover. 1994. [A5] Churchill, R. V., Operational Mathematics. 3rd ed. New York: McGraw-Hill, 1972. [A6| Coddington. F. A. and R. Carlson, Linear Ordinary Differential Equations. Philadelphia: SIAM, 1997. [A7] Coddington. E. A. and N. Levinson, Theory of Ordinary Differential Equations. Malabar, FL: Krieger, 1984. [A8] Dong, T.-R. et al., Qualitative Theory of Differential Equations. Providence, Rl: American Mathematical Society. 1992. [A9] Erdelyi, A. et al., Tables of Integral Transforms. 2 vols. New York: McGraw-Hill, 1954. IA10] Hartman, P.. Ordinary Differential Equations. 2nd cd. Philadelphia: SIAM, 2002. [AllJ Ince. E. L., Ordinary Differential Equations. New York: Dover, 1956. [A12] Schiff, J. L., The Ixiplace 'Transform: Theory and Applications. New York: Springer, 1999. [A 131 Watson, G. N., A Treatise on the Theory of Bessel Functions. 2nd cd. Reprinted. New York: Cambridge University Press. 1995. [A 14] Widder, D. V., The Laplace Transform. Princeton, NJ: Princeton University Press, 1941. [A 151 Zwillinger, D., Handbook of Differential Equations. 3rd cd. New York: Academic Press, 1997. Part B. Linear Algebra, Vector Calculus (Chaps. 7-10) For books on numeric linear algebra, see also Part E: Numeric Analysis. [B1J Bellman, R., Introduction to Matrix Analysis. 2nd ed. Philadelphia: SIAM, 1997. [B2] Chatelin. F., Eigenvalues of Matrices. New York: Wiley-Interscience, 1993. [B3] Gantmacher, F. R., The Theory of Matrices. 2 vols. Providence, RI: American Mathematical Society, 2000. IB4] Gohberg, I. P. ct al.. Invariant Subspaces of Matrices with Applications. New York: Wiley, 1986. [B5] Greub, W. H., Linear Algebra. 4th ed. New York: Springer, 1996. IB6] Herstein, I. N.. Abstract Algebra. 3rd ed. New York: Wiley, 1996. [B71 John. A. W., Matrices and. Tensors in Physics. 3rd ed. New York: Wiley, 1995. \ Al References [B8] Lang, S., Linear Algebra. 3rd ed. New York: Springer, 1996. [B9] Nef, W., Linear Algebra. 2nd ed. New York: Dover, 1988. IB1OJ Parlett, B., The Symmetric Eigenvalue Problem. Philadelphia: SIAM, 1997. Part C. Fourier Analysis and PDEs (Chaps. 10-11) For books on numerics for PDEs see also Part E: Numeric Analysis. [C11 Antimirov. M. Ya., Applied Integral Transforms. Providence, RI: American Mathematical Society, 1993. [C2] Braccwcll, R., The Fourier Transform and Its Applications. 3rd ed. New York: McGraw-Hill, 2000. [C31 Carslaw, H. S. and .1. C. Jaeger, Conduction of Heat in Solids. 2nd ed. Reprinted. Oxford: Clarendon, 1986. LC4J Churchill, R. V. and J. W. Brown, Fourier Series and Boundary Value Problems. 6th ed. New York: McGraw-Hill, 2000. [C5J DuChateau, P. and D. Zachmann. Applied Partial Differential Equations. Mineola, NY: Dover, 2001. [C6] Hanna, J. R. and J. H. Rowland, Fourier Series, Transforms, and Boundary Value Problems. 2nd ed. New York: Wiley, 1990. [C7] Jerri, A. J., The Gibbs Phenomenon in Fourier Analysis, Splines, and Wavelet Approximations. Boston: Kluwer, 1998. [C8] John, F., Partial Differential Equations. New York: Springer, 1995. [C9J Tolstov, G. P., Fourier Series. New York: Dover, 1976. [C10] Widder, D. V., The Heat Equation. New York: Academic Press, 1975. [Cll] Zauderer, E., Partial Differential Equations of Applied Mathematics. 2nd ed. New York: Wiley. 1989. [CI2] Zygmund. A. and R. Fcffcrman, Trigonometric Series. 3rd ed. New York: Cambridge University Press, 2003. Part D. Complex Analysis (Chaps. 13-18) [Dl] Ahlfors, L. V., Complex Analysis. 3rd ed. New-York: McGraw-Hill, 1979. [D2] Bieberbach, L., Conformed Mapping. Providence, RI: American Mathematical Society, 2000. [D31 Henrici, P., Applied and Computational Complex Analysis. 3 vols. New York: Wiley, 1993. [D4] Hille, E.. Analytic Function Theory. 2 vols. 2nd ed. Providence, RI: American Mathematical Society, 1997. |D5| Knopp, K., Elements of the Theory of Functions. New York: Dover, 1952. [D6] Knopp, K., Theory of Functions. 2 parts. New York: Dover, 1945, 1947. [D7] Krantz, S. G., Complex Analysis: The Geometric Viewpoint. Washington, DC: The Mathematical Association of America, 1990. [D8] Lang, S., Complex Analysis. 3rd cd. New York: Springer, 1993. [D9] Narasimhan, R., Compact Riemann Surfaces. New York: Springer, 1996. [D10] Nehari, Z„ Conformal Mapping. Mineola, NY: Dover, 1975. I Dil J Springer. G., Introduction to Riemann Surfaces. Providence, RI: American Mathematical Society, 2002. Part E. Numeric Analysis (Chaps. 19-21) [Ell Ames, W. F., Numerical Methods for Partial Differential Equations. 3rd ed. New York: Academic Press, 1992. [F.2] Anderson, E„ et al„ LAPACK User's Guide. 3rd ed. Philadelphia: SIAM, 1999. |E3J Bank, R. E., PLTMG. A Software Package for Solving Elliptic Partial Differential Equations: Users' Guide 7.0. Philadelphia: SIAM, 1994. [E4] Constanda, C, Solution Techniques for Elementary Partial Differential Equations. Boca Raton, FL: CRC Press, 2002. [E5] Dahlquist, G. and A. Björck, Numerical Methods. Mineola, NY: Dover, 2003. [E6] DeBoor. C..A Practical Guide to Splines. Reprinted. New York: Springer, 1991. |E7] Dongarra, J. J. et al., LIN PACK Users Guide. Philadelphia: SIAM, 1978. (See also at the beginning of Chap. 19.) [E81 Garbow, B. S. et al., Matrix Eigensystem Routines: EISPACK Guide Extension. Reprinted. New York: Springer, 1990. |F.9| Golub, G. H. and C. F. Van Loan, Matrix Computations. 3rd ed. Baltimore, MD: Johns Hopkins University Press, 1996. [RIO] Higham, N. .1., Accuracy and Stability of Numerical Algorithms. 2nd ed. Philadelphia: SIAM, 2002. [Ell] IMSL (International Mathematical and Statistical Libraries), FORTRAN Numerical Library. Houston, TX: Visual Numerics, 2002. (See also at the beginning of Chap. 19.) [El2] IMSL, IMSL for Java. Houston, TX: Visual Numerics, 2002. [E13] IMSL, C Library. Houston, TX: Visual Numerics, 2002. [E14] Kelley, C. T., Iterative Methods for Linear and Nonlinear Equations. Philadelphia: SIAM, 1995. [El5] Knabncr, P. and L. Angerman, Numerical Methods for Partial Differential Equations. New York: Springer, 2003. [EI6] Knuth, D. E., The Art of Computer Programming. 3 vols. 3rd ed. Reading, MA: Addison-Wesley, 2005. App. 1 [El 71 Kreyszig. E., Introductory Functional Analysis with Applications. New York: Wiley, 1989. [E18] Kreyszig, E., On methods of Fourier analysis in multigrid theory. Lecture Notes in Pure and Applied Mathematics 157. New York: Dekker, 1994, pp. 225-242. [El91 Kreyszig, E.. Basic ideas in modern numerical analysis and their origins. Proceedings of the Annual Conference of the Canadian Society for the History and Philosophy of Mathematics. 1997, pp. 34-45. [E20] Kreyszig, E., and J. Todd, QR in two dimensions. Elemente der Mathematik 31 (1976), pp. 109-114. |E21| Möllensen, M. E., Geometric Modeling. 2nd cd. New York: Wiley, 1997. [E221 Morion, K. W., and D. F. Mayers, Numerical Solution t>f Partial Differential Equations: An Introduction. New York: Cambridge University Press, 1994. [E23] Ortega, J. M., Introduction to Parallel and Vector Solution of Linear Systems. New York: Plenum Press, 1988. [E24] Overton, M. L., Numerical Computing with IEEE Floating Point Arithmetic. Philadelphia: SIAM, 2001. [E25] Press, W. H. et al., Numerical Recipes in C: The Art of Scientific Computing. 2nd ed. New York: Cambridge University Press, 1992. [E26] Shampine, L. F., Numerical Solutions of Ordinary Differential Equations. New York: Chapman and Hall. 1994. IE27] Varga. R. S., Matrix Iterative Analysis. 2nd ed. New York: Springer, 2000. [E28] Varga, R. S., Gersgorin and His Circles. New York: Springer, 2004. [E29] Wilkinson, J. H., The Algebraic Eigenvalue Problem. Oxford: Oxford University Press. 1988. Part F. Optimization, Graphs (Chaps. 22-23) [Fl] Bondy, J. A., Graph Theory with Applications. Hobokcn, NJ: Wiley-Interscience, 2003. [F2] Cook, W. J. ct al.. Combinatorial Optimization. New York: Wiley, 1993. [F3| Diestcl, R., Graph Theory. 2nd ed. New York: Springer. 2000. [F4] Diwekar, U.M., Introduction to Applied Optimization. Boston: Kluwer. 2003. [F5] Gass. S. L., Linear Programming. Method and Applications. 3rd ed. New York: McGraw-Hill, 1969. [F6] Gross. J. T., Handbook of Graph Theory and Applications. Boca Raton, FL: CRC Press. 1999. [F7] Goodrich, M. T., and R. Tamassia, Algorithm Design: Foundations, Analysis, and Internet Examples. Hoboken. NJ: Wiley, 2002. [F81 Harary, F., Graph Theory. Reprinted. Reading, MA: Addison-Wesley, 2000. [F9] Merris, R., Graph Theory. Hobokcn. NJ: Wiley-Interscience, 2000. [F10] Ralston, A., and P. Rabinowitz, ,4 First Course in Numerical Analysis. 2nd ed. Mineola, NY: Dover, 2001. [F11J Thulasiraman, K., and M. N. S. Svvarny. Graph Theory and Algorithms. New York: Wiley-Interscience, 1992. [F12] Tucker, A., Applied Combinatorics. 4th ed. Hoboken. NJ: Wiley, 2001. Part G. Probability and Statistics (Chaps. 24-25) [Gl] American Society for Testing Materials, Manual on Presentation of Data and Control Chart Analysis. 7th ed. Philadelphia: ASTM. 2002. |G2] Anderson, T. W., An Introduction to Multivariate Statistical Analysis. 3rd ed. Hoboken, NJ: Wiley. 2003. [G3] Cramer, H., Mathematical Methods of Statistics. Reprinted. Princeton. NJ: Princeton University Press. 1999. [G4] Dodge, Y., The Oxford Dictionary of Statistical Terms. 6th ed. Oxford: Oxford University Press, 2003. [G5] Gibbons, J. D., Nonparametric Statistical Inference. 4th ed. New York: Dekker, 2003. IG6] Grant, E. L. and R. S. Leavenworth, Statistical Quality Control. 7th ed. New York: McGraw-Hill. 1996. [G7] IMSL. Fortran Numerical Library. Houston, TX: Visual Numerics, 2002. [G8] Kreyszig, E., Introductory Mathematical Statistics. Principles and Methods. New York: Wiley. 1970. |G9] O'Hagan, T. et al.. Kendall's Advanced Theory of Statistics 3-Volume Set. Kent. U.K.: Hocldcr Arnold, 2004. [G10] Rohatgi, V. K. and A. K. MD. E. Saleh, An Introduction to Probability and Statistics. 2nd ed. Hoboken, NJ: Wiley-Interscience, 2001. APPENDIX 2 Answers to Odd-Numbered Problems Problem Set 1.1, page 8 1. (cos ttx)Itt + c 3. ex2'2 + c 5. First order 7. Second order 9. Third order 11. v = i tan (2x + nir), n = 0, ±1, ±2, • • • 13. v = 15. (A) No. (B) No. Only y = 0. 17. y" = g, y' = gt, y = gt2/2 19. y" = k, y' = kt + 6, y = |/tr2 + 6t, y(60) = 1800/: + 360 = 3000, k = 1.47, y'(60) = 1.47 • 60 + 6 = 94 [m/sec] = 210 [mph] 21. ekH =\,H= (In = (1011 In 2)/1.4 = 1570 [years] Problem Set 1.2, page 11 11. v = -(2/tt) cos \ttx + c 15. v = x(l - In x) + c 17. Verify the general solution y2 + i1 = c. Circle of radius 3V2 19. mtr- = mg - /w2, u' = 9.8 - u2, u(0) = 10. v' = 0 gives the limit V9!8 = 3.1 [meter/sec]. Problem Set 1.3, page 18 3. cos 2y dy = 2 dx, y = ^ arcsin {Ax + c) 5. y2 + 36.x2 = c, ellipses 7. dy/y = cot 7rx dx, y = c(sin -n-x)1'" 9. y = tan (c - e~"xlTf) ll.r=roe-*2 13. / = /0e-Rl/L 15. y = exNlx + 5 17. y = 4 In jc 19. y = Vln (x2 - 2x + e) 21. y' = (y - fo)/(.v - a), y - = c(x - a) 23. y0(?'c = 2y0, ek = 2 (1 week), '/x2 clx + \lx dy = d(ylx) = 0. A«.v. v = ex 13. -3y2/x4 rfx + 2ylxa dy = d(y2/x3) = 0. y = cv3/2 (semicubical parabolas) IS. Exact, a = e2** cos y + k(y), uy = —e2x sin v + k', k' = 0. Ans. e2x cos y = c, c = 1 17. Not exact. Try R. F — e~x, e~x(cos cox + •), Ma = ik' = -1 + ey, k = -y + ey. Ans. ex - y + ey = c 21. U = C, \Ax2 + Cxy + \Dy2 = c Problem Set 1.5, page 32 3. y = ce~z-5x + 0.8 5. y = 2.6e~1-2ax + 4 7. y = x + c (if it = 0). y = ce~fcT' + e2kxBk if ik # 0 9. Separate, y - 2.5 = c cosh4 1.5x 11. y = 2xec006*20 - 1/0.06] = 41.988732y0 = 178 076.12, y0 = k = $4241.05. 27. y' = 175(0.0001 - y/450), y(0) = 450 ■ 0.0004 = 0.18, y = 0.135 0, B > 0. Constant solutions y = 0, y = AIB. y' > 0 if y > AIB (unlimited growth), y < 0 if 0 < y < AIB (extinction). y = AI{ceAl + ß), y(0) > AIB if c < 0, y(0) < A/ß if c > 0. 33. y' = y - y2 - 0.2?, y = 1/(1.25 - 0.75e-afit), limit 0.8, limit 1 35. y' = y - 0.25y2 - O.ly = 0.25y(3.6 - y). Equilibrium harvest 3.6, y = 18/(5 + ce-0'9') 37. (y, + y2Y + p(yi + y2) = (yL' + pyj + (y2' + py2) = 0 + 0 = 0 39. (>'! + y2)' + p(yx + y2) = fa' + py,) + (y2' + Av2) = r + 0 = r 41. Solution of cyx' + pcyy — c(y/ + py\) = er App. 2 Answers to Odd-Numbered Problems 43. CAS Experiment (a) v = x sin (1/x) + ex. c = 0 if y(2/v) = 2/tt. y is undefined at x = 0, the point at which the "waves" of sin (1/x) accumulate; the factor x makes them smaller and smaller. Experiment with various ^-intervals, (b) y = xm[sin (1/x) + e]. y(2/ir) = (2/'tt)""'. n need not be an integer. Try n = |. Try n — 1 and see how the "waves" near 0 become larger and larger. 45. y = uy*, y + py = u y* + uy*' + puy* = u'y* + u(y*' + py*) = ay* + u • 0 = r, u = rly* = reSp *», u = J efp dx r dx + e. Thus, y = wyh gives (4). We shall see that this method extends to higher-order ODEs (Sees. 2.10 and 3.3). Problem Set 1.6, page 36 1. y = 4, y = -1/4, y = -x!4 + c* 3. ylx = c, y'lx = y/x2, y' = y/x, y = -xly, v2 + x2 = c*, circles 5. 2xy + x2y' = 0. v' = 2y/x. y = x/(2y), y2 - x2/2 = c*, hyperbolas 7. ye-*12 = c, / = xy, y' = -l/(xy), yy' = -1/x, y2/2 = -In |x| + C**, x = c*e~y2r2, bell-shaped curves (with x and y interchanged) 9. y = — 4x1 y, y' = y/4x, 4 In |y| = In \x\ + e**, x = c*y4, parabolas 11. xe-ylA = c, y' = 4/x, y = -x/4, y = -x2/8 + c* 13. Use dyldx = \l(dxldy). (y - 2x)ex = c, (y' - 2 + y - 2x)ex = 0, y' = 2 — y + 2x, dxldy = — 2 + y — 2x is linear, dxldy + 2x = y - 2, x = c*e_2y + y/2 - 5/4 15. m = c, m.,.Jx + Mydy = 0, y = —ujuy. Trajectories y = iiy/ux. Now v = c*, Ujudx + uac/y = 0, y' = —vx/vy. This agrees with the trajectory ODE in u if wx = vy (equal denominators) and uy = — 6'.r (equal numerators). But these are just __the Cauchy-Riemann equations. 17. 2x + 2yy' = 0, y = —xly. Trajectories y' = y/x, In |y| = In |x| + c**, y = c*x. 19. y' = -4x/9y. Trajectories y' = 9y/4x. y = c*xm (c* > 0). Sketch or graph these curves. Problem Set 1.7, page 41 1. In |x — Xq| < a; just take b in = /j/A^ large, namely, /? = aA". 3. No. At a common point (xi, y'i) they would both satisfy the "initial condition" yiXf) = yL, violating uniqueness. 5. y' = fix, y) = r(x) — p(x)y; hence df/dy = —p(x) is continuous and is thus bounded in the closed interval \x — x0| s a. 1. R has sides 2a and 2b and center (1,1) since y(l) = 1. In R, f = 2y2 g 2(b + l)2 = K, a = WJf = bl(2{b + l)2). rfa/rffo = 0 gives b= 1, and aopt = bIK = 1/8. Solution by dyly2 = 2 t/x, etc., y = 1/(3 - 2x). 9. |1 + y2| g ^ = 1 + b2, a = &/A", rfa/J/? = 0, fe = 1, a = 1/2. Chapter 1 Review Questions and Problems, page 42 11. dy/(y2 + |) = 4 r/x, 2 arctan 2y = 4x + c*, y = | tan (2x + c) 13. Logistic ODE. y = 1/m, y' = -u'/u2 = 4/k - 1/w2, u = c*e"te + | 15. dyl(y2 + 1) = x2 dx, arctan y = x3/3 + c, y = tan (x3/3 + c) 17. Bernoulli, y + xy = xly, u = y2, m = 2yy' = 2x — 2xu linear, u = e~xA (JexA 2x dx + c) = ] + ee~x\ y = vu. Or write yy = — x(y2 — 1) and separate. App. 2 Answers to Odd-Numbered Problems A7 19. Linear, y = * siaxdx + c) = cecos* + 1. Or by separation. 21. Not exact. Use Theorem 1, Sec. 1.4; R = 2/x, F = x2; the resulting exact ODE is 3x2 sin 2y dx + 2x3 cos 2y dy = d(x3 sin 2y), .v3 sin 2y = c. Or by separation, cot 2y dy = — 3/(2x) dx, etc., sin 2y = ca-3. 23. Exact, u = J M dx = sin at - x2 + k, «„ = x cos at + k' = N, k = y2, sin at - x2 + v2 = c. 25. Not exact, i?* - 1 in Theorem 2, Sec. 1.4, F* = Exact is '<*, '! = e~x, y2 = O.GOle* + e~x 35. Write £ = e~ax'2, c = cos wa, s = sin wx. Note that E' = -\aE, c = -w.v, s' = wc. Substitute, drop E, collect c-terms, then ^-terms, and use of = b — \a2, to get c(b - \a2 + \a2 - w2) + s{-a• = t'"3-^^ cos 2a- + B sin 2x) 13. y = A cos 4.2a>x + B sin 4.2g« Problem Set 2.4, page 68 1- y = )'o cos wo? + (.Vq/wq) sin w0r. At integer / (if cü0 = it), because of periodicity. 3. mLd" = —mg sin 0 = —ingO (tangential component of W = mg), ß" + co02ö = 0, co0/(2ir) = VglLlilTT). 5. No, because the frequency depends only on klm. 7. (i) Greater by a factor V'3. (ii) Lower 9. co* = [«02 - c2/(4m2)]U2 = oi0[] - c2/(4mk)]V2 « w0(l - c2/8mfe) = 2.9583 11. 2Wto* since Eq. (10) and y' = 0 give tan { Äerit = 2VI/C is Case I, etc. 11. 0 13. Cle-ZOi + c2e~lm + 16.5 sin 10? + 5.5 cos 10? 15. E' = -e'4t(7.605 cos \x + 1.95 sin \t), I - 'pl as yp in Example 1, yp2 = 24 sin 5x 17. u" + 1.1 — 0 by substitution of y = ?'i = ex cos x, y2 = ex sin x, W = 0, linearly dependent 13. Linearly independent 17. Lineaily independent 9. Linearly dependent 15. Linearly independent 19. Linearly dependent Problem Set 3.2, page 115 1. /" - 6y" + 11/ - 6y =0 3. >•"' - y = 0 5. yw + 4y" = 0 7. C\ + c2 cos x + c3 sin x 9. cxex + ic2 + csx)e-x 11. cxex + c2ca+v¥)* + c3el 13. e0-25* + 43e~°-lx +12.1 cos O.lx - 0.6 sin O.lx 15. 2.4 + f-lfia:(cos 1.5x - 2 sin 1,5*) 17. y = cosh 5.v — cos 4x 19. y = r,x 2 + c2x + c3x2. VV = 12/x2 Problem Set 3.3, page 122 1. (d + c2x)e2x + c3^2:r - 0.04e"3x + x2 + x + 1 3. c, cos |x + c2 sin |x + x(c3 cos \x + cA sin \x) — \e~x sin |x 5. cix05 + c2x + c3x15 + O.lx55 7. cx cos x + c2 sin x + c3 cos 3x + c4 sin 3x + 0.2 cosh 2x 9. y - (4 - x2)e3'' - 0.5 cos 3x + 0.5 sin 3x 11. x-2 - x2 + 5x4 + x(ln x + 1) 13. 3 + 9er2x cos 9x - (1.6 - \.5x)ex Chapter 3 Review Questions and Problems, page 122 7. ct + c2xV2 + c3x~1/2 9. c,e-^x + c2e05* + c3'[ = J2, y'z = 4}'!, >'! = cLe~2t + c2e2t = y, y2 = yi 13. yi = y2, y2 = y2, eigenvalues 0, 1, y, = c\ + c2ef, y2 = yi = y' 15. yi = y2, y2 = 0.109375.v! + 0.75y2 (divide by 64), yi = Cle_0125* + c2e0815' Problem Set 4.3, page 146 1. >, = Cle-6t + c2enl\ y2 = -2c,e-6' + 2c2em 3. yT = c\e2t + c2, y2 = c^e2' - c2 5. y, = c\em + c2e-4'* = (c1 + c2) cos 4/ + ř(ci - c2) sin At = /l cos 4? + B sin 4/, y2 = k^il - ic2e~4it — (fcj — ;'c2) cos 4r + /(/ci + ;V:2) sin At = B cos 4? — v4 sin At, A = c3 + c2, ß - í(Cl - c2) 7. v-! = 2C] + c2f"6t, v2 = -Cl + c3e"6t, y3 = -c, + 2(c2 + c3)e"6f 9. Vl = CleL8t + 2c2e^-91 + 2c2e~x%t, y2 = 2Clelst + c2e~0M - 2c3e-1M, y3 = 2c1eliit - 2c2e-°M + c^"18* 11. yx = 10 + 6e2', y2 = -5 + 3e2t 13. y1 = 2Ae.-t - 2e2M, y2 = 1.8c-ť + 2e25t 15. .Vl = 2e14'r"' + 10, y2 = 5e14J5i - 4 17. y2 = yi + yi, y'2 = yi + y[ = -yi - y2 = ->'i - (yi + yi), y" + 2y[ + 2yL = 0, y\ = e~\A cos / + B sin f), y2 = y'i + >'i = e~\B cos t - A sin t). Note that r2 = Vi2 + V22 = e"2t(^2 + B2). 19. I, = 4Cle-2m + c2e~50t, I2 = -clťT200í - 4c23t, y2 = —c2e* 4- c2ď3í 5. Stable and attractive node, yx = t^e-34 + c2e~r>t, y2 = c^-34 - c2e~54 7. Center, stable, y-, = A cos 4r + B sin 4i, y2 = -2B cos 4r + 2A sin 4/ 9. Saddle point, unstable, y1 = c1e3i + c2e~\ y2 = cx'! = y = c1efet 4- c2e"'rt, y2 = y', hyperbolas Z:2)'!2 - y2 = const 13. y = e_2404 cos ř + B sin f), stable and attractive spirals 17. For instance, (a) -2, (b) -1, (c) -\, (d) 1, (e) 4. Problem Set 4.5, page 158 1. (0, 0), yi = y2, y2 = 3yls saddle point; (0, -1), yt = % y2 = -1 + y2, y[ = -y2, % = 3yl3 center 3. (0, 0), yi = 4y2, y'2 = 2yt, saddle point; (2, 0), yx = 2 + yl5 y2 = y2, yi = 4y2, y2 = —2^, center 5. (0, 0), yi = -yj + y2, y2 = -y2 - y2, stable and attractive spiral point; (-2, 2), >'i = -2 + yi, y2 = 2 + y2, yi = -yx - 3y2, y2 = -y, - y2, saddle point App. 2 Answers to Odd-Numbered Problems 7- >'i = )'2- % = _.Vi() _ 4>'i)' (0, 0), y[ = y2, };2 = ~yi, center; (l, 0), >'! = | + ylt j2 = y2, = y2, % = (-| - y1)(-4j1), y:', = y1; saddle 9. (gTi ± 2n7T, 0) saddle points; (—\tt ± Inir, 0) centers. Use —cos (±f7T + Jt) = sin (±yi) ~ ±.yi-11. yi = y2, y2 = -v,(2 + y1)(2 - (0, 0), y'2 = -4yu center; (-2, 0), y2 = 8&, saddle point; (2, 0), y2 = 8yx saddle point 13. y"ly' + 2y'/y = 0, In y + 2 In y = c, y'y2 = y2y\2 = const 15. y = A cos f + £ sin t, radius Va2 + B2 Problem Set 4.6, page 162 3. y\ = A cos At + B sin At + §§, y2 = 5 cos At — A sin 4* - fr 5. yx = Cle4t + c2e~3t + A, y2 = ty4* - 2.5c2£>-3t - 10 7. yx = 2c1<^9t + c2e'^ - 90? + 28, y2 = c^-9' + c2e~4t - 126r + 14 9. v, = c-y + 4c2e2t 3t 4 2e~\ y2 = -cxel - 5c2e2t + 5/ + 7.5 + e~l 11. y, == 3 cos 2r - sin 2r + r + 1, y2 = cos It + 3 sin 2/ + 2/ - | 13. y1 = 4e_t - 4e* + e2t, y2 = -4(Tf + t 15. y'! = 7 - 2e2t + e3t - Ae~3\ y2 = -e2t + 3e~3t 17. /{ + 2.5(4 - 4) = 845 sin t, 2.5(72 - j[) + 25/2 = 0, lx = (95 + 162.5/)e_5t - 95 cos t + 312.5 sin t, I2 = (-30 - 162.5/)«"r" + 30 cos t + 12.5 sin t 19. /{ + 2(/i - I2) = 200, 2(/2 - 4) +8/2 + 2j/s*= 0. h = 2c ^ + 2c2e^t + 100, I2 = (1.1 + Va4T)c1eAlt + (1.1 - V0Al)c2e^\ A: = -0.9 + V041, A2 = -0.9 - V'0.41 Chapter 4 Review Questions and Problems, page 163 11. y: = c^ + c2e~*\ y2 = 2c!e8t - 2c2e"8t. Saddle point 13. j>] = Cje* + c2'i = 1 + — + — +■•• = - , y2 =- + — + — + ■■■= — 3! 5! x x 2! 4! x ~ . v2 1 144 36 x2 25x4 3-yi = 1 " 72 + ^84* ••••V-9V'lnA .v- - ^ + T T024 + 1 x y3 5. r(r - 1) + 4r + 2 = 0, r2 = -1, r2 = -2; v, =--- +---+ • ■ • ■ x 6 120 1 1 x2 x4 y2 = -Tj--H-----[--•■■ '2 x2 2 24 720 7. Euler-Cauchy equation with r = x + 3, y1 = (x + 3)5, y2 = y: In (x + 3) 9.b0= 1, c0 = 0, r2 = 0, yi = «T* y2 = e~* In x 11. y, = l/(x + 1), y2 = 1/x 13. b0 = h2,c0 = 0, n = i r2 = 0, y, = x1/2(l + 2x + 2x2 + |xB + ■••), y2 = 1 + 2x + 2x2 + • • • 15. yx = (x - 4)7, y2 = (x - 4)~5 (Euler-Cauchy with t = x - 4) I 6 17. y, = x + x3 - ix4 + 4x5 - ^x6 + • • y2 = l + 3x2 - ±x3 + fx4 ^-x6 - + • A14 App. 2 Answers to Odd-Numbered Problems 19. v = c^Xi i 5; x) + c2VxF(l, 1, |; x) 21. v = A(l - 4x + fx2) + ÄVxF(-§, §,1; x) 23. y = ClF(2, -2, -1; f - 2) + c2(» - 2)3/2F(l, -J, f; r - 2) Problem Set 5.5, page 197 1. Use (7b) in Sec. 5.2. 3. 0.77958 (exact 0.76520), 0.19674 (0.22389), -0.27651 (-0.26005), -0.39788 (-0.39715), -0.17038 (-0.17760), 0.15680 (0.15065), 0.30086 (0.30008), 0.16833 (0.17165) 5. y = dJu(te) + c2i_„(Ax), v * 0, ±1, ■ • • 7. y = ClJ„(Vx) + c2J_v{Vx).. v # 0, ±1, • ■ ■ 9. y — c1x/1(2x), Ju J_1 linearly dependent 11. y = x~v[CiJj[x) + c2J_v(x)\, v + 0, ±1, • • • 13. y = CiJJx3) + c2/_„(x3), v * 0, ±1, • • • 15. y = c1Vx/1(2Vx), JL, linearly dependent 17. y = x1/4/x(|x1/4), y1( /_! linearly dependent 19. y = x2/5(c1./8/5(4x1/4) + 62./_8/5(4x1/4)) 21. Use (24b) with v = 0, (24a) with v = 1, (24d) with v = 2, respectively. 23. Jn(x{) = JnXx2) = 0 implies x1_n/„(x1) = x2~nJn(x2) = 0 and [x_n/n(x)]' = 0 somewhere between xx and x2 by Rolle's theorem. Now use (24b) to get Jn+,(x) = 0 there. Conversely, Jn+](x3) = in+1(x4) = 0, thus x3K+1/n+|(xH) = x4*l+1/K+1(x4) = 0 implies Jn{x) = 0 in between by Rolle's theorem and (24a) with v = n + 1. 25. Integrate the formulas in (24). 27. Use (24a) with v = 1, partial integration, (24b) with v — 0, partial integration. 33. CAS Experiment, (b) x0 = 1, x, = 2.5, x2 = 20. approximately. It increases with n. (c) (14) is exact, (d) It oscillates, (e) Formula (24b) with v = 0 Problem Set 5.6, page 202 l.y = ClJ5(x) + c2Y5(x) 5. y = ClJ2(x2) + c2F2(x2) 9. y = x3(Cl/3(x3) + c2F3(x3)) 13. Set x = « in (1), Sec. 5.5, to get Sec. 5.5. 3.y = CjJ0(Vx) + c270(Vx) 7. y = x-5(Cl75(x) + c2Y5(x)) 11. Set H(1) = /t//<2), use (10). present ODE (12) in terms of s. Use (20), Problem Set 5.7, page 209 3. Set x = ct + k. 5. x = cos 9, dx = —sin 9 d9, etc. 7. Am = (m7r/5)2, m = 1, 2, • ■ • ; ym = sin (mirx/5) 9. Am = [(2m + l)ir/2L]2, m = 0, 1, • • • ; yTO(x) = sin [(2m + 1)ttx/2LJ 11. A,m = m2, m = 0,1, • * ■ ; y0 = 1, ym = cos mi, sin mx, m — 1, 2, ■ • ■ 13. k = k,m from tan k = —k. Am = &m2, m = 1, 2, ■ ■ • ; ym = sin fcmx 15. A„, = m2, m = 1, 2, • ■ ■ ; ym = x sin (m In |x|) 17. p = e8x, q — 0, r = e8x, Am = m2; ym = e~4x sin mx, m = 1, 2, ■ ■ • 19. A,re = {mir)2, y.m = x cos m7Tx, x sin m-7rx, m = 0, 1, • • • App. 2 Answers to Odd-Numbered Problems A15 Problem Set 5.8, page 216 1. 1.6P4(x) - 0.6P0(x) 3. §P3(x) - §P2(x) + g/^Cx) - 4P0(x) 7. -0.4775/51(x) - 0.6908P3(x) + 1.844P5(x) - 0.8234P7(x) + 0.1544P9(x) + ■ • • , n/0 = 9. Rounding seems to have considerable influence in Probs. 6-15. 9. 0.3799P2W + 1.673P4(x) - 1.397P6(x) + 0.3968P8(x) + • • • , m0 = 8 11. 1.175P0(x) + 1.104P1(x) + 0.3575P2(x) + 0.0700P3(x)----, = 3 or 4 13. 0.7855Po(x) - 0.3550P2(x) + 0.09()0P4(x) - • • - , m0 = 4 15. 0.1212PoW - 0.7955P2(x) + 0.9600P4(x) - 0.3360P6(x) + • ■ • , m0 = 8 17. (c) am = (2//12(a0.m))(./1(a0,m)/a()m) = 2/(a0,m./1(a0,n)) Chapter 5 Review Questions and Problems, page 217 11. e3x, e~3x, or cosh 3x, sinh 3x 13. ex 1 + x 15. e~:"'2, xe~x2 17. e~x, e~x In x 19. 1/(1 - x2), x/(l - x2) or 1/(1 - x), 1/(1 4- x) 21. y = c1Jvi(6x) 4- <:,/_x-(6.vl 23. y = c^.ix2) + c^Y^x2) 25. y = \rx[cvIv^kx2) + c2./_1/4(|fo-2)] 27. ATO = (2mTr)2, y0 = 1, ym = cos 2mirx, sin 2rmrx, m = 1, 2, • • ■ 29. y = C\J\(kx) + c^Y^kx), c2 = 0, v(l) = c1/1(fe) = 0, k — km = ax m (the positive zeros of /J, >•„,, = ./^a, mx) 31. 1.813P0(x) 4- 2.923f1(x) 4- 1.759P2(x) 4- 0.663P3(x) 4- 0.185P4(x) 4- • ■ • 33. 0.693P0(x) - 0.285P2(x) 4 0.144P4(x) - 0.091P6(x) 4- • • • 35. 0.25P0(x) + 0.5Pv(x) + 0.3l25P2(x) - 0.0938P4(x) 4- 0.0508P6(x) + • • ■ Problem Set 6.1, page 226 2 2 .v s-2 1 — — 3 - 5 - " s3 s2 ' s2 4- 4tt2 ' (s - 2)2 - 1 s cos 6 — co sin 0 e3a 1 7- -2-2- 9- - 11. -5- sz + oj2 s 4- 2/? 4- 4 19.--— 21. = 2(1) - 2 2a2 1 - 3 - 5 - (s - kf ' s(s2 + 4u>2) ' s(s2 - 4a2) its 1 - (s + -gTT ) 9. Use shifting. Use cos2 a = \ + § cos 2a; use cos2 a + sin2 a = 1. Am\ (2.?2 + l)/[2s(s2 + 1)] 11. 0 + |)F = -1 + 17 • 2/U2 + 4), y = 7e-m + 2 sin 2t - 8 cos 2? 13. (s2 - |)7 = 4s, y = 4 cosh I? 15. (s2 + 2s + 2)7 = s - 3 + 2 • 1, 7 = (s + 1 - 2)/[(s + I )2 + 1], y = e-t(cos f — 2 sin t) 17. (s2 + 7s + 12)7 = 3.5.v - 10 + 24.5 + 21/0 ~ 3), y = \eAt + §<>-4t + ^e~3' 19. + 1.5)2K = s + 31.5 + 3 + 54/.v4 + 64/.?, Y = 1/(5 + 1.5) + 1/(5 + 1.5)2 + 24/,v4 - 32/.v3 + 32/.v2, y = (1 + t)e-1M + 4t3 - \6t2 + 32t 21. t = t + 2, Y = 4l(s - 6), y = 4e6t, y = 4e6(T~2) 23. t = t + 1, (s - 1)0 + 4)Y - 4s + 17 + 670 - 2), y = 3et~1 + e2a_1) 25. (b) In the proof, integrate from 0 to a and then from a to x and see what happens, (c) Find ££(/) and it(j') by integration and substitute them into (1*). 27. 2 - 2e~m 29. (ekt - 1) - - 31. cosh V5/1 - 1 k k 33. | sinh 2? - |r 2s 5\ «. / 1 10 Problem Set 6.3, page 240 3. (1 - e2-2s)l(s - 1) (2 2 1 \ / 2 4 4, 7- (-e_s - ^ M;- •i) ' \s2 s, 11. T20*, (f_3s + e_6s) 13. —— (e~2s+2^ - eris 47T) s + tt s - tt 15. 0 if t < 4, t - 4 if t > 4 17. sin / if 2tt < f < 8tt, 0 elsewhere 19. 0 if t < 2, (t - 2)4/24 if t > 2 21. u(t - 3) cosh (2r - 6) 23. e~l sin t 25. e~2t cos 3r + 9 cos 2t + 8 sin 2? 27. sin 3f + sin t if 0 < t < tt and § sin 3t if r > tt 29. r - sin t if 0 < / < 1, cos (f - 1) + sin (r - 1) - sin t if f > 1 31. ef - sin / + u(t - 2ir)(sin r - g sin 2t) 33. t = 1 + J, y" + 4y = 8(1 + t )2(1 - w(? - 4)), cos 2t + 2t2 - 1 if r < 5, cos 2/* + 49 cos (2t - 10) + 10 sin (2t - 10) if t > 5 35. fty' + r//C = 0, Q = S£(q), q(0) = CV0, i = q'(t), R(sQ - CV0) + QIC = 0, q = CV0e-tma 100 100 , , / 1 1 \ 37. 10/ + -/ = e'2\ I = e_2s---, i = 0 if t < 2 and .9 s2 \ s s + 10 / 1 - e-10«-2)if/> 2 App. 2 Answers to Odd-Numbered Problems A17 39. i = e-20t + 20t - 1 + u(t - 2)[-20/ + 1 + 39e-20ct-2)] 41. 0.1/' + 25/ = 490e-5*[l - u(t - 1)]. i = 20(e_5t - 2 + 2,v + 2), / = e"* sin t - M(f - 2) e"t+2 sin (f - 2) 47. i = 27 cos / + 6 sin t - e~\21 cos 3f + II sin 3;) + u(t - 2tt) [-27 cos t - 6 sin t + e~ct_2ir)(27 cos 3/ + 11 sin 3?)] Problem Set 6.4, page 247 l. v = 10 cos / if 0 < t < 2tt and 10 cos t + sin f if f > 2tt 3. y = 5.5/ + 4.5 tt 23. 0 if 0 < t < 1, § [ sin (2(t - 1)) = -| cos (2/ - 2) + | if t > 1 25. y = 2e"2* - e"4' + (>! = 4 cos 5r + 6 sin 5/ - 2 cos r - 25 sin t, y2 = 2 cos 5t - 10 sin 5t + 20 sin f 11. Vj = -cos f + sin t + 1 + u(t - !)[-] + cos (/ - 1) - sin (f - 1)] y2 = cos t + sin t -1 + w(? - 1)[1 - cos (r - 1) - sin (t - 1)| e* + iu(t - 1)(- 13. Vi = 2«(f - 2)( = — e* l 49. yj = e2t, y2 = e2t + e< e l(cos 2/ ^ sin 2/) = e2t + e' 2s')/[^ + 10)1, i = 0.1(1 - e~w 15. e-™ I — 21. a — b (s — ä)(s — b) 27. u(t - 2)(5 + 4(1 - 2)) 77 33. -g ( = 300 + x4, x3 = 800 — x4, x4 arbitrary Problem Set 7.4, page 301 -2];[1 0 -3]T 4 0 7], [0 1]T 0 5], [0 3 0 4], [0 2 1. 1, [1 3. 3, [1 [0 0 5. 2, [3 7. 2, [8 9. 3, [1 11. 4, [1 13. No 19. Yes 29. No 35. 1, [5 -2 1 3], [0 0 5 1051; [~2 4 5JT, [0 1 5]T, 4]; [3 0 5]T, [0 3 0]; [8 0 4 0]T, [0 0 3 0 0 0], [0 0], [0 8 -37], [0 0 0 0], [0 0 1 15. No 21. (c) 1 31. 1, \-i 4]T 2 0 4]T 74 296]; same transposed 0], [0 0 0 11; same transposed 17. Yes 27. 2, [1 -1 0], [0 0 1] 1] 33. No Problem Set 7.7, page 314 5. 107 9. -66.88 13. u3 + v3 19.x = 23. 3 w — 3uvw 1.2, y = 0.8, z = 3. 7. cos (a + ß) 11. 0 15. 4 21. 1 App. 2 Answers to Odd-Numbered Problems Problem Set 7.8, page 322 I. 1.80 -2.32 _-0.25 0.60 1 0 0" -2 1 0 3 -4 1 cos 2d sin 2d -sin 26' cos 20 9. A~ '2 0 5 1 .0 1 11. No inverse 15. (A2)- (A )2 4 15 -1 -5 I -5 4 9. 19. AA"1 = I. (AA-1)-1 = (A-1)-^-1 = I. Multiply by A from the right. 21. del A 23. det A 1. C, C-12 — C21 — C: 1. c,9 33 — -2, Ca the other Q> are zero C 3, C -4, C 33 Problem Set 7.9, page 329 1. Yes, 2, [3 5 0]T, [2 0 -5|T 3. No 5. Yes, 2, [0 0 0 1 0]T, [0 0 0 0 1 ]T 0 1" 7. Yes, 1. 9. No .-1 0. 11. Yes, 2, xe~x, e~x 13. [1 0|T, [0 1]T: [1 1]T, [-1 If; [1 Of, [0 15. jtj = -0.6yx + 0.4y2 x2 = -0.8.V! + 0.2y2 19. xj = 5y1 + 3y2 — 3ya x2 = 3>i + 2y2 - 2y3 *3 = 2>'1 ~ >'2 + 2j3 21. V56 25. 2 If 17. X] = 2y1 + y2 x2 = 5\\ + 3y2 23. 16 V5 29. 4«! - 3y2 = 0, v = ± [| Chapter 7 Review Questions and Problems, page 330 11. x = 4. y = 7 I5.* = iy=-J,z = § 19. x = 2z, y = 4, - arbitrary 13. x = y + 6, z = y, y arbitrary 17. x = 7, y = -3 21. 0 8.0 -3.6 23. 638, 0, 0 25. -3.6 1.2 2.6 2.4 1.2 2.4 9.0. "12 0 6" ^-20 27. 14. 14, 28 0 14 29. [-20 9 -3], 9 _ 4 0 2_, -3 A22 App. 2 Answers to Odd-Numbered Problems 31. 2, 2 35. 2, 2 33. 2, 2 37 J- ■"• 51 4 _-5 "-5 10 5" " 72 -72 132 j^. 45 11 5 -2 *a* 12 31 -32 59 . 23 -10 4_ .-19 20 -35 43. Ix = 33 A, I2 = 11 A, Ig = 22 A 45. Ix = 12 A, /2 = 18 A, I3 = 6 A Problem Set 8.1, page 338 1. -2, [1 0]T; 0.4, [0 1]T 3. 4, 2xx + (-4 - 4)x2 = 0, say, xx = 4, x2 = 1; -4, [0 1]T 5. -4, [2 9]T; 3, [1 1]T 7. 0.8 + 0.6/, [1 -i]T; 0.8 - 0.6;, [1 ;]T 9. 5, [1 21T;0, [-2 1]T 11.4, [1 0 0]T; 0, [0 1 0]T; -1, [0 0 if 13. —(A3 - 18A2 + 99A - 162)/(A - 3) = -(A2 - 15A + 54); 3, |2 -2 1 ]T; 6, [1 2 2]T; 9, [2 1 -2]T 15. 1, [-3 2 10]T;4, [0 1 2]T; 2, [0 0 1]T 17. -(A3 - 7A2 - 5A + 75)/(A + 3) = -(A2 - 10A 5, [3 0 1]T, [-2 1 0]T 19. -(A - 9)3; 9, [2 -2 1]T; defect 2 21. A(A3 - 8A2 - 16A + 128)/(A - 4) = A(A2 - 4A - 32); 4, [-1 3 1 1]T; -4, [1 1 -1 -1]T;0, [1 1 1 1]T; 8, [1 -3 1 -3]T 23. 2, [8 8 -16 1]T; 1, [0 7 0 4]T; 3, [0 0 9 2]T, -6, [0 0 0 1]T 25. (A + 1)2(A2 + 2A - 15); -1, [1 0 0 0]T, [0 1 0 Of; -5, [-3 -3 1 1]T, 3, [3 -3 1 -1]T 29. Use that real entries imply real coefficients of the characteristic polynomial. 25); -3, [1 2 -1]T; Problem Set 8.2, page 343 ~-l 0 "l "0" 1. ; -1, ; l, 0 1_ _0_ .1 ; any point (x, 0) on the x-axis is mapped onto (—x, 0), so that [1 0]T is an eigenvector corresponding to A = —1. A point on the x-axis maps "1 0 "l "0" 3. (x, y) maps onto (x, 0). ; l, ; o, .0 0_ .0. .1. onto itself, a point on the y-axis maps onto the origin. 5. (x, >•) maps onto (5x, 5_y). 2X2 diagonal matrix with entries 5. 7. -2, [1 -1]T, -45°; 8, [1 11T, 45° 9.2,|3 -1JT, -18.4°; 7, [1 3]T, 71.6° 11. 1, [-1/V6 l], 112,2°; 8, [l 1/VrjJ, 22.2° 13. 1, [1 1]T, 45°; -5, [1 -1]T, -45° 15. c[15 24 50]T, c>0 17. x = (I - A)_1y = [0.73 0.59 1.04]T (rounded) 19. [1 1 1]T 21.1.8 23.2.1 App. 2 Answers to Odd-Numbered Problems A23 Problem Set 8.3, page 348 3. No 5. A-' = (-AT)_1 = -(A_1)T 7. No since del A = det (AT) = det (-A) = (-1)3 det A = -det A = 0. 9. Orthogonal, 0.96 ± 0.28* 11. Neither, 2, 2, defect 1 13. Symmetric, 9, 18, 18 15. Orthogonal, 1, i, -i 17. Symmetric, a + 2b, a — b, a — b Problem Set 8.4, page 355 "1 2" 7 0" 1. [1 2]\ [2 - if; x = 2 -1_ .0 2_ ^4 0" 3. [1 -1]T, [1 1]T, D = .0 6. 5. [2 -UT, [2 If: diag ( -2, 4) 7. [1 0 0]T, [1 -2 If . to 1 Of, diag (1, 2, 3) 9. [0 3 2f, [5 3 Of, [1 0 2f. diag (45, 9, -27) "16 -7 "1 "l -2 0" 13. ; -5, ; 2, ; x = _42 -19. _3_ _2_ 0_ _-l_ " -30 -72" 9 " 12 -3 n-3 15. ■ 2, ; 0, ; x = 40 3 32_ _-4j _-5_ - 6_ "-95 18 -144" r 4" r 6" " 3" 17. 24 -2 36 ; 4, — 2 ; -2, -1 ; 1, 0 66 -12 100_ .-3. _-4j _-2_ "-2" r o" " 0" x = -4 > — 2 » 0 -6 -4 -2 19. C 21. C 23. C 25. C = 27. C 1 12" .12 -6. 3 -4 .-4 -3 4 V3 V3 2 1 -6" -6 1. 12 16 16 12 , lO.v'!2 - 15y5 5, x y, hyperbola 5>'f - 5y22 = 0, x = >>'i2 + 5y22 = 10, x 7yj2 - 5y22 = 35, x 28y!2 - 4y22 = 112, x = 0.8 0.6 0.6 -0.8 2/V5 L/VŠ 1/V5 2/V5 1/2 V 3/2 -VŠ/2 1/2 l/V'2 1/V2" -1/V2 1/V2_ 1/V2 1/V2 1/V2 -1/V2 y, straight lines y, ellipse y, hyperbola y, hyperbola A24 App. 2 Answers to Odd-Numbered Problems Problem Set 8.5, page 361 V2, \ -i i + V2T 3. (ABC)T = CTBTAT = C1(B)A 5. Hermitian, 3 + V2, [-i 1 - V2T; 3 7. Hermitian, unitary, 1. [l < 9. Skew-Hermilian, 5i, [1 0 0]T, [0 I 1]T; -5i, [0 1 -1]T 11. Skew-Hermitian, unitary, i, [1 0 l]T, [0 1 0]T; -/. [1 0 -1]T 13. Skew-Hermitian, —66/ 15. Hermitian, 10 /V'2lT;-l,[l i + 1V2Y Chapter 8 Review Questions and Problems, page 362 3 2" -4 -3 15. 3 2 5 0" A = _-4 -3. _() -7_ 1/3 2/3" "1 2 ( 0" II. A = 2/3 -1/3. 2 l_ _0 9_ 2 9 l 9 o" 4 1 2 ~6 13. 1 y 0 9 A 1 -2 4 = 0 0 2 9 1 9_ -2 4 - 1_ 0 -1.0 4.8 0" 5.0 19. 1.1 Vi2 ,5,-1 1. ellipse 17. 0 -3 0 -2 0 0 0 0 0 0 -4.50' -2.75 4 , 4, 0, Problem Set 9.1, page 370 1. 2, -4, 0; V2Ö; [l/V'5, -2/V5, 0] 5. -8, -6, 0; 10: [-0.8, -0.6. 0] 9. (i §, |); V^7/8 13. [4, -2, 0], [-2, 1, 0], [-1. \ 17. [28, -14,-14] 23. (5.5, 5.5, 0). (i i f) 27. [-8, -2. 4]; V84 31. [-9, 0, 01, [0, -2, 01, [0, 0, -25 25 1 [ 20 V" 2 ' Vi J V 2 01 3. -1, 0, 5; V26; 7. (7, 5. 0); VlO 35 1J. Yes. 33. |p 20 \/2 11. (0, 1,|); V37/2 15. |10, -5, -15] 19. [-2, 1, 8], [6, -3 25. [0, 0, 9]; 9 29. v = [0, 0, -9] q + u 5 1/V26, 0, 5/V26] 24] 45 V2 V2 6. Nothing 37. |w|/(2 sin a) Problem Set 9.2, page 376 ''241 1.4 3. 5. [12, -8. 41, |-18, -9, -36J 9.-4,4 11.-24 17. |a + hf + |a - b|2 = a-a + 2a«b + b'b 19. 0 21. 15 7. 17 15. Use (1) and |cos -y| = 1-(a«a - 2a«b + b«b) = 2|a|2 + 2|b|2 23. Orthogonality. Yes App. 2 Answers to Odd-Numbered Problems 25. 2, 2, 0, -2 27. 79.11° 29. 82.45° 31. 54.74° 33. 54.79°, 79.11°, 46.10° 35. 63.43°, 116.57° 37. 3 39. 1.4 41. If |a| = |b| or if a and b are orthogonal Problem Set 9.3, page 383 1. [0, 0, -10J, [0, 0, 10] 3.|-4,-8,26| 5. [0, 0, -60] 7. -20, -20 9. 240 11. [19, -21, 24], V1378 13. [10, -5, -1] 15. 2 17. 30, -30 19. -20, -20 25. [-2. 2, 0| x [4, 4, 0] = -16k. 16 27. [1, -1, 2] x [1, 2, 3] = [-7, -1. 3], V59 29. [0, 10, 0] X [4, 3, 0] = [0. 0, -40], speed 40 31. |[7, 0, 0] x [1, 1, 0]| = 7 33. |V3 35. [18, 14, 26]; 9x + ly + 13z = c, 9 • 4 + 7 ■ 8 + 13 • 0 = 92 = c 37.16 39. c = 2.5 Problem Set 9.4, page 389 1. Hyperbolas 3. Hyperbolas 5. Circles 7. Ellipses; 288, 100, 409; elliptic ring between the ellipses x2 y2 x2 + —-—r = 1 and -s- + v2 = 1. (2/3r (1/2)2 (4/3 r 9. Ellipsoids 11. Cones 13. Planes 23. [8x. 0, yz], [0, 0, xz], [0, 18z. xy]; [0, z. y], [z, 0, xj, [y, x, 0] Problem Set 9.5, page 398 1. [4 + 3 cos /, 6 + 3 sin t] 3. [2 - t, 0, 4 + t\ 5. [3, -2 + 3 cos t, 3 sin f] 7. [a + 3/, b -It, c + 5f] 9. Wl cos t, sin f, sin f] 11. Helix on (x - 2)2 + (y - 6)2 = r2 13. Circle (x - if + (y + if = 1, z = 5 15. x4 + y4 = 1 17. Hyperbola xy = 1 23. r' = [-5 sin t, 5 cos t, 0], u = [-sin t, cos t, 0], q - [4 - 3w, 3 + 4w, 0] 25. r' = [sinh r, cosh t\, u = (cosh 2t)~V2 |sinh cosh /], q = [| + Aw, % + 5w] 27. Vr'-r' = cosh /, Í = sinh 1 = 1.175 29. Start from r(r) = [t, f(t)]. _ 33. v = r' = [l. It, 0], |v] = Vl + 4r2, a = [0, 2, 0| 35. v(0) = 2u,R\, a(0) = ~w2R\ 37. 1 year = 365 ■ 86400 sec, R = 30 ■ 365 • 86400/2tt = 151 • 106 [km]. |a| = a2R = \\f/R = 5.98 • 10"6 [km/sec2] 39. R - 3960 + 80 mi = 2.133 ■ 107 ft, g = |a| = ofR = \\\2/R, |v| = VgR = Vó.61 • 108 = 25700 [ft/sec] = 17500 [mph| 43. r(/) = [t, yd), 0], r' = [1, y\ 0], r'«r' = 1 + y'2, r" = [0, y", 0], etc. 47. 3/(1 + 9/2 + 9?4) App. 2 Answers to Odd-Numbered Problems Problem Set 9.6, page 403 1. w' = 2V2(sinh 4;)/(cosh 4t)m 3. w' = (cosh r)sinh t_1((cosh2 t) In (cosh t) + sinn2 f) 5. w' = 3(2r4 + r8)2(8f3 + 8r7) 7. e4" sin2 2v, |e4" sin 4u 9. -2(«2 + u2)"3«, -2(«2 + u2r3v Problem Set 9.7, page 409 1. [2x, 2yl 3. [1/y, -x/y2] 5. [y + 2, je - 2] 7. [6, 4, 4| 9. [-1.25, 0] 11. [0, -e] 13. [-4, 2J 15. [-18, 24] 17. [48, -36] 19. [6, 4] 21. [-6, -12] 23. [-0.0015, 0, -0.0020] 27. \a,_b, c] 29. [8, 6, 0] 31. [108, J08, 108] 33. V2/3 35. 7/3 37. 2e2/Vl3 39. fx2 + |v2 - 2z2 41. -4^'3 Problem Set 9.8, page 413 1. 3(x + v)2 3. 2(x + xz + z) 5. (v + x + 1) cos xy 7. 9x2y V 9. [Vl, v2, v3\ = r' = [x\ y', z'l = [y, 0, 0], z = 0, z = c3, y' = 0, y = c2, x = y = c2, x = c2r + cx. Hence as t increases from 0 to 1, this "shear flow" transforms the cube into a parallelepiped of volume 1. 11. div (w Xr)-0 because i\. v2, v3 do not depend on x, y, z, respectively. 13. (b) (JvOx + i.i'r,), + (fv3\ = f[(Vl)x + (v2)v + (t'3y + fxV! + fyv2 + fzv3, etc. (c) Use (b) with v = V^. 15. 4(x + y)/(y - x)3 17. 0 19. exyz(y2z2 + x2z2 + x2y2) Problem Set 9.9, page 416 1. [0, 0, 4x - 1] 3. [0, 0, 2ex sin y] 5. [0, 0, -4y/(x2 + y2)] 9. curl v = [ — 2z, 0, 0], incompressible, v = r' = fx', y', z'] — [0, z2, 0], ' 2 _ 2 2.i x = c1( z = c3, y = z - c3 , y = c3 t + c2 11. curl v - [0, 0, —2], incompressible, x = y, y' = —x, z = 0, z — ca, y dy + x dx = 0, x2 + y2 = c 13. Irrotational, div v = 1, compressible, r = [c^, c2e~f, c3ef] 17. 0, 0, [xy - zx, yz - xy, zx - yz.\ 19. 0, 0, 0, ~2yz2 - 2z.r2 - 2xy2 Chapter 9 Review Questions and Problems, page 416 11.1-1, 9, 24] 13.0,1-43, 54, 3], [43, -54, -3] 15. [0, 0, -740], [0, 0, -740] 17. [-24, 3, -398], [114, 95, -76] 19. -495, -495 21. 90°, 95.4° 23. If u X v = 0. Always 25. [vx, v2, -3] 27. 3.4 29. If y > \-n, \n A27 33. 45/6 35. No 37. 0, 2y2 + (z + xf 39. [-1, 1, -1], [-2z, ~2x, -2y] 41. 0, 2x2 + 4y2 + 2z2 + 4xz 43. 488/V3323 45. 0 Problem Set 10.1, page 425 1. F(r(f)) = [125f6, I3, 0], 16448/7 = 2350 3. 0 + 160 5. F(r(/)) = [cosh t sinh2 f, cosh2 t sinh f], 93.09 7. F(r(/)) = [?, cos t, sin r], 6tt 9. F(r(/)) = [cosh \t, sinh \t, em], 0.6857 11. F(r(f)) = [4)/(48(a - b)) Problem Set 10.4, page 444 1. 2x3y - 2xy3, 81 - 36 = 45 3. 3x2 + 3y2, 1875 77/2 = 2945 5. e*-» - ex+y, -±e3 + \e2 + e_1 - § 7. 2x - 2y, -56/15 9. 0 (why?) 11. Integrand 4. Ans. 40tt 13. y from 0 to \x, x from 0 to 2. Ans. cosh 2 — \ sinh 2 15. y from 1 to 5 - x2. Ans. 56 19. 4e4 - 4 Problem Set 10.5, page 448 1. Straight lines, k 3. x2la2 + y2/b2 = 1, ellipses, straight lines, [—b cos v, a sin u, 0] 5. z — (c/a)V x2 + y2, circles, straight lines, [—acu cos y, —acw sin u, «2m] 7. x2/9 + _y2/16 = z, ellipses, parabolas, [ — 8m2 cos v, —6u2 sin v, I2u\ 9. x2/4 + y2/9 + z2/16 = 1, ellipses, [12 cos2 v cos u, 8 cos2 v sin u, 6 cos v sin i>] 13. [10m, IOv, 1.6 - 4« + 2d], [40, -20, 100] 15. [—2 + cos v cos u, cos u sin u, 2 + sin u], [cos2 v cos «, cos2 u sin u, cos u sin v] App. 2 Answers to Odd-Numbered Problems 17. [u, v, 3o%[0, -6v, 1] 19. [it cos V, 3h sin v, 3u], [ —9u cos v, — 3u sin v, 3h] 21. Because r„ and r„ are tangent to the coordinate curves v = co/wť and w = const, respectively. 23. [m, V, ií2 + v2], Ň = [~2% -Iv, 1] Problem Set 10.6, page 456 1.-64 3.-18 5.-12877 7. 2t7 9. |a3 11. 17/7/4 15. 140V6/3 17. 12877V2/3 = 189.6 19. r^772(373/2 - 53/2) = 22.00 25. lirh 27. 77/z4/V/2 29. 77/2 + 2irh3ß Problem Set 10.7, page 463 1. 8a3/?3c3/27 3. 6 7. 23477 13. 77/!5/10 21. 0 9. 2a5/3 17. 108 77 23. 8 5. 42§77 11. /ia477/2 19. 21677 25. 38477 Problem Set 10.8, page 468 1. Integrals 4 ■ 1' 1 (x = 1), 4 • 1 • 1 (y = 1), 8 - 1 - 1 (z = 1), 0 {x = y = z = 0) 3. 2 (volume integral of 6y2), 2 (surface integral over x = 1). Others 0 5. Volume integral of 6y2 - 6x2 is 0. 2 (.v — 1), -2 (y = 1), others 0. 7. F = [jc, y, z], div F = 3, In (2), Sec. 10.7, F*n = |F||n| cos 4> = Vx2 + y2 + z2 cos 4> = r cos 4>. 9. F = [x, 0, 0], div F = 1, use (2*), Sec. 10.7, etc. Problem Set 10.9, page 473 1. [0, 8z, 16] «[0, -1, 1], ±12 3. [~ez, -ex, ev]'[-\, -1, 1], ±{e2 - 1) 5. S: [u, v, v2], (curl F)'N = -4ue2u2, ±(4 - 4e2) 7. (curl F)«n = 3/2, ±3ö2/2 9. The sides contribute a, 3a2/2, -a, 0. 11. curl F = [0, 0, 6], 2477 13. (curl F)«n = 2x - 2y, 1/3 15. -77/4 17. (curl F)*N = 77(cos tjx + sin Try), 2 19. Ft = L-sin B, cos 0] • [-sin 6, cos 0] = 1, 2tt, 0 Chapter 10 Review Questions and Problems, page 473 11. Exact, -542/3 17. By Stokes, ±1877 23. 0, Aal377 29. By Gauss, IOO77 35. Direct, 5(e2 - 1) 13. Not exact, 0 if co2 > 1 N 7. v = cY cos cot + c2 sin 0) gives A = I e K cos wf dv = -j , B 3. ,f(x) = We_x gives A = 1/(1 + w2). 2 ' 2\ 5. Use / = {it 12) cos v and (11) in App. 3.1 to get A = (cos (7w/2))/(l - w2) r03 sin aw cos xw 2 r -aw 9. — co r-\ co i • 1 2 r sin aw cos xw 2 r cos w + w sin w — 1 7. — -dw 9. — -=-cos xw dw tt ■>() w tt j0 w ttw —2 sin xw dw tt jq 1 — w 2 r cos 7TW +1 2 r sin 11. — -o— cos <^vv 15- — 1— 77 Jn 1 — w tt j0 1 - 2 r 77w — sin 7tw 2 r wa 17. — -s- sin xw dw 19. — — 77 j0 w tt j0 Problem Set 11.8, page 517 \ 77 \ w 7. V77/2 cos iv if 0 < w < 77/2 and 0 if w > tt/2 9. Yes, no / 2 / sin 2w — 2 sin w \ ,- 1. / — I- 5. Vtt/2 e~x (x > 0) 11. V(2/t7) w/(w2 + tt2) 13. W's(xe-x2'2) = 9^(-(e_*,/2)') = iv^c(2 1 cos 4t 16 of - 16 of - 1 4 o)z cos 2/ 1 cos 3/ - + 9 oß-9 33. 1 r00 (cos vv + w sin vv — I) cos vvx + (sin vi' — w cos vv) sin wx 1 77 Jn 2 sin w cos vf „2 2 r w 35. — - 77 J0 W 4 r°° sin 2w — 2vv cos 2vi 37. i/ sin wx dw cos k>jc dw tt ■>„ 39. . / \ 77 VT + 4 Problem Set 12.1, page 537 1. u = Cxix) cos 4y + cz(x) sin 4y 3. u = c\ix) + c2{x)y 5. u = ttx + — cos 5tt? sin 5ttx--h • • • 7. —2 I (V2 — 1) cos 77? sin ttx + — cos 2tt? sin 2ttx + — (V2 + 1) cos 3tt? sin 3t7x - ■ • • A34 App. 2 Answers to Odd-Numbered Problems (2 — V2) cos 77? sin ttx — —- (2 + V2) cos 3irt sin 3 ttx — (2 + v2) cos 577? sin 5ttx 25 17. m 8L2 tt COS / \2 ~1 c — * U/ _ ttx I I + —g COS 377 377X sin - + L L 3d 19. (a) w(0, t) = 0, (b) u{L, t) = 0, (c) 1^(0, r) = 0, (d) ux(L, t) = 0. C = -A, D = —B from (a), (b). Insert this. The coefficient determinant resulting from (c) (d) must be zero to have a nontrivial solution. This gives (22). Problem Set 12.4, page 552 3. c2 = 300/[0.9/(2 • 9.80)] = 80.832 [m2/sec2] 11. Hyperbolic, u = /i(x) + f2(x + y) 13. Elliptic, u = J\(y + 3ix) + f2(y — 3ix) 15. Parabolic, u = x/j(x — y) + f2(x — y) 17. Parabolic, « = xf^lx + y) + f2(2x + y) 19. Hyperbolic, u = (l/y)Mxy) + My) Problem Set 12.5, page 560 5. u = Sin 0.477X g-l-752-16^/100 7. ii = - (- sin 0.1 TTx e-o-ovna^t + I sin 02wc e-o.oiraaa.* 9. u tt \ tt 20V2 sin 0.1 TTx e-0-01752^ + - sin 0.3ttx e-o.mvat»^t _ 9 11. u = mt + mh, where wn = u — mt satisfies the boundary conditions of the text, so ^ mrx . „ 2 rL ft TTX that wn = 2j Rn sin —— e ""^> l^Bn= — [f(x) — ux(x)] sin -dx n=i L L 0 L 13. F = A cos px + B sin px, F'(0) = Bp = 0, B = 0, F'(L) = -Ap sin /?L = 0, p = rnrlL, etc. 15. m = 1 2tt2 17. u = —— + 41 cos x e 19. w tt ~12 1 cos 2x e 4t + — cos 3x - Ao = 24« l f(y) dy' An=nl f(y) C0S "2T * H77X «77(Ä — v) 2 r niTX 35. An sin - sinh -— , An = —;- f(x) sin -dx a a a sinh (mrb/a) J0 a Problem Set 12.6, page 568 2 sin ap 2 rx sin ap l sin ap ic sin ap 2 , 1. A = -, ß = 0, w = — - cos e p dp Tip TT J0 p 3. A = •'o 7. A = 2[cos /? + p sin p - l)/(77p2)|, ß = 0, « = A cos px e~cVt dp -Vi Problem Set 12.8, page 578 1. (a), (b) It is multiplied by V2. (c) Half 3. Bmn = 16/(mrc772) if m, n odd, 0 otherwise 5. Bmn = (-l)n 1 Hl(mmr2) if m odd, 0 if m even 7. fi„m = (-l)m+n4/(mmr2) 11. A cos V29 f sin 2x sin 5y 00 oo 13. —X X —cos (rV m2 + n2) sin mx sin «v m, n odd 17. C77V260 (corresponding cigenfunctions F418 and F1614), etc. 19. Bmn = 0 (m or n even), fimn = \6kl{rnmr2) (rn, n odd) 21. Bmra = (-l)m+n144fl3fo3/(m3rt3-rr6) / [~9~ 16 \ 3tta 4ttv 23. cos I 7TT — + — I sin - sin \ V a Problem Set 12.9, page 585 7. 3(V cos 0 + 1(V3 cos 30 220 / 1 „ 1 _ 9. 55 + - \ r cos 0--r3 cos 30 + - r5 cos 50 - + ■ ■ 7T \ 3 5 ^ 4 / 1 1 e 11. —--r cos 0 + - r3 cos 30 + — r5 cos 50 + ■ ■ ■ 2 tt \ 9 25 15. Solve the problem in the disk r < a subject to u0 (given) on the upper semicircle and — u0 on the lower semicircle. 4«o / r 1 „ 1 „ \ u = —- — sin 0 + —- r3 sin 30 + —F r5 sin 50 + ■ ■ • tt \a 3as 5a° ) 17. Increase by a factor V2 19. T = 6A2bpR2f2 App. 2 Answers to Odd-Numbered Problems 21. No 23. Differentiation brings in a factor 1/Am = Rl(cam). Problem Set 12.10, page 593 11. v = F(r)G(t), F" + k2F = 0, G + c2k2G = 0, Fn = sin {nrrrlR), 2 rR n Trr G„ = B.„ exp (-cWW), B„ = - /-/(/-) sin — dr R o « 13. u = 100 15. U = |f3P3(cos ) - frP^cos (h) 17. 64r4P4(cos 0) 21. Analog of Example 1 in the text with 55 replaced by 50 23. v = r(cos 6)lr2 = xl(x2 + y2), v = xy/(x2 + y2)2 Problem Set 12.11, page 596 c(s) x 5. W = —^r + -5:-- , W(0, s) = 0, c(s) = 0. vv(x. f) = x(t - 1 + O a" 5 (s + i) 7. w = f(x)g{f), xf'g + fg = xt, take /(x) = x to get g = ce~'' + f - 1 and c = 1 from w(x, 0) = x(c - 1) = 0. 9. Set x2/(4c2t) = z2. Use z as a new variable of integration. Use erf(o°) = 1. Chapter 12 Review Questions and Problems, page 597 19. u = cÁy)e* + c2{y)e-2* 21. K = g(x)(\ - 23. u — cos t sin x — \ cos 2t sin 2x 25. u = I cos / sin x — j cos 3t sin 3x 27. u sin (0.0277x) e-n004!S72t 29. ■'/ = ™ Ln JE ,-0.004572, _ tt2 v 50 1 3t7X sin „ e 9 50 -0.04115t + • 31. II = 100 cos4xe"16t 33. u = + • — — — 1 — cos 2x e -2 77 \4 1 1- — cos 6x e 36 -36f 1 1 Too cos lOxe"100* 37. u = h(y) + hix + >') 39. u = f\(y - 2/x) + fz(y + 2/x) 41. u = xfi(y - x) + f2(y - x) 49. u = (% — M0)(ln r)/ln (r^lr^) + (u0 In rx — «j In r0)/ln (r^r^f) Problem Set 13.1, page 606 5. x - iy = -(x + iy), x = 0 7. 484 9. -5/169 11. -7/13 -(22/13)/ 13. -273 + 136/ 15. -7/17 - (11/17)/ 17. x/(x2 + y2) 19. (x2 - >'2)/(x2 + y2)2 Problem Set 13.2, page 611 1. 3V2(cos (-í77) + / sin (-fir)) 3. 5 (cos 77 + / sin 77) = 5 cos 77 5. cos §77 + i sin 577 App. 2 Answers to Odd-Numbered Problems A37 7.1 Vol (cos arctan | + i sin arctan |) 9. -3tt/4 11. arctan (±3/4) 13. ±77/4 15. 3 77/4 17. 2.94020 + 0.59601/ 19. 0.54030 - 0.84147/ 21. cos (—377) + / sin (— jj.77), cos §77 + i sin |77 23. ±(1 ±/)/V2 25.-1, cos ^77 ± / sin 7A77, cos §77 ± i sin §77 27. 4 + 3/, 4 - 8/ 29. § - /, 2 + 4/ 35. |zi + z2|2 = (74 + z2)(Z! + z2) = (Zi + z2)(Xi + z2). Multiply out and use Re Z]z2 \zxz2\ (Prob. 32): ZiZi + Zi22 + z2Zi + z2z2 = hi2 + 2 Re ZiZ2 + |z2j2 S y2 + 2|Zl||z2| + = (kil + |z2|)2- Take the square root to get (6). Problem Set 13.3, page 617 1. Circle of radius §, center 3 + 2i 3. Set obtained from an open disk of radius 1 by omitting its center z = 1 5. Hyperbola xy =1 7. y-axis 9. The region above y = x 13. / = 1 - l/U + 1) = 1 - (x + 1 - iy)/[(x + l)2 + y2]; 0.9 - 0.1/ 15. (,v2 - y2 - 2/xy)/(A'2 + y2)2, -j/2 17. Yes since r2(sin 20)/r^ 0 19. Yes 21. 6z2(z3 + /) 23. 2/(1 - z)-3 Problem Set 13.4, page 623 1. Yes 3. No 5. Yes 7. No 9. Yes for z * 0 11. rx = xlr = cos 6, ry = sin ft ft; = -(sin 6>)/r, ft = (cos 0)/r, (a) 0 = Mj. — (Jy = «r cos 0 + mh( —sin ft/r — vr sin // — t;fl(cos &)/r. (b) 0 = ma + u,; = ur sin 0 + mw(cos 6)lr + vr cos ^ + v8( — sin f/)/r. Multiply (a) by cos ft (b) by sin ft and add. Etc. 13. z2/2 15. In \z\ + i Arg z 17. z3 19. No 21. No 23. c = 1, cos a- sinh v 27. Use (4), (5), and (1). Problem Set 13.5, page 626 3. -1.13120 + 2.47173/, e = 2.71828 5. -/, 1 7. ea8(cos 5 - i sin 5), 2.22554 9. e~2x cos 2y, -e~2x sin 2y 11. exp (a2 - y2) cos 2xy, exp (x2 - y2) sin 2yv 13. e177'4, e57rili 15. Vr exp [i(0 + 2kir)/n\, k = 0, • • % n - 1 17. 9em 19. z = hi 2 + m + 2nm (n = 0, ± 1, • ■ ■) 21. z = In 5 — arctan |/ ± 2»77/ (n = 0, 1, • ■ ■) Problem Set 13.6, page 629 3. Use (11), then (5) for ely, and simplify. 5. Use (11) and simplify. 7. cos 1 cosh 1 - i sin 1 sinh 1 = 0.83373 - 0.98890/ AJ8 App. 2 Answers to Odd-Numbered Problems 9. 74.203, 74.210 11. -3.7245 - 0.51182/ 13.-1 15. cosh 4 = 27.308 17. z = ±(2n + l)7rt'/2 19. z = |(2n + 1)tt - (-lfl.4436/ 21. z — ±nm 25. Insert the definitions on the left, multiply out, simplify. Problem Set 13.7, page 633 1. In 10 + Tri 3.1 In 8 - \ifi 5. In 5 + (arctan f - ff)i = 1.609 - 2.214/ 7. 0.9273/ 9. \ In 2 - \m 11. ±(2n + 1)77/, n = 0, 1, ■ ■ • 13. In 6 ± {In + Y)m, n = 0, 1, • • ■ 15. (tt - 1 ± 2mr)i, n = 0, 1, • • • 17. In (/2) = (±2n + 1)tt/, 2 In / = ±(4n + l)7ri, n = 0, 1, • ■ • 19. e03(cos 0.7 + / sin 0.7) = 1.032 + 0.870/ 21. e2{\ + /)/V2 23. 64(cos (In 4) + / sin (In 4)) 25. 2.8079 + 1.3179/ 27. (1 + /)/V2 Chapter 13 Review Questions and Problems, page 634 17. -32 - 24/ 19. -| - y 21. 5 - 3/ 23. 6V2e37r'/4 25. 12«"* 27. ±(2 + 2/) 29. (±1 ± /)/V2 31. f(z) = \lz 33. f(z) = (1 + i)z2 35. f(£) = e*1 37. (-X2 + y2)l2 39. No 41.0 43.0.6435/ 45.-1.5431 Problem Set 14.1, page 645 1. Straight segment from 1 + 3/ to 4 + 12/ 3. Circle of radius 3, center 4 4-/ 5. Semicircle, radius 1, center 0 7. Ellipse, half-axes 6 and 5 9. Parabola y = ^x3 from -1 - |i to 2 + 4/ 11. e-if (0g(§ 2tt) 13. t + i/t (1SfS4) 15. f + (4 - 4f2)/ (-IS(SI) 17. -a - ib + re~u (0 f ^ 2tt) 19. | + 1/ 21. 0 23. 77/ + \i sinh 277 25. ill 27. -1 + / tanh |t7 = -1 + 0.6558/ 29. 2 sinh \ Problem Set 14.2, page 653 1. 77/, no 3. 0, yes 5. 0, yes 7. 0, yes 9. 0, no 11. 0, yes 15. Yes, by the deformation principle 19. 77/ by path deformation 21. 77/ 23. 277/ 25. 0 27. (a) 0, (b) 77 29. 0 App. 2 Answers to Odd-Numbered Problems Problem Set 14.3, page 657 1. -47T 3. 4tt 5. 8777 7.0 9.-77/ 11. m 13.77 15. 2Tri Ln 4 = 8.710/ 17. 77/cosh2(l + /) = 77(-0.2828 + 1.6489/) Problem Set 14.4, page 661 1.277ri/4 3. -2mevl2 5. 77/a3/3 7. 277/ if \a\ < 2, 0 if \a\ > 2 9. tt/(cos \ - sin |) 11. 2tt2/ 13. 77/ 3 Chapter 14 Review Questions and Problems, page 662 17.-677/ 19.0 21.-77/ 23. \i sin 8 25. 0 27. 77 29. 0 Problem Set 15.1, page 672 1. Bounded, divergent, ±1 3. Bounded, convergent, 0 5. Unbounded 7. Bounded, divergent, ±1/V2 ± i, 0, 1, -2 9. Convergent, 0 13. \zn ~ l\< |e, \z*n ~ l*\ < \e {n > Me)), hence \zn + z*n - (I + <\e + \e 17. Convergent 19. Divergent 21. Conditionally convergent 23. Divergent by Theorem 3 27. n = 1100 + 75f| = 125 (why?); 1100 + 75/|125/125! = 125125/LV250t7 (125/e)125] = e,25/V'25077 = 6.91 ■ 1052 Problem Set 15.2, page 677 1. 2 anz2'" = 2 an(z2)n, \z2\ < r = Um \ajan+1\, hence \z\ R = °° 15. z3/(l !3) - z7/(3!7) + z1l/(5!ll) - + •••, R = so I n - i 1 -3 i 2 .5 i 17 _7 i n _ 1 _ 19. Z + 3z + igZ + gigZ + R ~ zTT Problem Set 15.5, page 697 1. Use Theorem 1. 5. \z"\ = 1 and X 1/n2 converges. 9. |z + 1 - 2i\ ^ r < /? = 4 13. Nowhere 3. R = l/\ 7t> 0.56 7. jtanh" |z|| g 1, l/(«2 + 1) < 1/n 11. \z\ ^ 2 - S (8 > 0) 15. |z| g V5 - 8 (S > 0) Chapter 15 Review Questions and Problems, page 698 11. », c3j 13. I, 1 Ln[(l + z)/(l - z)] 15. 17. IA/it, |1 - tt(z - 2/)2r1 19. 1/3 21.-1 " (z - 77() - (z - 77/)2/2! - ■ ■ -, /? = OO 23.1 + |(z + 1) + l(z + l)2 + ie(z + I)3 + • • •, R = 2 25. 1 + 3z + 6z2 + 10z3 + ••%/?= 1. Differentiate the geometric series 27. i + (z + 0 - i(z + 02 - (z + 03 + • ■ % R = 1 29. -(z - lir) + jj- (z - §tt)3 - ^ (z - ^tt)5 +----, /? = x Problem Set 16.1, page 707 1111 1. + — + -2+ — + 1+z + R = 1 11111 1 3'?~z¥+27~i+24z_t2ö z2 + R = 00 1 1 7. 1 1 -7 + -Q 4 2z7 6z9 ° (z - l)""1 l — e = e n=0 n+1 R = 2 R = oo 1 1 + 1 2! + t/2 1 / - + - + - (z - 0 : - i 4 8 16 R = OG (z - o2 - li. - 2 (z + o"'1 = - z + i (z + 0----,Ä = 1 13. + 2 + (z - 1) App. 2 Answers to Odd-Numbered Problems A4! n=0 n=0 " go co | 17.2 z4n+2.m< i. -2753. h>i TC-0 n-=0 z 1 19.---2 + -r + i + (z- i) (z - 0 z - 1 21. (i - 4z) 2 z*\ kl < 1. (4- - -t) 2 4r. W > 1 n=0 ' v' 4 ' n-() 23. 2, — , |z + 1^1 > 0 n=0 \-">- Problem Set 16.2, page 711 1. ±5, ±|, ■ • • (poles of 2nd order), 00 (essential singularity) 3. 0, ±Vtt, ±V2~77, • • ■ (simple poles), 00 (essential singularity) 5. °c (essential singularity) 7. ±1, ±1 (fourth-order poles), 30 (essential singularity) 9. ±i (essential singularities) 13. — 16z (fourth order) 15. ±1, ±2, • • • (third order) 17. + //V3 (simple) 19. ±2i (simple), 0, ±27rt, ±4m, • • ■ (second order) 21. 0, ±2tt, ±4tt, • • • (fourth order) 23. f(z) = (z- ZoTgd), g(zo) * 0, hence f(z) = (z - Zo)2ng2(z). Problem Set 16.3, page 717 I. i, 4i 5. 1/5! (at z = 0) 9. -{(atz = 1), 3 (at z = -1) 15. eUz = 1 + 1/z + • • •, Ans. 2m 19. —477/ sinh \tj 23. 0 3. -\i (at z = 2z), \i (at -2;) 7. 1 (at ±«77) 11. -1 (at z = ±§77, ±f 77, • ■ •) 17. Simple poles at ±\. Ans. —4i 21. —4i sinh 5 25.1 (at z = 1), 2 (at z = §). Ans. 5m Problem Set 16.4, page 725 1. 2tt/V13 3. 27t-/35 5. 2tt/3 7. 0 9. 77 11. 277/3 13. 77/16 15. 0 17. 77/2 19. 0 21. 0 23. 77 25. 0 27. - 77/2 Chapter 16 Review Questions and Problems, page 726 17. 2777.73 19. 577 21. \tt cos 10 23. 77(74 25. 0 (n even), (-l)cn_1)/2 2m/(n - 1)! (n odd) 27. 677; 29. 2tt/7 31. 4t7/V3 33. 0 35. 77/2 A42 App. 2 Answers to Odd-Numbered Problems Problem Set 17.1, page 733 3. Only the size S. x = c, w = —y + ic, y = k, w = —k + ix 7. -3tt/4 < Arg w < 3tt/4, \w\ < 1/8 9. \w\ > 3 11. \w\ S 16, v g 0 13. Annulus 3 < |vv| < 5 15. In 2 = « = In 3, tt/4 = v = it/2 17. ±1, ±i 19. 0, ±1, ±2, - ■ ■ 21. -a/2 23. a and 0, V7a 25. M = ex = 1 when x = 0, / = e2:l 27. = l/|z| = 1 on the unit circle,./ = l/|z|2 Problem Set 17.2, page 737 —iw 5i 5'Z ~ -2w + 3 7' Z ~ Aw - 2 9. - = 0 11. z = \ + i ± Vi + i 13. z = ±i 15. w = 4/z, etc. 17. a - d = 0, fe/c = 1 by (5) 19. vv = azld (a ŕO,<íŕO) Problem Set 17.3, page 741 5. Apply the inverse g of f on both sides of n = f(Zi) to get = g(/(Zi)) = Zi- 7. vv = (z + 2i)/(z - 20 9. w = z - 4 11. vv = l/z 13. w = (3iz + l)/z 15. w = (z + l)/(-3z + 1) 17. w = (2z - i)/(-iz - 2) 19. w = (z4 - 0/(-íz4 + O Problem Set 17.4, page 745 1. Annulus 1 g |vv| š e2 3. 1/Vď < |vv| < Ve, 3rr/4 < arg vv < 5tt/4 5. 1 < |vv| < e, v > 0 7. vv-plane without 0 9. «2/cosh2 1 + y2/sinh2 1 ž 1. u š 0 11. Elliptic annulus bounded by «2/cosh2 1 + t;2/sinh2 1 = 1 and M2/cosh2 5 + y2/sinh2 5 = 1 13. ±(2n + 1)77/2, n = 0, 1, • ■ ■ 15. 0 < Im t < 77 is the image of R under t = z2. Ans. el = e~z 17. 0, ±L ±2i, ■ ■ ■ 19. «2/cosh2 1 + i>2/sinh2 1 ž 1, v < 0 21. u < 0 23. -1 š u š 1, ľ = 0 (c = 0), M2/cosh2 c + t/2/sinh2 r = 1 (c ^ 0) 25. In 2 š m š ln 3, tt/4 š « i 77/2 Problem Set 17.5, page 747 1. w moves once around the unit circle. 5. —5/3, 2 sheets 7. -ill, 3 sheets 9. 0, 2 sheets App. 2 Answers to Odd-Numbered Problems Chapter 17 Review Questions and Problems, page 747 11. u = \v2 ~ 1, \v2 - 1 13. \w\ = 20.25, |arg w\ < tt/2 15. The domain between u = | — v2 and u = \ — \v2 17. |w + i| =| 19. w — 1 21. [arg vv| < tt/4 23. 0, (±1 ±_i)/V2 25. tt/8 ± nirll, n = 0,1, ■ • • 27. 0, ±(7V'2 29. w = iz 31. w = 1/z 33. vv = z/(z + 2) 35. ±V2~ 37. 2 ± V6 39. 1 + j ± VI + 2i 41. vv = e3z 43. z2/2/c 45. iz3 + 1 Problem Set 18.1, page 753 1. 20.v + 200. 20z + 200 3. 110 - 50xy, 110 + 25/'z2 5. F = (110/ln 2)Ln z 7. F = 200 - (100/ln 2) Ln z 13. Use Fig. 388 in Sec. 17.4 with the z- and vv-planes interchanged, and cos z = sin (z + \tt). 15. * = 220 - 11 Oxy Problem Set 18.2, page 757 1. u2 - v2 = e2x(cos2 y - sin2 y), 4>.CT = 4e2x(cos2 y - sin2 y) = -y?y, V2* = 0 3. Straightforward calculation, involving the chain rule and the Cauchy-Riemann equations 5. See Fig. 389 in Sec. 17.4. = sin2 x cosh2 y - cos2 x sinh2 y. 9. (i) *=[/!(] - xy). (ii) w = /'z2 maps R onto -2 g u g 0, thus = t/iCl + |«) = f/x(l + i(-2xy)). 11. By Theorem 1 in Sec. 17.2 13. = 10| 1 - (1/tt) Arg (z - 4)], F = 10[1 + (Hit) Ln (z - 4)] 15. Corresponding rays in the w-plane make equal angles, and the mapping is conformal. Problem Set 18.3, page 760 3. (I00kl)y. Rotate through tt/2. 5. 100 - 2400/tt 7. Re F(z) = 100 + (200/tt) Re (arcsin z) 9. (240/77) Arg z 11. T0 + (2/7r)(F1 - T0) Arg z 13. 50 + (400/tt-) Arg z Problem Set 18.4, page 766 1. V = iV2 = iK, W = -Kx = const, $ = Ky = const 3. F(z) = A"z (A' positive real) 5. V = (1 + 2i)K, F = (1 - 20#z 7. F(z) = z3 9. Hyperbolas (.v + 1 )y = const. Flow around a corner formed by x = — 1 and the x-axis. 11. y/(x2 + y2) = c or a'2 + (>• - k)2 = k2 13. F(z) = z/r0 + r0/z App. 2 Answers to Odd-Numbered Problems 15. Use that w = arccos z is w = cos z with the roles of the z- and vt'-planes interchanged. Problem Set 18.5, page 771 5. 1 - r2 cos 20 7. 2(r sin 6 - \r2 sin 26 + |r3 sin 30 - + ■ ■ ■) 9. |r2 sin 20 - \r6 sin 60 11. ^tt2 - 4(r cos 0 - Jr2 cos 20 + £r3 cos 30 - + • • ■) 4 / 1 , I = \ 13. r sin 0--r3 sin 30 + — r5 sin 50 - + ■ • • 77 \ 9 25 / Problem Set 18.6, page 774 1. No; \z\2 is not analytic. 3. Use (2). F(|) = f 5. 0>(4, -4) = -12 7. Use (3). 4>(1, 1) = -2 11. \F(ei&)\2 = 2 - 2 cos 20, 0 = 77/2, Max = 2 13. |F(z)| = [cos2 2x + sinh2 2y]1/2, z = ±(, Max = [1 + sinh2 2]1/2 = cosh 2 = 3.7622 15. No Chapter 18 Review Questions and Problems, page 775 11. 4> = 10(1 - x + v), F = 10 - 10(1 + i)z 13. (20/ln 10) Ln z 15. (10/ln 10)(ln 100 - In r) 17. Arg z = const 19. {-Hit) Ln z 23. T(x, y) = x(2y + 1) = const 25. Circles (x - cf + y2 = c2 27. F(z) = ^- Ln (z - 5), Arg (2 - 5) = c 277 80 / 1 , 1 , 29. 20 + — [r sin 0 + - r3 sin 30 + - rJ sin 50 + ■ 77 \ 3 5 Problem Set 19.1, page 786 1. 0.9817- 102, -0.1010- 103, 0.5787- I0-2, -0.1360- 105 3.0.36443/(17.862 - 17.798) = 0.36443/0.064 = 5.6942,0.3644/(17.86 - 17.80) 0.3644/0.06 = 6.073, 0.364/(17.9 - 17.8) = 3.64, impossible 0.36443(17.862 +17.798) 0.36443-35.660 12.996 5. -—n-5—- = - = - = 5.7000, 17.8622 - 17.7982 319.05 - 316.77 2.28 13.00 13.0 13 10 5.702. —— = 5.70, — = 5.7, — = 5 2.28 2.28 2.3 2 7. 19.95, 0.049, 0.05013; 20, 0, 0.05 9. In the present calculation, (b) is more accurate than (a). 11. -0.126- 10-2, -0.402- 10"3; -0.267- HP6, -0.847- 10-7 13. Add first, then round. . Oi = gi + e1 = ax + eY i _ e2 + e22 _ + a2 a2 + e2 a2 \ a2 ) a2 a2 a2 a2 App. 2 Answers to Odd-Numbered Problems A45 Ol *2 \Ö2 «2 «2 hence 19. (a) 19/21 = 0.904761905, echop = erm er,ehop er,round = 0.2106" 10 5, etc. = krll + k2| = &1 + ßr-Z 0.1905 • IO-f>, Problem Set 19.2, page 796 1. g = 1.4 sin a", 1.37263 (= xs) 5. g = a4 + 0.2, 0.20165 (= x3) 7. 2.403 (= x5, exact to 3S) 9. 0.904557 (= x3) 11. 1.834243 (= .v4) 13. jc0 = 4.5, x4 = 4.73004 (6S exact) 15. (a) 0.5. 0.375, 0.377968, 0.377964; (b) 1/V7 = 0.377964 473 I7.xn+i = (2xn + 7/jen2)/3, 1.912931 (= x3) 19. (a) Algorithm Bisect (/, a0, b0, N) Bisection Method This algorithm computes an interval [an, /?„] containing a solution of fix) = 0 (/ continuous) or it computes a solution cn, given an initial interval [a0, b0] such that /(ao)/(fc0) < 0. Here N is determined by (b - a)llN g B, )S the required accuracy. INPUT: Initial interval \a0, b0], maximum number of iterations N. OUTPUT: Interval [aN, bN] containing a solution, or a solution cn. For n = 0, 1, ■ • •, N - I do: Compute c.„ = \(an + bn). If f(cn) = 0 then OUTPUT cn. Stop. [Procedure completed] Else continue. If f(an)f(cn) < 0 then an+1 = an and bn+1 = cn. and b. n + 1 Else set an+1 = c End OUTPUT [aN, bN]. Stop. [Procedure completed] End BISECT Note that [aN, bN] gives (aN + bN)/2 as an approximation of the zero and (bN as a corresponding error bound. (b) 0.739085; (c) 1.30980, 0.429494 21.1.834243 23.0.904557 aN)l2 Problem Set 19.3, page 808 1. L0(x) = -2x + 19, Lx(x) = 2a - 18, Plix) = 0. 1082a + 1.2234, Pi(9.4) = 2.2405 3. 0.9971, 0.9943, 0.9915 (0.9916 4D), 0.9861 (0.9862 4D), 0.9835, 0.9809 5. p2(x) = -0.44304a2 + 1.30906a - 0.02322, /?2(0.75) = 0.70929 7.p2(x) = -0.1434a2 + 1.0895a, p2(0.5) = 0.5089, p2(l.5) = 1.3116 9. L0 = -i(x - l)(.v - 2)(a - 3), Lx = \x(x - 2)(x - 3), L2 = -\x(x - 1)(a - 3), L3 = \x(x - l)(x - 2); p3(x) = 1 4- 0.039740a- - 0.335187x2 + 0.060645a3; p3(0.5) = 0.943654 (6S-exact 0.938470), p3(L5) = 0.510116 (0.511828), p3(2.5) = -0.047993 (-0.048384) A46 App. 2 Answers to Odd-Numbered Problems Í3.p2(x) = 0.9461a - 0.2868x(x - l)/2 = -0.1434x2 + 1.0895a; 15. 0.722, 0.786 17. S/1/2 = 0.057 839, 8fm = 0.069 704, etc. Problem Set 19.4, page 815 9. [-1.39(x - 5)2 + Q.58(;c - 5)3]" = 0.004 at x = 5.8 (due to roundoff; should be 0). 11. 1 - fx2 + |x4 13. 4 - 12x2 - 8x3, 4 - 12x2 + 8x3. Yes 15. 1 - x2, -2(x - 1) - (x - l)2 + 2(x " l)3, -1 + 2(x - 2) + 5(x - 2)2 - 6(x " 2)3 17. Curvature /7(1 + /'2)3/2 » f" if |/'| is small 19. Use that the third derivative of a cubic polynomial is constant, so that g" is piecewise constant, hence constant throughout under the present assumption. Now integrate three times. Problem Set 19.5, page 828 1. 0.747131 3. 0.69377 (5S-exact 0.69315) 5. 1.566 (4S-exact 1.557) 7. 0.894 (3S-exact 0.908) 9. Jh/2 + em = 1.55963 - 0.00221 = 1.55742 (6S-exact 1.55741) 11. Jm + em = 0.90491 + 0.00349 = 0.90840 (5S-exact 0.90842) 13. 0.94508, 0.94583 (5S-exact 0.94608) 15. 0.94614588, 0.94608693 (8S-exact 0.94608307) 17. 0.946083 (6S-exact) 19. 0.9774586 (7S-exact 0.9774377) 21.x - 2 = t, 1.098609 (7S-exact 1.098612) 23. x = Jif + 1), 0.7468241268 (10S-exact 0.7468241330) 25. (a) M2 = 2, M2* = \, \KM2\ = 2/(1 In2), n = 183, (b) /(iv) = 24/x5, 2m = 14 27. 0.08, 0.32, 0.176, 0.256 (exact) 29. 5(0.1040 - \ ■ 0.1760 + | • 0.1344 - \ ■ 0.0384) = 0.256 Chapter 19 Review Questions and Problems, page 830 17. 4.266, 4.38, 6.0, impossible 19. 49.980, 0.020; 49.980, 0.020008 21. 17.5565 g s ě 17.5675 23. The same as that of a. 25. -0.2, -0.20032, -0.200323 27. 3, 2.822785, 2.801665, 2.801386, 2.801386 29. 2.95647, 2.96087 31. 0.26, M2 = 6, M2* = 0, -0.02 šešO, 0.24 š a Š 0.26 33. 1.001005, -0.001476 šešO Problem Set 20.1, page 839 1. .v, = —2.4, x2 = 5.3 5. Xi = 2, x2 = 1 3. No solution l.x1 = 6.78, x2 = -11.3,% = 15.82 App. 2 Answers to Odd-Numbered Problems 9. xx - 0, x2 = tx arbitraiy, x3 = 5tx + 10 11. Xx — ti, x2 = ?2, both arbitrary, x3 = 1.25^ — 2.25?2 13. xx = 1.5, x2 = -3.5, x3 = 4.5, x4 = -2.5 Problem Set 20.2, page 844 1. 9. 11. 13. I L a 0" 1_ 0 1 6 0 1 9 0 3 1 0 5 8 0 1 i 9 xx = 4.2 x2 = 1.3 1 0 r4 -6^ xx = -4 _2 1_ .0 5_ x2 = 3 6 0 2 3 0 3 5 0 '2 0 0 L0 Xx = 0.5 , x2 = -2.5 x3 = 3.0 Xx = -1/30 ,x2= 2/15 x3 = 1/5 Xx = 3 , x2 = 6 x3 = 2 xx = 4 . x2 = 0 x3 = -2 x"! = 6 x„ = -2 3 x4 0 14 "-2 4 -i1 341 6 -10 17. -2 3 0 19. 15 2 4 3 7 -12 2 -17 3 1 3 J Problem Set 20.3, page 850 7. Exact 0.5, 0.5, 0.5 3. Exact 21.5, 0, -13.8 5. Exact 2, 1,4 9. (a) x(3)T = [0.49982 0.50001 0.50002], (b) x(3)T = [0.50333 0.49985 0.499681 11. 6, 15, 46, 96 steps; spectral radius 0.09, 0.35, 0.72, 0.85, approximately 13. [1.99934 1.00043 3.99684]T (Jacobi, step 5); [2.00004 0.998059 4.00072]1" (Gauss-Seidel) _ 17. V306 = 17.49, 12, 12 19. Vl8/c2 = 4.24|fc|, 4|fe|, 4|/c| App. 2 Answers to Odd-Numbered Problems Problem Set 20.4, page 858 1. 12, V62 = 7.87, 6, [| -1 g] 5. 1.9, VT35 = 1.16, 1, |0.3 -0.1 7. 6, \ 6. 1, [1 1 1 1 1 1] 13. k = 100 • 100 17. 46 g 6 - 17 or 7- 17 21. [-0.6 2.8]T 3. 14, V5Ö = 7.07, 4, [-1 1 § -f] 0.5 1.0] 11. II A II ! = 17, (J A"1 II j = 17, K = 289 15. K — 1.2 ■ ^ = 1.469 19. [0 1]T, [1 -0.4]T, 289 23. 27, 748, 28375, 943656, 29070279 Problem Set 20.5, page 862 1. -11.4 + 5Ax 3. 8.95 - 0.388x 5. s = -675 + 90/, vm = 90 km/h 9. 4 - 0,75x - 0.125x2 11. 5.248 + 1.543a-, 3.900 + 0.532Ix + 2.021x2 13. -2.448 + 16.23a, -9.114 + 13.73x + 2.500x2, -2.270 + 1.466* -1.778X2 + 2.852xB Problem Set 20.7, page 871 1. 5 s A s 9 3. 5, 0, 7; radii 4, 6, 6 5. |A - 4i\ g V2 + 0.1, |A| g 0.1, |A - 9i\ g V2 7. tu = 100, t22 = f33 = 1 9. They lie in the intervals with endpoints ± (« — 1)10-6. (Why?) 11. 0 lies in no Gerschgorin disk, by (3) with >; hence det A = A2 • ■ • Xn + 0. 13. p(A) g Row sum norm || A\\x = max 2 lajfcl = f"1^* (lajjl + Gerschgorin radius) 15. VT53 = 12.37 17. Vl22 = 11.05 19. 6 g A g 10, 8 g A g 8 Problem Set 20.8, page 875 1. q = 4, 4.493, 4.4999; |e| g 1.5, 0.1849, 0.0206 3. g = 8, 8.1846, 8.2252; |e| g 1, 0.4769, 0.2200 5. q = 4, 4.786, 4.917; |e| g 1.63, 0.619, 0.399 l.q = 5.5, 5.5738, 5.6018; |e| § 0.5, 0.3115, 0.1899; eigenvalues (4S) 1.697, 3.382, 5.303, 5.618 9. y = Ax = Ax, yTx = AxTx, yTy = A2xTx, e2 g yTy/xTx - (yTx/xTx)2 = A2 - A2 = 0 11. q = —2.8993 approximates —3 (0 of the given matrix), 1e| g 1-633, ■ • •, 0.7024 (Step 8) Problem Set 20.9, page 882 3.500000 -1.802776 0 1. -1.802776 6.730769 1.846154 . 0 1.846154 1.769230 0.980000 -0.441814 0 -0.441814 0.870164 0.371803 0" 0.371803 0.489836 App. 2 Answers to Odd-Numbered Problems A49 5. Eigenvalues 8, 3, 1 5.64516 -2.50867 7. -2.50867 0 7.91494 -0.646602 0 "18.3171 0.881767 . 0 '18.3910 0.177669 - 0 '7.00224 0.0571287 .0 '7.00322 0.0186419 ,0 5.307219 0.374953 -0.646602 3.08458 0 0.374953 1.04762 0 0.0312469 7.45139 -1.56325 0 0.0312469 1.000482 0.881767 8.29042 0.360275 0.177669 8.23540 0.0100214 0.0571287 4.00088 0.0249333 0.0186419 4.00011 0.00154782 0 0.360275 1.39250 J 0 0.0100214 1.37363 18.3786 0.39651 0 0.0249333 0.996875 . 0 0.00154782 0.996669 -1.56325 3.544142 0.0983071 0' 0.0983071 1.00446. '7.00298 0.0326363 .0 0.396511 8.24727 0.0600924 0.0326363 4.00034 0.00621221 0 0.0600924 1.37414 . 0 0.00621221 0.996681 Chapter 20 Review Questions and Problems, page 883 17. [4 -1 2]T 19. [6 -3 1|T 21. All nonzero entries of the factors are 1. 2.8193 -1.5904 -0.0482" 23. -1.5904 1.2048 - 0.0241 (4D-values) .-0.0482 - 0.0241 0.1205. 27. 15, V89, 8 29. 7, V2T, 4 25. Exact |-2 1 2]T 33. 6 35. 9 39. y = 1.98 + 0.9&X 41. Centers 1, 1, 1, radii 2.5, 1, 2.5 (A = 2.944, 0.028 ± 0.290;'. 3D) 43. Centers 5, 6, 8; radii 2, 1, 1, (A = 4.1864, 6.4707, 8.3429, 5S) 31. 14, V78, 7 37. 11.5-4.4578 = 51.2651 45. 1.5 -2.23607 0 ' -2.23607 5.8 -3.1 0 -3.1 6.7 , Step 3: 9.44973 -1.06216 0 -1.06216 4.28682 -0.00308 0 0.00308 0.26345. App. 2 Answers to Odd-Numbered Problems Problem Set 21.1, page 897 1. y = ex, 0.0382, 0.1245 (error of *5, x10) 3. y = x - tanh x (set y - x = u), 0.009292, 0.0188465 (error of x5, x10) 5. y = e1, 0.001275, 0.004200 (error of x5, xw) 7. y = 1/(1 - x2/2), 0.00029, 0.01187 (error of x5, xw) 9. y = 1/(1 - x2/2), 0.03547, 0.28715 (error of x5, x1Q) 11. y = 1/(1 - a-2/2); error -10~8, -4- 10"H, 10-7, +10"5; about 1.3 • 10"5 by (10) 13. y = xe*\ error - 105 (for x = 1, • • •, 3) 19, 46, 85, 139, 213, 315, 454, 640, 889, 1219 15. y = 3 cos x — 2 cos2 x; error - 107: 0.18, 0.74, 1.73, 3.28, 5.59. 9.04, 14.33, 22.77, 36.80, 61.42 17. v = l/(x5 + 1), 0.000307, -0.000259 (error of x5, x10) 19. The errors are for E.-C. 0.02000, 0.06287, 0.05076, for Improved E.-C. -0.000455. 0.012086. 0.009601, for RK 0.0000012. 0.000016, 0.000536. Problem Set 21.2, page 901 3. y = e-01x2; errors 10"6 to 6 • 10"6 5. y = tan a; y4, ■ • •, y10 (error - 105): 0.422798 (-0.48), 0.546315 (-1.2), 0.684161 (-2.4), 0.842332 (-4.4), 1.029714 (-7.5), 1.260288 (-13), 1.557626 (-22) 7. RK-error smaller, error • 105 = 0.4, 0.3, 0.2, 5.6 (for x = 0.4, 0.6, 0.8, 1.0) 9. y4 = 4.229 690, y5 = 4.556 859, y6 = 5.360 657, y7 = 8.082 563 11. Errors between -6 ■ 10"7 and +3 • 10"7. Solution ex - x - 1 13. Errors ■ 105 from x = 0.3 to 0.7: —5, —11, —19, —31, —47 15. (a) 0, 0.02, 0.0884, 0.215 848, y4 = 0.417 818, y5 = 0.708 887 (poor), (b) By 30-50% Problem Set 21.3, page 908 3. Vj = ex, y2 = —ex. errors range from ±0.02 to ±0.23, monotone. 5. y\ = y2, y'2 = -4ylt y = yJ = 1, 0.84, 0.52, 0.0656, -0.4720; y = cos 2x 7. yx = 4e-' sin x, y2 = 4e~x cos x; errors from 0 to about ±0.1 9. Errors smaller by about a factor 104 11. y = 0.198669, 0.389494, 0.565220, 0.719632, 0.847790; y' = 0.980132, 0.922062, 0.830020, 0.709991, 0.568572 13. y, = e-** - e-5x, y2 = e 'Ax + e'5*; y\ = 0.1341, 0.1807, 0.1832, 0.1657, 0.1409; y2 = 1.348, 0.9170, 0.6300, 0.4368, 0.3054 17. You get the exact solution, except for a roundoff error [e.g., yx = 2.761 608. y(0.2) = 2.7616 (exact), etc.]. Why? 19. y = 0.198669, 0.389494, 0.565220, 0.719631, 0.847789; y' = 0.980132, 0.922061, 0.830019, 0.709988, 0.568568 Problem Set 21.4, page 916 3. 105, 155, 105, 115; Step 5: 104.94, 154.97, 104.97, 114.98 5. 0.108253, 0.108253, 0.324760, 0.324760; Step 10: 0.108538, 0.108396, 0.324902, 0.324831 App. 2 Answers to Odd-Numbered Problems A51 7. 0, 0, 0, 0. All equipotential lines meet at the corners (why?). Step 5: 0.29298, 0.14649, 0.14649, 0.073245 9. -3«u + m,2 = -200, uu - 3m12 = -100 11. «12 = u-32 ' 31.25, h21 = m23 = 18.75, «;7c = 25 at the others 13. m21 = m23 = 0.25, «12 = h32 = -0.25, ujk = 0 else 15. (a) uu = — h]2 = —66. (b) Reduce to 4 equations by symmetry. Mn — "31 = ~Mis = ~u35 = —92.92, «21 = —m2s = —87.45, l'i2 = "s2 = ~~Mi4 = ~~"34 = —64.22, m22 = — m24 = —53.98, «13 = «23 = «33 = 0 17. V3, wn = m21 = 0.0849, «12 = m22 = 0.3170. (0.1083, 0.3248 are 4S-values of the solution of the linear system of the problem.) Problem Set 21.5, page 921 5. un = 0.766. «21 = 1.109, «12 = 1.957, m22 = 3.293 7. A as in Example 1, right sides -2, -2, -2, -2. Solution uu = m21 = 1.14286, "i2 = "22 = 1.42857 11. — 4«n + h21 + h12 = —3, mu — 4m21 + h22 = —12, uXi - 4«12 + m22 = 0, 2m21 + 2h12 - 12h22 = -14, mu = = 2, «21 = 4, m12 = 1. Here -14/3 = -|(1 + 2.5) with 4/3 from the stencil. 13. b = [-380 -190, -190, Of; uu = 140, u21 = «12 = 90, «22 = 30 Problem Set 21.6, page 927 5. 0.1636, 0.2545 (t = 0.04, x = 0.2, 0.4), 0.1074, 0.1752 (f = 0.08), 0.0735, 0.1187 (r = 0.12), 0.0498, 0.0807 (t = 0.16), 0.0339, 0.0548 (t = 0.2; exact 0.0331, 0.0535) 7. Substantially less accurate, 0.15, 0.25 (t = 0.04), 0.100, 0.163 (t = 0.08) 9. Step 5 gives 0, 0.06279, 0.09336, 0.08364, 0.04707, 0. 11. Step 2: 0 (exact 0), 0.0453 (0.0422), 0.0672 (0.0658), 0.0671 (0.0628), 0.0394 (0.0373), 0 (0) 13. 0.1018, 0.1673, 0.1673, 0.1018 (/ = 0.04), 0.0219, 0.0355, •■•(? = 0.20) 15. 0.3301, 0.5706, 0.4522, 0.2380 (r = 0.04), 0.06538, 0.10604, 0.10565, 0.6543 (t = 0.20) Problem Set 21.7, page 930 1. For* = 0.2, 0.4 we obtain 0.012, 0.02 (t = 0.2), 0.004, 0.008 (t = 0.4), -0.004, -0.008 (t = 0.6), etc. 3. u{x, 1) = 0, -0.05, -0.10, -0.15, -0.075, 0 5. 0.190, 0.308, 0.308, 0.190 (0.178, 0.288, 0.288, 0.178 exact to 3D) 7. 0, 0.354, 0.766, 1.271, 1.679, 1.834, -••(/ = 0.1); 0, 0.575, 0.935, 1.135, 1.296, 1.357, •••(?= 0.2) Chapter 21 Review Questions and Problems, page 930 17. y = tan x; 0 (0), 0.10050 (-0.00017), 0.20304 (-0.00033), 0.30981 (-0.00047), 0.42341 (-0.00062), 0.54702 (-0.00072) A52 App. 2 Answers to Odd-Numbered Problems 19. 0.1003349 (0.8 • 10"7) 0.2027099 (1.6 • 10"7), 0.3093360 (2.1 ■ 10"7), 0.4227930 (2.3 • 10-7), 0.5463023 (1.8 • 10"7) 25. y(0.4) = 1.822798, y(0.5) = 2.046315,y(0.6) = 2.284161, y(0.7) = 2.542332, y(0.8) = 2.829714, y(0.9) = 3.160288, y(1.0) = 3.557626 27. ya = 3e~9x, y2 = -5e'9x, [1.23251 -2.05419], [0.506362 -0.843937], • • % [0.035113 -0.058522] 29. 1.96, 7.86, 29.46 31. h(Pn) = u(P31) = 270, u(P21) = u(P13) = u(P23) = m(P33) = 30, u(P12) = u(P32) = 90, u(P22) = 60 35. 0.06279, 0.09336, 0.08364, 0.04707 37. 0, -0.352, -0.153, 0.153, 0.352, 0 if t = 0.12 and 0, 0.344, 0.166, -0.166, -0.344, 0 if / = 0.24 39. 0.010956, 0.017720, 0.017747, 0.010964 if t = 0.2 Problem Set 22.1, page 939 3. / = 3(xa - 2)2 + 2(x2 + 4)2 - 44. Step 3: [2.0055 -3.9975]T 5. / = 0.50cj - l)2 + 0.7(x2 + 3)2 - 5.8, Step 3: [0.99406 -3.0015]T 7. / = 0.2(x1 - 0.2)2 + x22 - 0.008. Step 3: [0.493 -0.011]T, Step 6: [0.203 0.004]T Problem Set 22.2, page 943 1. x3, x4 unused time on Mx, M2, respectively 3. No U. /max = f(0, 5) = 10 13. /max = /(9, 6) = 36 15. /mi„ = /(3.5, 2.5) = -30 17. xx/3 + x2/2 100, Xj/3 + x2/6 g 80, / = 150x, + 100x2, /max = /(210, 60) = 37500 19. 0.5xt + 0.75x2 S 45 (copper), O.Sjcj + 0.25x2 S 30, / = 120x, + 100x2, /max = /(45, 30) = 8400 Problem Set 22.3, page 946 / 2100 200 \ 1. /(120/11, 60/11) = 480/11 3. /(—p : -^-J = 78000 5. Matrices with Rows 2 and 3 and Columns 4 and 5 interchanged 7./(0,^) = -10 9. ,f(5, 4, 6) = 478 Problem Set 22.4, page 952 1. ,f(4, 4) = 72 3. .f(20, 30) = 50 5. /(10, 5) = 5500 7. /(I, 1, 0) = 12 9. f(i 0,|) = 3 Chapter 22 Review Questions and Problems, page 952 11. Step 5: [0.353 -0.028]7 Slower 13. Of course! Step 5: [-1.003 1.897|T 21. f(2, 4) = 100 23. f(3, 6) = -54 25. f(50, 100) = 150 App. 2 Answers to Odd-Numbered Problems A53 Problem Set 23.1, page 958 9. "0 0 1 r "0 1 1" "0 1 1 1 0 0 0 i 11. 0 0 1 13. 0 0 0 0 1 0 0 0 _1 1 o. 1 0 0 0 J 1 0 0. .0 0 0 0 15.©- ®- ßi 1 " 1 2 0 <ä 3 > 1 4 . 0 Edge l 0 0 1 0 Vertex Incident Edges 1 -elt ~e2, e3, -e4 2 ei 3 e2- -e-i 4 ?4 23. 1 X S 2 > 3 L 0 1 -1" 0 0 1 1. Problem Set 23.2, page 962 1. 4 3. 5 5. 4 9. The idea is to go backward. There is a vk_1 adjacent to vk and labeled k — 1, etc. Now the only vertex labeled 0 is ,v. Hence A(i>0) = 0 implies v0 = s, so that v0 — vx - • • ■ — ufe_x — vk is a path s vk that has length jfc. 15. No; there is no way of traveling along (3, 4) only once. 21. From m to 100m, 10m, 2.5m, m + 4.6 Problem Set 23.3, page 966 1. (1, 2), (2, 4), (4, 3); L2 = 6, L3 = 18, L4 = 14 3. (1, 2), (1, 4), (2, 3); L2 = 2, L3 = 5, L4 = 5 5. (1, 4), (2, 4), (3, 4), (3, 5); L2 = 4, L3 = 3, L4 = 2, L5 = 8 7. (1, 5), (2, 3), (2, 6), (3, 4), (3, 5); L2 = 9, L3 = 7, L4 = 8, L5 = 4, L6 = 14 Problem Set 23.4, page 969 2 1 8 1. 1 ( 3 L = 12 3. 4 f 2 L = 10 5. 1 - 2 ; 5 L = 28 4( 3-5 3^6-4 5 7 App. 2 Answers to Odd-Numbered Problems 2 9. 1 - 3 - 4 ( L = 38 11. Yes 5 — 6, 15. G is connected. If G were not a tree, it would have a cycle, but this cycle would provide two paths between any pair of its vertices, contradicting the uniqueness. 19. If we add an edge (u, v) to T, then since T is connected, there is a path u —> v in T which, together with (u, v), forms a cycle. Problem Set 23.5, page 972 1. (1, 2), (1, 4), (3, 4), (4, 5), L = 12 3. (1, 2), (2, 8), (8, 7), (8, 6), (6, 5), (2, 4), (4, 3), L = 40 5. (1, 4), (3, 4), (2, 4), (3, 5), L = 20 7. (1, 2), (1, 3), (1, 4), (2, 6), (3, 5), L = 32 11. If G is a tree 13. A shortest spanning tree of the largest connected graph that contains vertex 1 Problem Set 23.6, page 978 1. 1 - 2 - 5, A/ = 2; 1 - 4 - 2 - 5, A/ = 2, etc. 3. 1 - 2 - 4 - 6, Af = 2; 1 - 2 - 3 - 5 - 6, A/ = 1, etc. 5. /la = 4, f13 = 1, /14 = 4, f42 = 4, /43 = 0, f25 = 8. f35 = 1, / 7. f12 = 4, f13 = 3, f24 = 4, /35 = 3, f54 = 2, f46 = 6, ,f56 = 1, / 9. {4, 5, 6}, 28 11. {2, 4, 6}, 50 13. 1-2-3-7, 4f = 2; 1-4-5-6 -7, A/=1; 1-2-3-6-7, A/= l;/max= 14 15. {3, 5, 7}, 22 17. 5 = {1, 4}, cap (S, T) 19. If/,, < ,u as well as 0 Problem Set 23.7, page 982 3. (2, 3) and (5, 6) 5. 1 - 2 - 5, At = 2; 1 - 4 - 2 - 5, At = 1; / = 6 + 2+l=9 7. 1-2-4-6, At = 2; I - 3 - 5 - 6, At = 1; f = 4 + 2+1=7 9. By considering only edges with one labeled end and one unlabeled end 17. S = {1, 2, 4, 5}, T = {3, 6), cap (S, T) = 14 Problem Set 23.8, page 986 1. No 3. No 5. Yes, S = {1,4, 5, 8} 7. Yes; a graph is not bipartite if it has a nonbipartite subgraph. 9. 1 - 2 - 3 - 5 11. (1, 5), (2, 3) by inspection. The augmenting path 1—2 — 3 — 5 gives 1-2-3-5, that is, (1, 2), (3, 5). 13. (1, 4), (2, 3), (5, 7) by inspection. Or (1, 2), (3, 4), (5, 7) by the use of the path 1-2-3-4. 15. 3 19. 3 23. No: K5 is not planar. 25. K3 = 6 + 8 = 14 App. 2 Answers to Odd-Numbered Problems Chapter 23 Review Questions and Problems, page 987 13. A55 17. "0 1 tr 0 0 l J 0 0_ "0 1 0 r 0 0 1 0 1 0 f) i -0 0 0 0- '0 1 0 r 1 0 1 0 15. 0 1 0 1 -1 0 1 0- 21. Vertex Incident Edges 23. 4 1 2 -By, e3 3 'ei, -ca 25.4 27. L 29. 1 - 4 - 3 - 2, L = 16 33./ 15, L4 13 Problem Set 24.1, page 996 1. qL = 19, qM = 20, qu = 20.5 5. qL = 69.7, qM = 70.5, qv = 71.2 9. qL = 399, qM = 401, qv = 401 13. x = 70.49, s = 1.047, IQR = 1.5 17.0 0 300 3. qL = 38, qM = 44, qv = 54 7. to = 2.3, qM = 2.4, 1200) = \ 6[0.25 - (x ■ 11. k = 2.5; 50% 17. X > b, X g b, X < c, X S c, etc. Problem Set 24.6, page 1019 1. 2/3, 1/18 5. 4, 16/3 9. fi = I/O = 25; P = 20.2% 13. 750, 1, 0.002 Problem Set 24.7, page 1025 1. 0.0625, 0.25, 0.9375, 0.9375 5. 0.265 7. f(x) = 0.5xe-°3/x\, /(0) + f(l) 9. 1 - e'0'2 = 18% it 120 13s 1j- 286- 28«' 286' 286 Problem Set 24.8, page 1031 1. 0.1587, 0.6306, 0.5, 0.4950 5. 16%. 9. About 23 13. t= 1084 hours 5. (23°) = 1140 9. 9!/(2!3!4!) = 1260. Ans. 1/1260 13. 1/84, 5/21 3. k = 1/8 by (10) 7. 1 - PCX S 3) = 0.5 = 0.896. Ans. 0.8963 = 72% = 1.1565; 26.9% 3. 3.5, 2.917 7. $643.50 11. i JL, (X - |)V20 15. 15c - 500c3 = 0.97, c = 0.0855 3. 64% c_0-5(1.0 + 0.5) = 0.91. Ans. 9% 11. 0.99100 = 36.6% 3. 17.29, 10.71, 19.152 7. 31.1%, 95.5%. 11. About 58% - 1.5)2] dx 13. k Problem Set 24.9, page 1040 1.1/8,3/16,3/8 3.2/9,2/9,1/2 5- fz(y) = l/(/32 ~ a2) if «2 < >' < fh. anu 0 elsewhere 7. 27.45 mm, 0.38 mm 9. 25.26 cm, 0.0078 cm App. 2 Answers to Odd-Numbered Problems A57 13. Independent, fx(x) = 0.1 0, f2(y) = 0.le~0-ly if v > 0, 36.8% 15. 50% 17. No Chapter 24 Review Questions and Problems, page 1041 21. QL = 22.3, QM = 23.3, Q0 = 23.5 23. x = 22.89, s = 1.028, s2 = 1.056 25. H, TH, TTH, etc. 27. /(0) = 0.80816, f(l) = 0.18367, ,f(2) = 0.00816 29. Always B C A U fi. If also ACJ), then fi = AUB, etc. 31. 7/3, 8/9 33. 118.019, 1.98, 1.65% 35. 0, 2 37. p = 100/30 39. 16%, 2.3% (see Fig. 520 in Sec. 24.8) Problem Set 25.2, page 1048 3. / = pk(\ — pj"~k, p = kin, k = number of successes in n trials 5. 11/20 7. / = f(x), d(ln /)% = L/p - (a- - 1)/(1 -p) = 0, p = 1/a- 9. fl = x 11. 0 = w/2 a- = l/.v 13. 0 = 1 15. Variability larger than perhaps expected Problem Set 25.3, page 1057 1. CONF0.95 {37.47 g /a S 43.87} 3. Shorter by a factor V2 5. 4, 16 7. Cf. Example 2. n = 166 9. CONF0.99{ 20.07 20.33} 11. CONF0.99{ 63.71 g ^ g 66.29} 13. c = 1.96, x = 87, ^2 = 87 • 413/500 = 71.86, k - cslVn = 0.743, CONFa9g {86 S m = 881- CONF095{0.17 0.18} 15. CONF095{ 0.00045 £(r2s 0.00131} 17. CONF095{0.73 o-2 g 5.19} 19. CONF()95{23 g a2 ^ 548}. Hence a larger sample would be desirable. 21. Normal distributions, means —27, 81, 133, variances 16, 144, 400 23. Z = X + Y is normal with mean 105 and variance 1.25. Am. P(104 g Z ^ 106) = 63% Problem Set 25.4, page 1067 1. t = V7(0.286 - 0)/4.31 = 0.18 < c = 1.94; do not reject the hypothesis. 3. c = 6090 > 6019; do not reject the hypothesis. 5. cP'ln = I, c — 28.36; do not reject the hypothesis. 7. fi < 28.76 or ft > 31.24 9. Alternative p * 1000, t = V20(996 - 1000)/5 = -3.58 < c = - 2.09 (Table A9, 19 degrees of freedom). Reject the hypothesis p = 1000 g. 11. Test p. = 0 against p. + 0. t = 2.11 < c = 2.36 (7 degrees of freedom). Do not reject the hypothesis. 13. a = 5%, c = 16.92 > 9 • 0.52/0.42 = 14.06; do not reject hypothesis. 15. l0 = VlO-9 - 17/19 (21.8 - 20.2)/V'9 ■ 0.62 + 8 • 0.52 = 6.3 > c = 1.74 (17 degrees of freedom). Reject the hypothesis and assert that B is better. App. 2 Answers to Odd-Numbered Problems 17. v0 = 50/30 = 1.67 < c = 2.59 [(9. 15) degrees of freedom]; do not reject the hypothesis. Problem Set 25.5, page 1071 1. LCL = 1 - 2.58 • 0.03/V6 = 0.968, UCL = 1.032 3. n = 10 5. Choose 4 times the original sample size (why?). 7. 2.58V0.024/V2 = 0.283, UCL = 27.783, LCL = 27.217 11. In 30% (5%) of the cases, approximately 13. UCL = np + 3Vnp{\ - p), CL = np, LCL = np - 3\'np(l - p) 15. CL = fi = 2.5, UCL = fi + 3\-> = 7.2, LCL = p. - 3V> is negative in (b) and we set LCL = 0. Problem Set 25.6, page 1076 1. 0.9825, 0.9384, 0.4060 3. 0.8187, 0.6703, 0.1353 Problem Set 25.7, page 1079 1. xo2 = (30 - 50)2/5() + (70 - 50)2/50 = 16 > c = 3.84; no 3. 41 5. xoZ = 2.33 < c = 11.07. Yes 7. ej = npj = 370/5 = 74, x02 = 984/74 = 13.3, c = 9.49. Reject the hypothesis. 9- Xo2 = 1 < 3.84; yes 13. Combining the results for x = 10, 11, 12, we have K — r — 1 = 9 (r = 1 since we estimated the mean, 1^ == 3.87). Xo2 = 12-98 < c = 16.92. Do not reject. 15. = 49/20 + 49/60 = 3.27 < c = 3.84 (1 degree of freedom, a = 5%), which supports the claim. 17. 42 even digits, accept. Problem Set 25.8, page 1082 3. (|)18(1 + 18 + 153 + 816) = 0.0038 5. Hypothesis: A and B are equally good. Then the probability of at least 7 trials favorable to A is |a + 8 • g8 = 3.5%. Reject the hypothesis. 7. Hypothesis p, = 0. Alternative p. > 0, x = 1.58, t = V'10- 1.58/1.23 = 4.06 > c = 1.83 (a = 5%). Hypothesis rejected. 9. x = 9.67, s = 11.87, t0 = 9.67/(11.87/VT5) = 3.15 > c = 1.76 (a = 5%). Hypothesis rejected. 11. Consider Vj = Xj — pl0. 13. P(T S 2) = 0.1% from Table A12. Reject. 5. P(A\ 9) « e_30s(l + 300) 7. P(A; 0) « e-50fi 11. (1 - 0f, (1 - 0)5 + 500 - Of 15. 4>((9 - 12 + |)/V'12(1 - 0.12)) 17.(1 - if + 3-1(1 -|)2 =1 9. 19.5%, 14.7% 13. Because n is finite 0.22 (if c = 9) App. 2 Answers to Odd-Numbered Problems 15. P(T g 15) = 10.8%. Do not reject. 17. P(T S2)= 2.8%. Reject. Problem Set 25.9, page 1091 1. y = 1.9 + x 3. y = 6.7407 + 3.068 x S.y = 4 + 4.8.y, 172 ft 7. y = -1146 + 4.32.1- 9.y = 0.5932 + 0.1138x, R = 1/0.1138 11. q0 = 76,K= 2.36V76/(7 • 944) = 0.253, CONF(),85{ -1.58 g ^ g -1.06) 13. 3.s.,2 = 500. 3sxy = 33.5, kx = 0.067, 3sv2 = 2.268, q0 = 0.023, K = 0.021 CÖNF0.9S{0.046 sKlg 0.088} Chapter 25 Review Questions and Problems, page 1092 21. ß. = 5.33, a2 = 1722 23. lt will double. 25. CONF099{ 19.1 S fjL g 33.7) 27. CONF0 95{0.726 S/ii 0.751} 29. CONF0.95( 1.373 S/iS 1.451) 31. CONF099{0.05 § er2 g 10) 37. 30.14/3.8 = 7.93 < 8.25. Reject. 39. u0 = 2.5 < 6.0 [(9, 4) degrees of freedom); accept the hypothesis. 41. Decrease by a factor V2. By a factor 2.58/1.96 = 1.32. 33. c = 14.74 > 14.5; reject /jl0. 43. 0.9953, 0.9825. 0.9384, etc. 45. y = 1.70 + 0.55.v APPENDIX 3 Auxiliary Material A3.1 Formulas for Special Functions For tables of numeric values, see Appendix 5. Exponential function ex (Fig. 544) e = 2.71828 18284 59045 23536 02874 71353 (!) = ex+v, eJTev = ex~v, {exf = exlJ Natural logarithm (Fig. 545) (2) In (xy) = In x + In y, In (x/y) = In x — In y. In (xa) = a In x In x is the inverse of ex, and cln a; = x, e_ln x = e1" n/x) = 1/x. Logarithm of base ten log10x or simply log x (3) log x = M In x, M = log e = 0.43429 44819 03251 82765 11289 18917 1 1 (4) lnx = — logx, — = In 10 = 2.30258 50929 94045 68401 79914 54684 M M logx is the inverse of 10x, and 10logx = x, 10-logx = 1/x. Sine and cosine functions (Figs. 546, 547). In calculus, angles are measured in radians, so that sin x and cos x have period 2tt. sin x is odd, sin (—x) = —sin x, and cos x is even, cos (—x) = cos x. -2 0 2 x Fig. 544. Exponential function e* Fig. 545. Natural logarithm In x A60 SEC. A3.1 Formulas for Special Functions A61 15) (6) (7) (8) (9) (10) (11) (12) y l Y 1 / ' -l V K ./ 2k Fig. 546. sin x Fig. 547. cos x sin2 x + cos2 x 1° = 0.01745 32925 19943 radian 1 radian = 57° 17' 44.80625" = 57.29577 95131° 1 sin (x + y) = sin x cos y + cos x sin y sin (x — y) = sin x cos y — cos x sin y cos (x + y) = cos x cos y — sin a' sin y cos (a — y) - cos a cos y 4- sin x sin y sin 2a = 2 sin x cos a, cos 2a = cos2 x — sin2 x cos a ") (" tJ = cos (t (" t) = sin 2 sin (tt — a) = sin x, ,2 , _ 1, cos (77 — a) = —cos a cosza = 5(1 + cos 2a), sin2 a = §(1 — cos 2a) sin a sin)' = I [ — cos (a + y) + cos (a — y)] cos a cos y = |[cos (a + y) + cos (x — y)] sin a cos y = ||sin (a + y) + sin (a — y)| « + v u — v sin h + sin v = 2 sin-cos- 2 2 cos // + cos v = 2 cos ■ v k — i,' — cos- 2 u k — V cos (J — cos U sin - 2 (13) A Cosa + ß sinx = Va2 + B2 cos (a ± 5), (14) a cos a + B sin a = va2 + S2 sin (a ± 8), tan 8 = tan 8 sin 5 cos 5 sin 8 cos S ß A~ A R A62 APP. 3 Auxiliary Material Fig. 548. tan x Tangent, cotangent, secant, cosecant (Figs. 548, 549) Fig. 549. cot x (15) tanx sin x cos x cotx sin x sec x cos x (16) tanix • v) tanx + tan y tan (x — y) esc x tanx — tan y tan x tan y 1 + tan x tan y Hyperbolic functions (hyperbolic sine sinhx, etc.; Figs. 550, 551) (17) (18) (19) (20) (21) sinhx = g (ť' - e x), cosh x = |(e* + e x) tanh x sinh x coth x cosh x cosh x sinh x cosh x + sinh x = ex, cosh x — sinh x = e~x cosh2 x — sinh2 x = 1 sinh2x = |(cosh2x — 1), cosh2x = |(cosh2x + 1) j_l j_L 2 x (dashed) and cosh x Fig. 551. tanh x (dashed) and coth x SEC. A3.1 Formulas for Special Functions (22) sinh (x ± y) = sinh x cosh y ± cosh x sinh y cosh (x ± y) = cosh x cosh y ± sinh x sinh y tanh x ± tanh y (23) tanh (x ± y) = —---~- 1 ± tanh x tanh y Gamma function (Fig. 552 and Table A2 in App. 5). The gamma function T(a) is defined by the integral rx (24) T(a) = e_tr_1 dr (a > 0) which is meaningful only if a > 0 (or, if we consider complex a, for those a whose real part is positive). Integration by parts gives the important functional relation of the gamma function, (25) T(a + 1) = aT(a). From (24) we readily have F(l) = 1; hence if a is a positive integer, say k, then by repeated application of (25) we obtain (26) T(k + 1) = k\ (k = 0, 1, • • •)■ This shows that the gamma function can be regarded as a generalization of the elementary factorial function. [Sometimes the notation (a — 1)1 is used for F(a), even for noninteger values of a, and the gamma function is also known as the factorial function.] By repeated application of (25) we obtain F(a + 1) T(a + 2) T(a + k + 1) Y{a) = - a(a + 1) a(a + ])(a + 2) ■ ■ ■ (a + k) Via) i I)' ! i i i i i -4 A h 2 4a --2 --4 Fig. 552. Gamma function APP. 3 Auxiliary Material and we may use this relation T(a + k + 1) (27) r(a) = -(tt*0, -1,-2, •••) a(a + 1) • • ■ (a + k) for defining the gamma function for negative a (# —1, —2, ■ ■ ■), choosing for k the smallest integer such that a + k + 1 > 0. Together with (24), rftis then gives a definition of T {a) for all a not equal to zero or a negative integer (Fig. 552). It can be shown that the gamma function may also be represented as the limit of a product, namely, by the formula (28) r(o) = lim - -——— (a * 0, -1, > ■ ■). n-^x a(a + \)(a + 2) • • • (a + n) From (27) or (28) we see that, for complex a, the gamma function i(a) is a meromorphic function with simple poles at a = 0, — 1, — 2, ■ * * . An approximation of the gamma function for large positive a is given by the Stirling formula (29) T(a + 1) = Vlwa \ — where e is the base of the natural logarithm. We finally mention the special value (30) r(|) = Vtt. Incomplete gamma functions X oo (31) P(a,x) = J e-*?"-1 dt, Q(a, x) = J e'Y'-1 dt (a > 0) 0 x (32) T(a) = P(a, x) + Q(a, x) Beta function (33) B(x, y) = f ;x_1(l - tf~x dt (x > 0, y > 0) Representation in terms of gamma functions: r(x)r(v) (34) B(x, y) = V- r(x + >■) Error function (Fig. 553 and Table A4 in App. 5) SEC. A3.1 Formulas for Special Functions A65 Fig. 553. Error function erf (°°) = 1, complementary error function (37) 2 f _,2 erlc.T = 1 — err* = —— e dt Vtt jx Fresnel integrals1 (Fig. 554) (38) C(*) = | cos (t2) dt, S(*) = f sin (f2) dt C(°°) = Vtt/8, S(cc) = Vtt/8, complementary functions CW = ,7 - C(*) = I cos (r2) J/ oo ) = [ sin (t2) dt X Sine integral (Fig. 555 and Tabic A4 in App. 5) f sin t (40) Si(*) = - dt (39) 0.5 c (x) .__. ^ - - //\" - Z~7 / ^ — / / \ ~y / \_. / / -—— - - / ~A I I a- 4~f I I I 1 I I i i' 1 I I i I 1 , i i i 1 i i i 1 i [ i 1 1 1 1 1 1 1 2 3 Fig. 554. Fresnel integrals 4 x 1 AUGUSTUS! FRESNEL (1788-1827), French physicist and mathematician. For tables see Ref. [GRIJ. A66 APP. 3 Auxiliary Material Site) 2 n/2 1 j_i_i_l 0 5 Fig. 555. Sine integral 10 x Si(oo) = 7j-/2, complementary function (41) si(x) = — - Si(% f sin / ) = J — * Cosine integral (Tabic A4 in App. 5) cos t (42) ci(x) = - dt Exponential integral (43) Ei(x) = f — dt Logarithmic integral rx dt (44) \i(x) = — Jo In t (x > 0) (x > 0) A3.2 Partial Derivatives For differentiation formulas, see inside of front cover. Let z = f(x, y) be a real function of two independent real variables, x and y. If we keep y constant, say, y = \\, and think of x as a variable, then f(x, yx) depends on x alone. If the derivative of f(x, \\) with respect to x for a value x = xx exists, then the value of this derivative is called the partial derivative of f(x, y) with respect to x at the point (xlt y-f) and is denoted by dx tel.yi) Other notations are or by and dz dX Czi,vO zx(x\, v,): these may be used when subscripts are not used for another purpose and there is no danger of confusion. SEC. A3.2 Partial Derivatives A67 We thus have, by the definition of the derivative, (1) 01 dx (a-i.Bi) = lim f JXX + Ax, \\) - f(xx, yx) Ax The partial derivative of z, = ,f(x, y) with respect to y is defined similarly; we now keep x constant, say, equal to xlt and differentiate f(xx, y) with respect to v. Thus (2) 3/ 8y (xi.yi) dZ ; lim Ay ^0 /(xx, yx + Ay) - f (xx, yx) Av Other notations are fy(xx, yx) and Zj,^, >'i). It is clear that the values of those two partial derivatives will in general depend on the point [xx, y{). Hence the partial derivatives dz/dx and c>z/dy at a variable point (x, y) are functions of x and y. The function dz/<)x is obtained as in ordinary calculus by differentiating z = fix, y) with respect to x, treatingy as a constant, and dz/dy is obtained by differentiating z with respect to y, treating x as a constant. EXAMPLE 1 Let fix, y) = .v y + x sin v. Then 2xy + sin y, dy x + x cos v. The partial derivatives dz/dx and dz/dy of a function z = fix, y) have a very simple geometric interpretation. The function z = fix, y) can be represented by a surface in space. The equation y = yx then represents a vertical plane intersecting the surface in a curve, and Ihe partial derivative dz/dx at a point (xx, \\) is the slope of the tangent (that is, tan a where a is the angle shown in Fig. 556) to the curve. Similarly, the partial derivative dz/dy at (Xj, yx) is the slope of the tangent to the curve x = xx on the surface z = fix,, y) at (xi, yj). Fig. 556 A68 The partial derivatives dz/dx and dzldy are called first partial derivatives or partial derivatives of first order. By differentiating these derivatives once more, we obtain the four second partial derivatives (or partial derivatives of second order)1 J XX fyx fxy fyy (3) a2/ 8 (df_ dx2 ' dx \ dx d2f d dx dy dx \ >. By differentiating the second partial derivatives again with respect to x and y, respectively, we obtain the third partial derivatives or partial derivatives of the third order of /, etc. If we consider a function fix, y, z) of three independent variables, then we have the three first partial derivatives fx(x, y, z), fy(x, y, z), and fz(x, y, z). Here fx is obtained by differentiating f with respect to x, treating both y and z as constants. Thus, analogous to (1), we now have df_ dx = lim (.ti,öi,zi) Ax—>0 f(Xi + AX, >'!, ZX) ~ f(Xj_, >'!, Zl) Ax etc. By differentiating fy, fz again in this fashion we obtain the second partial derivatives of /, etc. EXAMPLE 3 Let fix, y, z) = x2 + y2 + 2 + xy ez. Then fx = 2x + y f', fy 2y i x e\ fz = 2z + xy ez. fxx — 2. fxy ~ fyx ~~ ^ » fxz ~ fzx ~ X ^ > fyy = 2. fyz = fzy = x ez, fzl = 2 + xy ez. ■ 2 CAUTION! In the subscript notation the subscripts are written in the order in which we differentiate, whereas in the "d" notation the order is opposite. SEC. A3.3 Sequences and Series A69 A3.3 Sequences and Series See also Chap. 15. Monotone Real Sequences We call a real sequence xv .v2, • • ■, xn, • • ■ a monotone sequence if it is either monotone increasing, that is, We call xu x2, • ' • a bounded sequence if there is a positive constant K such that \xn\ < K for all n. PROOF Let xx, x2, ■ ■ • be a bounded monotone increasing sequence. Then its terms are smaller than some number B and, since xv S xn for all n, they lie in the interval xx — xn § B, which will be denoted by I0. We bisect /0; that is. we subdivide it into two parts of equal length. If the right half (together with its endpoints) contains terms of the sequence, we denote it by IL. If it does not contain terms of the sequence, then the left half of/0 (together with its endpoints) is called Ix. This is the first step. In the second step we bisect /l7 select one half by the same rule, and call it J2, and so on (see Fig. 557 on p. A70). In this way we obtain shorter and shorter intervals /0, Ilt I2, • • • with the following properties. Each Im contains all /,„ for n > m. No term of the sequence lies to the right of I.m, and, since the sequence is monotone increasing, all xn with n greater than some number N lie in Im\ of course, N will depend on in, in general. The lengths of the im approach zero as m approaches infinity. Hence there is precisely one number, call it L, that lies in all those intervals,3 and we may now easily prove that the sequence is convergent with the limit L. In fact, given an e > 0, we choose an m such that the length of /„, is less than e. Then L and all the x,„ with n > N(m) lie in Im, and, therefore, \xn — L\ < e for all those n. This completes the proof for an increasing sequence. For a decreasing sequence the proof is the same, except for a suitable interchange of "left" and "right" in the construction of those intervals. ■ 3This statement seems to be obvious, but actually il is not; it may be regarded as an axiom of the real number system in the following form. Let yx, J2. ■ ■ ■ be closed intervals such that each Jm contains all Jn with n > m, and the lengths of the Jm approach zero as m approaches infinity. Then there is precisely one real number that is contained in all those intervals. This is the so-called Cantor-Dcdekind axiom, named after the German mathematicians GEORG CANTOR (1845-1918), the creator of set theory, and RICHARD DEDEKIND (1831-1916). known for his fundamental work in number theory. For further details sec Ref. [GR2] in App. 1. (An interval / is said lo be closed if its two endpoints are regarded as points belonging to /. It is said to be open if the endpoints are not regarded as points of /.) or monotone decreasing, that is, x2 g x3 s • • - . THEOREM 1 If a real sequence is bounded and monotone, it converges. A70 APP. 3 Auxiliary Material x, x„ x., i i i i i mini B j h-h- Fig. 557. Proof of Theorem 1 THEOREM 2 Real Series Leibniz Test for Real Series Let X], x2, • ' " be real and monotone decreasing to zero, that is, (1) (a) .q ' x2 ' x3 " (b) lim xm — 0. 77?en ?/jc series with terms of alternating signs X\ A*2 ~t~ X3 A"4 "T converges, and for the remainder Kn after the nth term we have the estimate (2) \Rn\ g xn,!. PROOF Let s„ be the nth partial sum of the series. Then, because of (la), S\ = X], S2 = Xi — x2 is sx, SZ = s2 + x3 = s2> s3 = sl ~ (x2 ~ x3) — 's'l> so that s2 ' s3 ' Si, Proceeding in this fashion, we conclude that (Fig. 558) (3) sl = s3 = s5 which shows that the odd partial sums form a bounded monotone sequence, and so do the even partial sums. Hence, by Theorem 1, both sequences converge, say, lim s2r, = s, lim s2n "2 "A *3 "I Fig. 558. Proof of the Leibniz test SEC. A3.4 Grad, Div, Curl, V2 in Curvilinear Coordinates ATI Now, since s2n + l — s2n = Xj^+j, we readily see that (lb) implies s - s* = lim s2n+i ~ lim s2n = lim (s2n+i - s2n) = lim x2n+1 = 0. Hence s* = 5, and the series converges with the sum s. We prove the estimate (2) for the remainder. Since sn —* s, it follows from (3) that •Sfcn+l = S§ 52n and als0 *2n-l = = s2n- By subtracting j2n and s2n-i> respectively, we obtain s2n l-l — s2n = s ~ s2n = 0 1? S — •'>'2ri-l = ^2»! ~~ *2n-l' In these inequalities, the first expression is equal to x2n+1, the last is equal to — x2n, and the expressions between the inequality signs are the remainders R2n and R2n-i- Thus the inequalities may be written -x2n+l — R2rl = 0, 0 i£ R2fl_1 = — X2„ and we see that they imply (2). This completes the proof. ■ A3.4 Grad, Div, Curl, V2 in Curvilinear Coordinates To simplify formulas we write Cartesian coordinates x = xx, y = x2, z, = x3. We denote curvilinear coordinates by qlt q2, q3. Through each point P there pass three coordinate surfaces qx = const, q2 = const, q3 = const. They intersect along coordinate curves. We assume the three coordinate curves through P to be orthogonal (perpendicular to each other). We write coordinate transformations as (1) % = Xifai, q2, q3), x2 = x2{qx, q2, q3), x3 = x3(qu q2, q3). Corresponding transformations of grad, div, curl, and V2 can all be written by using (2) h^ii^2 k=l '~'Cll Next to Cartesian coordinates, most important are cylindrical coordinates c/, = r, q2 = 6. q3 = z. (Fig. 559a on p. A72) defined by (3) xx = qx cos q2 = r cos 6, x2 — qx sin q2 = r sin 6, x3 — q3 = z and spherical coordinates qL = r, q2 = 0, q3 = <]) (Fig. 559b) defined by4 ^ X] = qx cos q2 sin c/3 = r cos 6 sin , x2 = qv sin q2 sin q3 = r sin 0 sin x3 = qY cos q3 = r cos (p. 4This is the notation used in calculus and in many other books. It is logical since in it, 0 plays the same role as in polar coordinates. CAUTION! Some books interchange the roles of 0 and 4>. APP. 3 Auxiliary Material ? (r, 0,z) (a) Cylindrical coordinates (6) Spherical coordinates Fig. 559. Special curvilinear coordinates In addition to the general formulas for any orthogonal coordinates qu q2, we shall give additional formulas for these important special cases. Linear Element ds. In Cartesian coordinates, ds2 = dxf + dx22 + dxi For the ^-coordinates, (5) ds2 = lh2 dqx2 + h22 dq22 + h2 dq32. i.Vi (Sec. 9.5). For polar coordinates set dz = 0. (5") ds2 = dr2 + r2 sin2 $ dt? + r2 dd>2 (Cylindrical coordinates). (Spherical coordinates). Gradient, grad f = V/ = [fx, fZz, fxJ (partial derivatives; Sec. 9.7). In the (/-system, with u, v, w denoting unit vectors in the positive directions of the qlt q2, q3 coordinate curves, respectively, (6) (o\) (6") i df ia/ i of grad / = Vf =---:— u + — -— v +---;— w hx aqx h2 dq2 df 1 df df grad / V/ = — u + - — v + — dr r do dz Of 1 df 1 3/ grad / = V/ = — u + —— — v--— w dr r sin d (sin 4> Fs) (Spherical coordinates). Laplacian V2/ = V«Vf = div (grad /) = /.,. Xi t + / (Sec. 9.8) (8) V2/ 1 _ \ ftj c'Jt/! / dq2 \ h2 dq. + 2 92/ 1 0/ 1 r*2/ a2/ (8') V2/- — + (-• hyh2 df_ dq3 \ h3 8q3 (Cylindrical coordinates) V2/ a,-2 r 9r r2 sin2 SO2 ' 2 d^2 1 a2/ 1 d2f cot a/ Curl (Sec. 9.9): (Spherical coordinates). (9) curl F = V x F /ijU h2\ h3w 1 8 d d h\h2h3 h2F2 h3F3 For cylindrical coordinates we have in (9) (as in the previous formulas) h1 = hT = 1, h2 = h„ = qx — r, h3 = hz = 1 and for spherical coordinates we have li1 = hr =1, hz = h„ = c/i sin q3 = r sin <\>, hs = h$ = q, = r. » APPENDIX 4 Additional Proofs Section 2.6, page 73 PROOF OF THEOREM 1 Uniqueness1 Assuming that the problem consisting of the ODE (1) y" + p(x)y' + q(x)y = 0 and the two initial conditions (3) y(x0) = K0, y'(x0) = Kx has two solutions yt(x) and y2(x) on the interval / in the theorem, we show that their difference y(x) = y,(x) - y2(x) is identically zero on /; then yx = y2 on /, which implies uniqueness. Since (1) is homogeneous and linear, y is a solution of that ODE on /, and since yx and y2 satisfy the same initial conditions, y satisfies the conditions (10) We consider the function and its derivative From the ODE we have y(.r0) = 0, y'(x0) = 0. z(x) = y{xf + y'{xf z = 2yy + 2y y . y = -py - qy- By substituting this in the expression for z we obtain (11) z = 2yy' - 2py'2 - 2qyy . Now, since y and y are real, l.v ± y')2 - v2 • 2yv' • y'2 : ' 0. 'This proof was suggested by my colleague. Prof. A. D. Ziebur. In this proof we use formula numbers that have not yet been used in Sec. 2.6. A74 APP. 4 Additional Proofs A75 From this and the definition of z we obtain the two inequalities (12) (a) 2yy' g y2 + y'2 = z, (b) -2yy' g y2 + y'2 = z. From (12b) we have 2yy' § —z. Together, \2yy'] g z. For the last term in (11) we now obtain -2qyy' g |-2^yy'| = |-?||2yy'| § \q\z. Using this result as well as — p : ]/?| and applying (12a) to the term 2yy in (II), we find Z g Z + 2|/?]y'2 + k|z. Since y'2 = y2 + y'2 = z, from this we obtain z g (1 + 21/?] + |?|)z or, denoting the function in parentheses by h, (13a) z = hz for all x on A Similarly, from (11) and (12) it follows that -z = -2yy' + 2py'2 + 2 0). We use the method of reduction of order (see Sec. 2.1), that is, we determine u{x) such that y2(x) = uixjy^ix) is a solution of (1). By inserting this and the derivatives I i . r n li i ~ I l , II y2 = u y, + m>'i, y2 = u >>! + 2« .vy + uy\ into the ODE (1') wc obtain a-2(m"v1 + 2u'y[ + uy") + xb(u' \\ + uy[) + cuyx = 0. APP. 4 Additional Proofs A77 Since yx is a solution of (1'), the sum of the terms involving u is zero, and this equation reduces to 2 It I m 2 ' ' i / ' r\ x yxu + 2x y\U + xbj\U = 0. By dividing by x2yx and insetting the power series for b we obtain " 4. (n yi j. b° \ ' n u +12 — + — + •••!« =0. Here and in the following the dots designate terms that are constant or involve positive powers of x. Now from (7) if follows that y[ _ xr-x[ra0 + (r + l)ajx + • • ■] yx XT[a0 + a\x + ' ' 'J 1 / ra0 + (r + \)avx + ■ ■ • \ r ein -t- a-iX + ■ ■ ■ I x Hence the previous equation can be written , 2r + b0 (A) Since r = (1 — b(l)/2, the term (2r + b0)/x equals l/x, and by dividing by u we thus have u u' x By integration we obtain In u - -In x + ■ • • , hence u = (l/x)e('' °. Expanding the exponential function in powers of x and integrating once more, we sec that u is of the form u = Inx + + kzx2 + ■ ■ ■ . Inserting this into y2 = uylt we obtain for v2 a representation of the form (8). Case 3. Roots Differing by an Integer. We write rx = r and r2 = r — p where p is a positive integer. A first solution (9) yx(x) = xr*(a0 + axx + a2x2 + • • ■) can be determined as in Cases 1 and 2. We show that a second independent solution is of the form (10) y2(x) = kyx(x) Inx + xrHA0 + Axx + A2x2 + • - •) where we may have k =h 0 or k = 0. As in Case 2 we set y2 = uyx. The first steps are literally as in Case 2 and give Eq. (A), I 2> + b° + ■ ■ A78 APP. 4 Additional Proofs Now by elementary algebra, the coefficient b0 — 1 of r in (4) equals minus the sum of the roots, b0 - 1 = -fa + r2) = -(r + r - p) = -2r + p. Hence 2r + b0 = p + 1, and division by u gives P + 1 The further steps are as in Case 2. Integrating, we find In u = —(p + 1) ln.v + • • • , thus r -(P+l) (• • 0 u — x e where dots stand for some scries of nonnegative integer powers of x. By expanding the exponential function as before we obtain a series of the form I In A-P + 1 + xp ■ + x x We integrate once more. Writing the resulting logarithmic term first, we get // = kp In x kp-i px' kp+1x Hence, by (9) we get for y2 = uy\ the formula y2 = kpy\ In x + x*1 ■ ■ ■ - fcp_jXP 1 + • • • 1 (a0 + aLx + ■ ■ ■) But this is of the form (10) with k = kp since r± — p = r2 and the product of the two series involves nonnegative integer powers of x only. ■ Section 5.7, page 205 THEOREM Reality of Eigenvalues Ifp, q, r, and p in the Sturm-Li ouville equation (1) of Sec. 5.7 are real-valued and continuous on the interval a = x g b and r(x) > 0 throughout that interval (or r(x) < 0 throughout that interval), then all the eigenvalues of the Sturm—Liouville problem (1), (2), Sec. 5.1, are real. PROOF Let A = a + iB be an eigenvalue of the problem and let y(x) - u(x) + iv(x) be a corresponding eigenfunction; here a, B, u, and u are real. Substituting this into (1), Sec. 5.7, we have (pu + ipv')' + (q + ar + iBr)(u + iv) = 0. APP. 4 Additional Proofs A79 This complex equation is equivalent to the following pair of equations for the real and the imaginary parts: (pu'Y + (q + ar)u — Brv = 0 (pv'Y + (q + ar)v + Bru = 0. Multiplying the first equation by v, the second by —u and adding, we get -B(u2 + v2)r = uipu')' - v(pu')' = [(pv')u - (pu')u]'. The expression in brackets is continuous on a g x =s b, for reasons similar to those in the proof of Theorem 1, Sec. 5.7. Integrating over x from a to b, we thus obtain B J (h2 + v2)r dx p(uv — u v) Because of the boundary conditions the right side is zero; this is as in that proof. Since y is an eigenfunction, u2 + v2 # 0. Since y and r are continuous and r > 0 (or r < 0) on the interval a g x "s b, the integral on the left is not zero. Hence, B = 0, which means that A = a is real. This completes the proof. ■ Section 7.7, page 308 THEOREM Determinants The definition of a determinant «n «12 a1n «21 «22 • - • «2n. (7) D = det A = «n1 as given in Sec. 7.7 is unambiguous, that is, it yields the same value of D no matter which rows or columns we choose in developings. PROOF In this proof we shall use formula numbers not yet used in Sec. 7.7. We shall prove first that the same value is obtained no matter which row is chosen. The proof is by induction. The statement is true for a second-order determinant, for which the developments by the first row ana22 + «i2(~«2i) and by the second row «2i( — «12) + «22«n Sive tne same value a11a22 — a12a21. Assuming the statement to be true for an (n — l)st-order determinant, we prove that it is true for an nth-order determinant. APP. 4 Additional Proofs For this purpose we expand D in terms of each of two arbitrary rows, say, the /th and the 7'th, and compare the results. Without loss of generality let us assume i < j. First Expansion. We expand D by the /th row. A typical term in this expansion is (19) alkClk = alk-(-\fkMlk. The minor Mik of aik in D is an (n — l)st-order determinant. By the induction hypothesis we may expand it by any row. We expand it by the row corresponding to the /th row of D. This row contains the entries ajt (/ ¥= k). It is the (j — l)st row of Mik, because Mik does not contain entries of the /th row of D, and / < j. We have to distinguish between two cases as follows. Case I. If / < k, then the entry aj: belongs to the /th column of Mik (see Fig. 560). Hence the term involving aiX in this expansion is (20) afl ■ (cofactor of ajL in Mik) = a.}l ■ (- 1 )°~1'1 lMikjl where Mi(y7 is the minor of a^ in Mik. Since this minor is obtained from Mlk by deleting the row and column of Ujh it is obtained from D by deleting the /th and j th rows and the kth and /th columns of D. We insert the expansions of the Mik into that of D. Then it follows from (19) and (20) that the terms of the resulting representation of D are of the form (21 a) alkan ■ ( -1 )bMikj, (1 < k) where b = i + k + j + I - 1. Case II. If / > k, the only difference is that then aj7 belongs to the (/ — I )st column of Mik, because Mik does not contain entries of the kth column of D, and k < I. This causes an additional minus sign in (20), and, instead of (21a), we therefore obtain (21b) -aikajr(-\)bMikjl (/ > k) where b is the same as before. Fig. 560. Cases I and II of the two expansions of D APP. 4 Additional Proofs Second Expansion. We now expand D at first by the /th row. A typical term in this expansion is (22) = ajr(-\y+lM:jl. By the induction hypothesis we may expand the minor of in D by its /th row, which corresponds to the /th row of D, since j > i. Case I. If /c > /, the entry aik in that row belongs to the (A — 1 )st column of Mj(, because Mji does not contain entries of the /th column of D, and / < A (see Fig. 560). Hence the term involving aik in this expansion is (23) alk ■ (cofactor of aik in Mfl) = aih • ( -1)« Kfc-1)MjkiJ, where the minor of aik in M^j is obtained by deleting the /th and 7 th rows and the Ath and /th columns of Z) [and is, therefore, identical with MikjL in (20), so that our notation is consistent|. We insert the expansions of the into that of D. It follows from (22) and (23) that this yields a representation whose terms are identical with those given by (21a) when / < k. Case II. If k < I, then aik belongs to the Ath column of MjL, we obtain an additional minus sign, and the result agrees with that characterized by (21b). We have shown that the two expansions of D consist of the same terms, and this proves our statement concerning rows. The proof of the statement concerning columns is quite similar; if we expand D in terms of two arbitrary columns, say, the Ath and the /th, we find that the general term involving Ojiaik is exactly the same as before. This proves that not only all column expansions of D yield the same value, but also that their common value is equal to the common value of the row expansions of D. This completes the proof and shows that our definition of an nth-order determinant is unambiguous. ■ Section 9.3, page 377 PROOF OF FORMULA [2] We prove that in right-handed Cartesian coordinates, the vector product v = a X b = [alt a2, a3] x [/;,, b2, b3] has the components (2) vt = a2bd - aab2, v2 = a3bx - axb%, v3 = axb2 - a2bv We need only consider the case v 0. Since v is perpendicular to both a and b, Theorem 1 in Sec. 9.2 gives a • v = 0 and b • v = 0: in components [see (2), Sec. 9.2], a1v1 + a2v2 + a3v3 = 0 (3) bxvx + b2v2 + b3v3 = 0. APP. 4 Additional Proofs Multiplying the first equation by b3, the last by a3. and subtracting, we obtain («3/?! - fli/?3)i'i = (a2b3 - a3b2)v2. Multiplying the first equation by bx, the last by ax, and subtracting, we obtain {ayb2 - a%bx)v2 = (a3bL - axb3)v3. We can easily verify that these two equations arc satisfied by (4) Vi = c(a2b3 - a3b2), v2 = c(a3b1 - axb3), v3 = c{axb2 - a2bx) where c is a constant. The reader may verify by inserting that (4) also satisfies (3). Now each of the equations in (3) represents a plane through the origin in L'^Dg-space. The vectors a and b are normal vectors of these planes (see Example 6 in Sec. 9.2). Since v # 0, these vectors are not parallel and the two planes do not coincide. Hence their intersection is a straight line L through the origin. Since (4) is a solution of (3) and, for varying c, represents a straight line, we conclude that (4) represents L, and every solution of (3) must be of the form (4). In particular, the components of v must be of this form, where c is to be determined. From (4) we obtain |v|2 = v* + v.2 + v* = c2[(a2b3 - a3b2f + (a^ - axb3f + (axb2 - a2bx)2}. This can be written |v|2 = c2[{ax2 + a22 + a32)(b2 + b2 + b2) - (a,b, + a2b2 + a3b3f], as can be verified by performing the indicated multiplications in both formulas and comparing. Using (2) in Sec. 9.2, we thus have |v|2 = c2[(a • a)(b • b) - (a • b)2]. By comparing this with formula (12) in Team Project 24 of Problem Set 9.3 we conclude that c = ± 1. We show that c = +1. This can be done as follows. If we change the lengths and directions of a and b continuously and so that at the end a = i and b = j (Fig. 186a in Sec. 9.3), then v will change its length and direction continuously, and at the end, v = i x j = k. Obviously we may effect the change so that both a and b remain different from the zero vector and are not parallel at any instant. Then v is never equal to the zero vector, and since the change is continuous and c can only assume the values +1 or —1, it follows that at the end c must have the same value as before. Now at the end a = i, b = j, v = k and, therefore, ax = 1, b2 = 1, v3 = 1, and the other components in (4) are zero. Hence from (4) we see that v3 = C = +1. This proves Theorem 1. For a left-handed coordinate system, i X j = —k (see Fig. 186b in Sec. 9.3), resulting in c = — 1. This proves the statement right after formula (2). ■ APP. 4 Additional Proofs A83 Section 9.9, page 416 PROOF OF THE INVARIANCE OF THE CURL This proof will follow from two theorems (A and B), which we prove first. THEOREM A Transformation Law for Vector Components For any vector v the components vx, v2, v3 and v2*, v3* in any two systems of Cartesian coordinates xt, x3, x3 and jc,*, x2*. x3", respectively, are related by (1) and conversely (2) with coefficients (3) satisfying (4) V* = CllVl + cL2v2 + cl3v3 V2* = c21v1 + c22v2 + c23v3 v3 = c31vl c32v2 + c33v3, Vl = C'llUl* + c21v2* + c3lv3* CísPl c22v2* 4 C32U3 Ci3fl* + C23V2* + <;;;í'':í:í: cn = i*«i c2\ = j*'Í c31 = k*«i c22 = C32 = k*«j 2 ckjc» c13 = i*«k C23 = j**k c33 = k*k (/c, m = 1,2, 3), where the Kroneckcr delta2 is given by 8u km (k + m) (A = m) and i, j, k and i*, j*, k* denote the unit vectors in the positive xx-, x2-, x3- and Xi*-, x2*-, x3*'-directions, respectively. 2LEOPOLD KRONECKKR (1823-1891), German mathematician at Berlin, who made important contributions to algebra, group theory, and number theory. We shall keep our discussion completely independent of Chap. 7, but readers familiar with matrices should recognise that we are dealing with orthogonal transformations and matrices and that our present theorem follows from Theorem 2 in Sec. 8.3. A84 APP. 4 Additional Proofs PROOF The representation of v in the two systems are (5) (a) x = Vl\ + v2\ + v3k (b) v = u^i* + u2*j* + v3*k*. Since i* • i* = 1, i* • j* = 0, i* • k:I= = 0, we get from (5b) simply i* • v = vx* and from this and (5a) Vl* = j* • y = i* • uyi + i* • u2j + i* • v3k = vxi* • i + v2\* • j + v3i* • k. Because of (3), this is the first formula in (1), and the other two formulas are obtained similarly, by considering j* • v and then k* • v. Formula (2) follows by the same idea, taking i • v = vx from (5a) and then from (5b) and (3) Uj = i ■ v = vt*i ' i* + u2*i • j* + rj3*i • k* = CuBj* + c2Xv2* + ^3^3*, and similarly for the other two components. We prove (4). We can write (1) and (2) briefly as 3 3 (6) (a) vj = 2 cmjVm*, (b) L>fc* = 2 c^Vj. m = l j=l Substituting Vj into vk*, we get 33 3/3 \ vk* = 2 ckj 2 cmjV„* = 2 "m* 2 .•'"••V I ' j — l m—l m—1 v=l / where k = 1, 2, 3. Taking k = 1, we have Vr* = vx* 12 ''•''•j + is" 12 ri''> j + y3* 12 ri,--3,J • For this to hold for every vector v, the first sum must be 1 and the other two sums 0. This proves (4) with k = 1 for m = 1, 2, 3. Taking k: = 2 and then A: = 3, we obtain (4) with k = 2 and 3, form = 1, 2, 3. ■ THEOREM B Transformation Law for Cartesian Coordinates The transformation of any Cartesian x1x2x3-coordinale system into any other Cartesian xx*x2*x3*-coordinate system is of the form (7) tvz'f/7 coefficients (3) aW constants bx, b2, b3; conversely, (8) k = 1, 2, 3. APP. 4 Additional Proofs A85 Theorem B follows from Theorem A by noting that the most general transformation of a Cartesian coordinate system into another such system may be decomposed into a transformation of the type just considered and a translation; and under a translation, corresponding coordinates differ merely by a constant. PROOF OF THE INVARIANCE OF THE CURL We write again x,, x2, x3 instead of x, y, z, and similarly xt*, x2*, .v3* for other Cartesian coordinates, assuming that both systems are right-handed. Let al5 a2, a3 denote the components of curl v in the x^^-coordinates, as given by (1), Sec. 9.9, with x = x-i, y — xz, z = x3. Similarly, let ax*, a2*, a3* denote the components of curl v in the x1*x2*x3*-coordinate system. We prove that the length and direction of curl v are independent of the particular choice of Cartesian coordinates, as asserted. We do this by showing that the components of curl v satisfy the transformation law (2), which is characteristic of vector components. We consider av We use (6a), and then the chain rule for functions of several variables (Sec. 9.6). This gives dv3 dv2 / dvin" dvrn* «i = —---—- = 2 \L ™3 ~^z— L™2 dx? 5x3 , \ "x2 ax3 ^ m. — 1 * » / dvm* dx* _ dvm* dx* m=1 j= 1 From this and (7) we obtain dXj* dx2 dXj* dx3 3 dv, * m=lj—1 3 dVi* dv-r* J2 ~ (c33c22 ~ C3zC23)ai^ + (t'l3c'32 — c12c33)a2 + (c23c'l2 — C22c13)a3*- Note what we did. The double sum had 3X3 = 9 terms, 3 of which were zero (when m — j), and the remaining 6 terms we combined in pairs as we needed them in getting a2*, a3*. We now use (3), Lagrange's identity (see Team Project 24 in Problem Set 9.3) and k* x j* = -i* and k X j = -i. Then C33C22 ~ c32<-'23 = (k* * k)(j* • j) - (k* • j)(j* • k) = (k* x j*) • (k x j) = i* • i = cn, etc. A86 APP. 4 Additional Proofs Hence a, = cna,* + c2la2* + c3ia3*- This is of the form of the first formula in (2) in Theorem A, and the other two formulas of the form (2) are obtained similarly. This proves the theorem for right-handed systems. If the x^Xy-coordinates are left-handed, then k X j = +i, but then there is a minus sign in front of the determinant in (1), Sec. 9.9. ■ Section 10.2, pages 426-427 PROOF OF THEOREM 1, PART (b) We prove that if (1) f F(r) • dr = \ (Fx dx + F2 dy + F3 dz) Jc Jc with continuous Fx, F2, F3 in a domain D is independent of path in D, then F = grad / in D for some f; in components (2') Fi = ~ , F2 = ^ ■ F3 = ~ . dx dy dz We choose any fixed A: (x0, y0, Zq) in D and any B: (x, y, z) in D and define / by (3) f(x, y, z) = f0 + f (F, dx* + F2 dy* + F3 dz*) JA with any constant f0 and any path from A to in D. Since A is fixed and we have independence of path, the integral depends only on the coordinates x, y, z, so that (3) defines a function f(x, y, z) in D. We show that F = grad / with this /, beginning with the first of the three relations (2r). Because of independence of path we may integrate from A to Bv: (xL, y, z) and then parallel to the x-axis along the segment BXB in Fig. 561 with Bx chosen so that the whole segment lies in D. Then B, B f(x, y, z) = f0 + (Fx dx* + F2 dy* + F3 dz*) + (Ft dx* + F2 dy* + F3 dz*). JA JB1 We now take the partial derivative with respect to x on both sides. On the left we gel dfldx. We show that on the right we get Fx. The derivative of the first integral is zero because A: (x0, y0, z0) and (-^l; y, z) do not depend on x. We consider the second integral. Since on the segment B^B, both y and z are constant, the terms F2dy* and Fig. 561. Proof of Theorem 1 APP. 4 Additional Proofs A87 F3 dz* do not contribute to the derivative of the integral. The remaining part can be written as a definite integral, f /•, dx* = | / jf.vv, z) dx*. Hence its partial derivative with respect to x is Fx(x, y, z), and the first of the relations (2') is proved. The other two formulas in (2') follow by the same argument. Section 13.4, page 620 PROOF OF THEOREM 1 Cauchy-Riemann Equations We prove that Cauchy-Riemann equations (1) It, = Vy, Uy = ~VX are sufficient for a complex function f(z) = u(x, y) + iv{x, y) to be analytic; precisely, if the real part u and Ihe imaginary part v of f(z) satisfy (1) in a domain D in the complex plane and if the partial derivatives in (1) are continuous in D, then f(z) is analytic in D. In this proof we write Az = Ax + /Ay and A/ = f(z + Az) — f(z). The idea of proof is as follows. (a) We express Af in terms of first partial derivatives of u and v, by applying the mean value theorem of Sec. 9.6. (b) We get rid of partial derivatives with respect to y by applying the Cauchy-Riemann equations. (c) We let Az approach zero and show that then A//Az as obtained approaches a limit, which is equal to ux + ivx, the right side of (4) in Sec. 13.4, regardless of the way of approach to zero. (a) Let P: (x, y) be any fixed point in D. Since D is a domain, it contains a neighborhood of P. We can choose a point Q: (x + Ax, y + Ay) in this neighborhood such that the straight-line segment PQ is in D. Because of our continuity assumptions we may apply the mean value theorem in Sec. 9.6. This yields u(x + Ax, y + Ay) - u(x, y) = (Ax^Mj) + (Ay)uy{Mx) v(x + Ax, y + Ay) - v(x, y) = (Ax)vJM2) + (Ay)vy(M2) where Mx and M2 (# M1 in general!) are suitable points on that segment. The first line is Re Af and the second is Im Af, so that Af = {Ax)ux(M1) + (Ay)Uy(M,) + i[(Ax)vx(M2) + (Ay)vy(M2)]. (b) uy = — vx and vy = ux by the Cauchy-Riemann equations, so that Af = (Ax)Mt(Mx) - (Ay)vx(M-d + i[(Ax)vx(M2) + {Ay)ux{M2)]. A88 APP. 4 Additional Proofs Also Az = Ax + /Ay, so that we can write Ax = Az - I Ay in the first term and Ay = (Az — Ax)/i — —/(Az — Ax) in the second term. This gives Af = (Az - iAy)uT(Mx) + '(Az - Ax)vx(M1) + i[(Ax)vx(M2) + i A v)//,(,!/,) ]. By performing the multiplications and reordering we obtain Af = (Aztn,(Mi) - /Ay !„,.<>/,) ~ uT(M2)} + i[(Az)vx(Mi) ~ A.vj r (V/,) - r.(A/,i]|. Division by Az now yields Af iAy iAx (A) — = ux(Mi) + iuJMJ ~ -j- (n,(iWi) - «,(M2)} - — [vJMJ ~ vx(M2)}. (c) We finally let Az approach zero and note that |Av/Az| = 1 and |Ajc/Az| g 1 in (A). Then Q: (x + A.v, y + Ay) approaches P: (x, v), so that M, and M2 must approach P. Also, since the partial derivatives in (A) arc assumed to be continuous, they approach their value at P. In particular, the differences in the braces {• • ■} in (A) approach zero. Hence the limit of the right side of (A) exists and is independent of the path along which Az 0. We see that this limit equals the right side of (4) in Sec. 13.4. This means that f(z) is analytic at every point z in D, and the proof is complete. ■ Section 14.2, pages 647-648 GOURSAT'S PROOF OF CAUCHY'S INTEGRAL THEOREM Goursal proved Cauchy's integral theorem without assuming that /'(z) is continuous, as follows. We start with the case when C is the boundary of a triangle. We orient C counterclockwise. By joining the midpoints of the sides we subdivide the triangle into four congruent triangles (Fig. 562). Let CIs Cn, Cm, CIV denote their boundaries. We claim that (sec Fig. 562). Indeed, on the right we integrate along each of the three segments of subdivision in both possible directions (Fig. 562), so that the corresponding integrals cancel out in pairs, and the sum of the integrals on the right equals the integral on the left. We now pick an integral on the right that is biggest in absolute value and call its path Cj. Then, by the triangle inequality (Sec. 13.2), m, and (2) dp fdz ^4" j> fdz n = 1, 2, Let Zq be the point that belongs to all these triangles. Since / is differentiable at z the derivative / (z0) exists. Let (3) Kz) ./'■) - f(Z0) f'(Zo). Solving this algebraically for f(z) we have f(z) = /(Z0) + Cz - 2o)/'(zo) + Kz)(z - Zo). Integrating this over the boundary Cn of the triangle Tn gives 0 we can find a 8 > 0 such that (4) \h(z)\ < e when < 8. We may now take n so large that the triangle Tn lies in the disk |z - z0| < 8. Let Ln be the length of Cn. Then |z - z.ol < Ln for all z on Cn and z0 in Tn, From this and (4) we have \h(z)(z - z0)| < eLn. The A/L-iinequality in Sec. 14.1 now gives (5) /(z) dz h(z)(z ~ Zo) * Now denote the length of C by L. Then the path Cx has the length Lx = L/2, the path C2 has the length L2 = Lx/2 = LI A, etc., and C,, has the length Ln = L/2". Hence Lr2 = L2/4n. From (2) and (5) we thus obtain fdz 4n 4 A90 By choosing e (> 0) sufficiently small we can make the expression on the right as small as we please, while the expression on the left is the definite value of an integral. Consequently, this value must be zero, and the proof is complete. The proof for the case in which C is the boundary of a polygon follows from the previous proof by subdividing the polygon into triangles (Fig. 563). The integral corresponding to each such triangle is zero. The sum of these integrals is equal to the integral over C, because we integrate along each segment of subdivision in both directions, the corresponding integrals cancel out in pairs, and we are left with the integral over C. The case of a general simple closed path C can be reduced to the preceding one by inscribing in C a closed polygon P of chords, which approximates C "sufficiently accurately," and it can be shown that there is a polygon P such that the integral over P differs from that over C by less than any preassigned positive real number e, no matter how small. The details of this proof are somewhat involved and can be found in Ref. [D6| listed in App. 1. ■ Fig. 563. Proof of Cauchy's integral theorem for a polygon Section 15.1, page 667 PROOF OF THEOREM 4 Cauchy's Convergence Principle for Series (a) In this proof we need two concepts and a theorem, which we list first. 1. A bounded sequence sy, s2, • • • is a sequence whose terms all lie in a disk of (sufficiently large, finite) radius K with center at the origin; thus |sn| < K for all n. 2. A limit point a of a sequence Sy, s2, • • • is a point such that, given an e > 0, there are infinitely many terms satisfying \sn — a\ < e. (Note that this does not imply convergence, since there may still be infinitely many terms that do not lie within that circle of radius e and center a.) Example: |, |, §, a> Tfi' it- ' ' ' nas me limit points 0 and 1 and diverges. 3. A bounded sequence in the complex plane has at least one limit point. (Bolzano-Weierstrass theorem; proof below. Recall that "sequence" always mean infinite sequence.) (b) We now turn to the actual proof that Zy + z% + " ■ ■ converges if and only if for every e > 0 we can find an N such that (1) \Zn+i + ' ' ' + Zn t pi < e ror every n > N and p = 1, 2, • • • . Here, by the definition of partial sums, $n l p ->n <-n+1 "f ' "f Zn+p' APP. 4 Additional Proofs A91 Writing n + p = r, we see from this that (1) is equivalent to (1*) \sr - sn\ < e for all r > N and n > N. Suppose that sx, .v2, ■ ■ ■ converges. Denote its limit by s. Then for a given e > 0 we can find an N such that € \sn ~ s\ < y f°r every n > N. Hence, if r > N and a > N, then by the triangle inequality (Sec. 13.2), i i i i i i e e K ~ an\ = \(sr - s) - (sn - s)\ '■' \sr - s\ + \sn - s\ < — + — = e, that is, (1*) holds. (c) Conversely, assume that Aj, s2, • • ■ satisfies (1*). We first prove that then the sequence must be bounded. Indeed, choose a fixed e and a fixed n = n0 > N in (1*). Then (1 *) implies that all sr with r > N lie in the disk of radius e and center s„o and only finitely many terms slt • ■ • , sN may not lie in this disk. Clearly, we can now find a circle so large that this disk and these finitely many terms all lie within this new circle. Hence the sequence is bounded. By the Bolzano-Weierstrass theorem, it has at least one limit point, call it s. We now show that the sequence is convergent with the limit s. Let e > 0 be given. Then there is an N* such that \sT - sj < ell for all r > N* and n > N*, by (1*). Also, by the definition of a limit point, \sn — s\ < ell for infinitely many n, so that we can find and fix an n > N* such that |s„ — s\ < ell. Together, for every r > A/*, \sr - s\ = \(sr - sn) + [sn - 5)| ' |sr - sn\ + \sn - s\ < — + — = e; that is, the sequence sh s2, • ■ ■ is convergent with the limit s. ■ THEOREM Bolzano-Weierstrass Theorem3 A bounded infinite sequence z-y, Z2, z3, ■ ■ ■ in the complex plane has at least one limit point. PROOF It is obvious that we need both conditions: a finite sequence cannot have a limit point, and the sequence 1, 2, 3, ■ ■ ■ , which is infinite but not bounded, has no limit point. To prove the theorem, consider a bounded infinite sequence z±, z2, • • • and let K be such that \zn\ < K for all «. If only finitely many values of the zn are different, then, since the sequence is infinite, some number z must occur infinitely many times in the sequence, and, by definition, this number is a limit point of the sequence. We may now turn to the case when the sequence contains infinitely many different terms. We draw a large square Q0 that contains all z„. We subdivide Q0 into four congruent squares, which we number 1, 2, 3, 4. Clearly, at least one of these squares (each taken 3BKRNARD BOLZANO (1781-1848), Austrian mathematician and professor of religious studies, was a pioneer in the study of point sets, the foundation of analysis, and mathematical logic. For W'eierstrass, see Sec. 15.5. A92 APP. 4 Additional Proofs with its complete boundary) must contain infinitely many terms of the sequence. The square of this type with the lowest number (1, 2, 3, or 4) will be denoted by <2t- This is the first step. In the next step we subdivide (2i into four congruent squares and select a square Q2 by the same rule, and so on. This yields an infinite sequence of squares Q0, Qi> Q-2i' ' ' > Qm ' ' ' wrm the property that the side of Qn approaches zero as n approaches infinity, and Qm contains all Q„ with n > m. It is not difficult to see that the number which belongs to all these squares,4 call it z = a, is a limit point of the sequence. In fact, given an e > 0, we can choose an N so large that the side of the square QN is less than e and, since QN contains infinitely many zn, we have \zn — a\ < e for infinitely many n. This completes the proof. ■ Section 15.3, pages 681-682 PART (b) OF THE PROOF OF THEOREM 5 We have to show that 2 ««. (Z + At/ A; nz thus, 2 an Az[(z + Az)""2 + 2z(z + A;)"" (z + Azf - z + ••• + («- 1)Z"-2], n— 1 If we set (7a) Az = Az[(z + Az)""2 + 2z(z + Az)"-3 + ••• + («- l)zre"2l - Az = b and z = a, thus Az = b — a, this becomes simply bn - an = (b - a)Ar b — a where An is the expression in the brackets on the right, (7b) An = b'1-'2 + 2abn~3 + ?>a2bn-4 +••• + (« (n = 2, 3, IK thus, A2 = 1, A3 since then b + 2a, etc. We prove (7) by induction. When n — 2, then (7) holds, (b + a)(b - a) 2a 2a (h - a)A2 Assuming that (7) holds for n = k, we show that it holds for n = k + 1. By adding and subtracting a term in the numerator and then dividing we first obtain uk i 1 baK bak - a k+l = b bk - ak b — a + ak. 4 The fact that such a unique number z = a exists seems to be obvious, but it actually follows from an axiom of the real number system, the so-called Cantor-Dedekind axiom: see footnote 3 in App. A3.3. APP. 4 Additional Proofs A93 By the induction hypothesis, the right side equals b[(b — a)Ak + kak 1] + ak. Direct calculation shows that this is equal to (b - a){bAk + lea*'1} + cika*'1 + ak. From (7b) with n — k we see that the expression in the braces {• * •} equals b*'1 + 2abk~2 + ■ ■ • + (k - \)bak~2 + kak~l = Akl. Hence our result is bk+l - ak+1 --- =(b- a)Ak,! + (k + l)ak. b — a Taking the last term to the left, we obtain (7) with n = k + 1. This proves (7) for any integer n g 2 and completes the proof. ■ Section 18.2, page 754 ANOTHER PROOF OF THEOREM 1 without the use of a harmonic conjugate We show that if w = U + iv = f(z) is analytic and maps a domain D conformally onto a domain D* and $>*(u, v) is harmonic in D*, then (1) ®(x, y) = ®*(u(x, y), v(x, y)) is harmonic in 73, that is. V2 = 0 in D. We make no use of a harmonic conjugate of <£*, but use straightforward differentiation. By the chain rule, Wc apply the chain rule again, underscoring the terms that will drop out when we form V24): *„ + (cb* u + * v )u ^XX x U 1/1XX ' \^UUvlX Uu WX' llx + '!», -r,,,. + ($*,«,, + «»'*. rx)v,. <$yy is the same with each x replaced by y. We form the sum V2<1>. In it, 4>*„ = *„ is multiplied by UXVX + UyVy which is 0 by the Cauchy-Riemann equations. Also V2u = 0 and V2i> = 0. There remains Y24> = QZJm* + u2) + $Uvx2 + v2). By the Cauchy-Riemann equations this becomes V2<£ = (0>*u + *.)(h,2 + v2) and is 0 since * is harmonic. APPEN Tables D I X 5 For Tables of Laplace transforms see Sees. 6.8 and 6.9. For Tables of Fourier transforms see Sec. 11.10. If you have a Computer Algebra System (CAS), you may not need the present tables, but you may still find them convenient from time to time. Table Al Bessel Functions For more extensive tables see Ref. [GR1] in App. 1. X J0(x) X Mx) X 4C&) 0.0 1.0000 0.0000 3.0 -0.2601 0.3391 6.0 0.1506 -0.2767 0.1 0.9975 0.0499 3.1 -0.2921 0.3009 6.1 0.1773 -0.2559 0.2 0.9900 0.0995 3.2 -0.3202 0.2613 6.2 0.2017 -0.2329 0.3 0.9776 0.1483 3.3 -0.3443 0.2207 6.3 0.2238 -0.2081 0.4 0.9604 0.1960 3.4 -0.3643 0.1792 6.4 0.2433 -0.1816 0.5 0.9385 0.2423 3.5 -0.3801 0.1374 6.5 0.2601 -0.1538 0.6 0.9120 0.2867 3.6 -0.3918 0.0955 6.6 0.2740 -0.1250 0.7 0.8812 0.3290 3.7 -0.3992 0.0538 6.7 0.2851 -0.0953 0.8 0.8463 0.3688 3.8 -0.4026 0.0128 6.8 0.2931 -0.0652 0.9 0.8075 0.4059 3.9 -0.4018 -0.0272 6.9 0.2981 -0.0349 1,0 0.7652 0.4401 4.0 -0.3971 -0.0660 7.0 0.3001 -0.0047 1.1 0.7196 0.4709 4.1 -0.3887 0.1033 7.1 0.2991 0.0252 1.2 0.6711 0.4983 4.2 -0.3766 -0.1386 7.2 0.2951 0.0543 1.3 0.6201 0.5220 4.3 -0.3610 -0.1719 7.3 0.2882 0.0826 1.4 0.5669 0.5419 4.4 -0.3423 -0.2028 7.4 0.2786 0.1096 1.5 0.5118 0.5579 4.5 -0.3205 -0.2311 7.5 0.2663 0.1352 1.6 0.4554 0.5699 4.6 -0.2961 -0.2566 7.6 0.2516 0.1592 1.7 0.3980 0.5778 4.7 -0.2693 -0.2791 7.7 0.2346 0.1813 1.8 0.3400 0.5815 4.8 -0.2404 -0.2985 7.8 0.2154 0.2014 1.9 0.2818 0.5812 4.9 -0.2097 -0.3147 7.9 0.1944 0.2192 2.0 0.2239 0.5767 5.0 -0.1776 0.3276 8.0 0.1717 0.2346 2.1 0.1666 0.5683 5.1 -0.1443 -0.3371 8.1 0.1475 0.2476 2.2 0.1104 0.5560 5.2 -0.1103 -0.3432 8.2 0.1222 0.2580 2.3 0.0555 0.5399 5.3 -0.0758 -0.3460 8.3 0.0960 0.2657 2.4 0.0025 0.5202 5.4 -0.0412 -0.3453 8.4 0.0692 0.2708 2.5 -0.0484 0.4971 5.5 -0.0068 -0.3414 8.5 0.0419 0.2731 2.6 -0.0968 0.4708 5.6 0.0270 -0.3343 8.6 0.0146 0.2728 2.7 -0.1424 0.4416 5.7 0.0599 -0.3241 8.7 -0.0125 0.2697 2.8 -0.1850 0.4097 5.8 0.0917 -0.3110 8.8 -0.0392 0.2641 2.9 -0.2243 0.3754 5.9 0.1220 -0.2951 8.9 -0.0653 0.2559 J0(x) = 0 for J! = 2.40483, 5.52008, 8.65373, 11.7915, 14.9309, 18.0711, 21.2116, 24.3525, 27.4935, 30.6346 Jt(x) = 0 for* = 3.83171, 7.01559, 10.1735, 13.3237. 16.4706, 19.6159, 22.7601, 25.9037, 29.0468, 32.1897 A94 Table Al (continued) X Yt(x) X Y0(x) Yi(x) X Y0(x) Yiix) 0.0 (-co) (-*) 2.5 0.498 0.146 5.0 -0.309 0.148 0.5 -0.445 -1.471 3.0 0.377 0.325 5.5 -0.339 -0.024 1.0 0.088 -0.781 3.5 0.189 0.410 6.0 -0.288 -0.175 1.5 0.382 0.412 4.0 - 0.017 0.398 6.5 -0.173 -0.274 2.0 0.510 -0.107 4.5 -0.195 0.301 7.0 -0.026 -0.303 Table A2 Gamma Function [see (24) in App. A3.1J a T(a) a T(a) a r» a I» a i» 1.00 1.000 000 1.20 0.918 169 1.40 0.887 264 1.60 0.893 515 1.80 0.931 384 1.02 0.988 844 1.22 0.913 106 1.42 0.886 356 1.62 0.895 924 1.82 0.936 845 1.04 0.978 438 1.24 0.908 521 1.44 0.885 805 1.64 0.898 642 1.84 0.942 612 1.06 0.968 744 1.26 0.904 397 1.46 0.885 604 1.66 0.901 668 1.86 0.948 687 1.08 0.959 725 1.28 0.900 718 1.48 0.885 747 1.68 0.905 001 1.88 0.955 071 1.10 0.951 351 1.30 0.897 471 1.50 0.886 227 1.70 0.908 639 1.90 0.961 766 1.12 0.943 590 1.32 0.894 640 1.52 0.887 039 1.72 0.912 581 1.92 0.968 774 1.14 0.936 416 1.34 0.892 216 1.54 0.888 178 1.74 0.916 826 1.94 0.976 099 1.16 0.929 803 1.36 0.890 185 1.56 0.889 639 1.76 0.921 375 1.96 0.983 743 1.18 0.923 728 1.38 0.888 537 1.58 0.891 420 1.78 0.926 227 1.98 0.991 708 1.20 0.918 169 1.40 0.887 264 1.60 0.893 515 1.80 0.931 384 2.00 1.000 000 Table A3 Factorial Function and Its Logarithm with Base 10 n n\ log («!) n n\ log (n!) n n\ log («!) 1 1 0.000 000 6 720 2.857 332 11 39 916 800 7.601 156 2 2 0.301 030 7 5 040 3.702 431 12 479 001 600 8.680 337 3 6 0.778 151 8 40 320 4.605 521 13 6 227 020 800 9.794 280 4 24 1.380 211 9 362 880 5.559 763 14 87 178 291 200 10.940 408 5 120 2.079 181 10 3 628 800 6.559 763 15 1 307 674 368 000 12.116 500 Table A4 Error Function, Sine and Cosine Integrals [see (35), (40), (42) in App. A3.11 X erf x Si(x) ci(.r) X erf x Si(*) ciOi) 0.0 0.0000 0.0000 00 2.0 0.9953 1.6054 -0.4230 0.2 0.2227 0.1996 1.0422 2.2 0.9981 1.6876 -0.3751 0.4 0.4284 0.3965 0.3788 2.4 0.9993 1.7525 -0.3173 0.6 0.6039 0.5881 0.0223 2.6 0.9998 1.8004 -0.2533 0.8 0.7421 0.7721 -0.1983 2.8 0.9999 1.8321 0.1865 1.0 0.8427 0.9461 -0.3374 3.0 1.0000 1.8487 -0.1196 1.2 0.9103 1.1080 -0.4205 3.2 1.0000 1.8514 -0.0553 1.4 0.9523 1.2562 - 0.4620 3.4 1.0000 1.8419 0.0045 1.6 0.9763 1.3892 -0.4717 3.6 1.0000 1.8219 0.0580 1.8 0.9891 1.5058 -0.4568 3.8 1.0000 1.7934 0.1038 2.0 0.9953 1.6054 -0.4230 4.0 1.0000 1.7582 0.1410 A96 APP. 5 Tables Table A5 Binomial Distribution Probability function f(x) [see (2), Sec. 24.7] and distribution function F(x) P = = 0.1 P = = 0.2 P '■ = 0.3 P = 0.4 P = = 0.5 n ,t /w F(x) fix) Fix) m m m F(x) m Fix) 0. 0. 0. ü. 0. 1 0 9000 0.9000 8000 0.8000 7000 0.7000 6000 0.6000 5000 0.5000 1 1000 1.0000 2000 1.0000 3000 1.0000 4000 1.0000 5000 1.0000 0 8100 0.8100 6400 0.6400 4900 0.4900 3600 0.3600 2500 0.2500 2 1 1800 0.9900 3200 0.9600 4200 0.9100 4800 0.8400 5000 0.7500 2 0100 1.0000 0400 1.0000 0900 1.0000 1600 1.0000 2500 1.0000 0 7290 0.7290 5120 0.5120 3430 0.3430 2160 0.2160 1250 0.1250 1 2430 0.9720 3840 0.8960 4410 0.7840 4320 0.6480 3750 0.5000 j ? 0270 0.9990 0960 0.9920 1890 0.9730 2880 0.9360 3750 0.8750 3 0010 1.0000 0080 1.0000 0270 1.0000 0640 1.0000 1250 1.0000 0 6561 0.6561 4096 0.4096 2401 0.2401 1296 0.1296 0625 0.0625 1 2916 0.9477 4096 0.8192 4116 0.6517 3456 0.4752 2500 0.3125 4 2 0486 0.9963 1536 0.9728 2646 0.9163 3456 0.8208 3750 0.6875 3 0036 0.9999 0256 0.9984 0756 0.9919 1536 0.9744 2500 0.9375 4 0001 1.0000 0016 1.0000 0081 1.0000 0256 1.0000 0625 1.0000 0 5905 0.5905 3277 0.3277 1681 0.1681 0778 0.0778 0313 0.0313 1 3281 0.9185 4096 0.7373 3602 0.5282 2592 0.3370 1563 0.1875 2 0729 0.9914 2048 0.9421 3087 0.8369 3456 0.6826 3125 0.5000 J 3 0081 0.9995 0512 0.9933 1323 0.9692 2304 0.9130 3125 0.8125 4 0005 1.0000 0064 0.9997 0284 0.9976 0768 0.9898 1563 0.9688 5 0000 1.0000 0003 1.0000 0024 1.0000 0102 1.0000 0313 1.0000 0 5314 0.5314 2621 0.2621 1176 0.1176 0467 0.0467 0156 0.0156 1 3543 0.8857 3932 0.6554 3025 0.4202 1866 0.2333 0938 0.1094 2 0984 0.9841 2458 0.9011 3241 0.7443 3110 0.5443 2344 0.3438 6 3 0146 0.9987 0819 0.9830 1852 0.9295 2765 0.8208 3125 0.6563 4 0012 0.9999 0154 0.9984 0595 0.9891 1382 0.9590 2344 0.8906 5 0001 1.0000 0015 0.9999 0102 0.9993 0369 0.9959 0938 0.9844 6 0000 1.0000 0001 1.0000 0007 1.0000 0041 1.0000 0156 1.0000 0 4783 0.4783 2097 0.2097 0824 0.0824 0280 0.0280 0078 0.0078 1 3720 0.8503 3670 0.5767 2471 0.3294 1306 0.1586 0547 0.0625 2 1240 0.9743 2753 0.8520 3177 0.6471 2613 0.4199 1641 0.2266 7 3 0230 0.9973 1147 0.9667 2269 0.8740 2903 0.7102 2734 0.5000 / 4 0026 0.9998 0287 0.9953 0972 0.9712 1935 0.9037 2734 0.7734 5 0002 1.0000 0043 0.9996 0250 0.9962 0774 0.9812 1641 0.9375 6 0000 1.0000 0004 1.0000 0036 0.9998 0172 0.9984 0547 0.9922 7 0000 1.0000 0000 1.0000 0002 1.0000 0016 1.0000 0078 1.0000 0 4305 0.4305 1678 0.1678 0576 0.0576 0168 0.0168 0039 0.0039 1 3826 0.8131 3355 0.5033 1977 0.2553 0896 0.1064 0313 0.0352 2 1488 0.9619 2936 0.7969 2965 0.5518 2090 0.3154 1094 0.1445 3 0331 0.9950 1468 0.9437 2541 0.8059 2787 0.5941 2188 0.3633 8 4 0046 0.9996 0459 0.9896 1361 0.9420 2322 0.8263 2734 0.6367 5 0004 1.0000 0092 0.9988 0467 0.9887 1239 0.9502 2188 0.8555 6 0000 1.0000 0011 0.9999 0100 0.9987 0413 0.9915 1094 0.9648 7 0000 1.0000 0001 1.0000 0012 0.9999 0079 0.9993 0313 0.9961 8 0000 1.0000 0000 1.0000 0001 1.0000 0007 1.0000 0039 1.0000 Table A6 Poisson Distribution Probability function f(x) [see (5), Sec. 24.7] and distribution function Fix) X P /« = 0.1 Fix) fix) = 0.2 Fix) P = fix) = 0.3 Fix) P = fix) = 0.4 F(x) P fix) = 0.5 Fix) 0. 0. 0. 0. 0. 0 9048 0.9048 8187 0.8187 7408 0.7408 6703 0.6703 6065 0.6065 1 0905 0.9953 1637 0.9825 2222 0.9631 2681 0.9384 3033 0.9098 2 0045 0.9998 0164 0.9989 0333 0.9964 0536 0.9921 0758 0.9856 3 0002 1.0000 0011 0.9999 0033 0.9997 0072 0.9992 0126 0.9982 4 0000 1.0000 0001 1.0000 0003 1.0000 0007 0.9999 0016 0.9998 5 0001 1.0000 0002 1.0000 P '■ = 0.6 p' = 0.7 P = 0.8 P ' 0.9 P = 1 X fix) Fix) fix) Fix) fix) Fix) fix) Fix) fix) Fix) 0. 0. 0. 0. 0. 0 5488 0.5488 4966 0.4966 4493 0.4493 4066 0.4066 3679 0.3679 1 3293 0.8781 3476 0.8442 3595 0.8088 3659 0.7725 3679 0.7358 2 0988 0.9769 1217 0.9659 1438 0.9526 1647 0.9371 1839 0.9197 3 0198 0.9966 0284 0.9942 0383 0.9909 0494 0.9865 0613 0.9810 4 0030 0.9996 0050 0.9992 0077 0.9986 0111 0.9977 0153 0.9963 5 0004 1.0000 0007 0.9999 0012 0.9998 0020 0.9997 0031 0.9994 6 0001 1.0000 0002 1.0000 0003 1.0000 0005 0.9999 7 0001 1.0000 P m = 1.5 Fix) P fix) = 2 Fix) P fix) = 3 Fix) P fix) = 4 Fix) P fix) = 5 Fix) 0. 0. 0. 0. 0. 0 2231 0.2231 1353 0.1353 0498 0.0498 0183 0.0183 0067 0.0067 1 3347 0.5578 2707 0.4060 1494 0.1991 0733 0.0916 0337 0.0404 2 2510 0.8088 2707 0.6767 2240 0.4232 1465 0.2381 0842 0.1247 3 1255 0.9344 1804 0.8571 2240 0.6472 1954 0.4335 1404 0.2650 4 0471 0.9814 0902 0.9473 1680 0.8153 1954 0.6288 1755 0.4405 5 0141 0.9955 0361 0.9834 1008 0.9161 1563 0.7851 1755 0.6160 6 0035 0.9991 0120 0.9955 0504 0.9665 1042 0.8893 1462 0.7622 7 0008 0.9998 0034 0.9989 0216 0.9881 0595 0.9489 1044 0.8666 8 0001 1.0000 0009 0.9998 0081 0.9962 0298 0.9786 0653 0.9319 9 0002 1.0000 0027 0.9989 0132 0.9919 0363 0.9682 10 0008 0.9997 0053 0.9972 0181 0.9863 11 0002 0.9999 0019 0.9991 0082 0.9945 12 0001 1.0000 0006 0.9997 0034 0.9980 13 0002 0.9999 0013 0.9993 14 0001 1.0000 0005 0.9998 15 0002 0.9999 16 0000 1.0000 A98 APP. 5 Tables Table A7 Normal Distribution Values of the distribution function (z) [see (3), Sec. 24.8]. (-z) = 1 - <3?(z) z z z (z) z z. 0. 0. 0. 0. 0. 0. 0.01 5040 0.51 6950 1.01 8438 1.51 9345 2.01 9778 2.51 9940 0.02 5080 0.52 6985 1.02 8461 1.52 9357 2.02 9783 2.52 9941 0.03 5120 0.53 7019 1.03 8485 1.53 9370 2.03 9788 2.53 9943 0.04 5160 0.54 7054 1.04 8508 1.54 9382 2.04 9793 2.54 9945 0.05 5199 0.55 7088 1.05 8531 1.55 9394 2.05 9798 2.55 9946 0.06 5239 0.56 7123 1.06 8554 1.56 9406 2.06 9803 2.56 9948 0.07 5279 0.57 7157 1.07 8577 1.57 9418 2.07 9808 2.57 9949 0.08 5319 0.58 7190 1.08 8599 1.58 9429 2.08 9812 2.58 9951 0.09 5359 0.59 7224 1.09 8621 1.59 9441 2.09 9817 2.59 9952 0.10 5398 0.60 7257 1.10 8643 1.60 9452 2.10 9821 2.60 9953 0.11 5438 0.61 7291 1.11 8665 1.61 9463 2.11 9826 2.61 9955 0.12 5478 0.62 7324 1.12 8686 1.62 9474 2.12 9830 2.62 9956 0.13 5517 0.63 7357 1.13 8708 1.63 9484 2.13 9834 2.63 9957 0.14 5557 0.64 7389 1.14 8729 1.64 9495 2.14 9838 2.64 9959 0.15 5596 0.65 7422 1.15 8749 1.65 9505 2.15 9842 2.65 9960 0.16 5636 0.66 7454 1.16 8770 1.66 9515 2.16 9846 2.66 9961 0.17 5675 0.67 7486 1.17 8790 1.67 9525 2.17 9850 2.67 9962 0.18 5714 0.68 7517 1.18 8810 1.68 9535 2.18 9854 2.68 9963 0.19 5753 0.69 7549 1.19 8830 1.69 9545 2.19 9857 2.69 9964 0.20 5793 0.70 7580 1.20 8849 1.70 9554 2.20 9861 2.70 9965 0.21 5832 0.71 7611 1.21 8869 1.71 9564 2.21 9864 2.71 9966 0.22 5871 0.72 7642 1.22 8888 1.72 9573 2.22 9868 2.72 9967 0.23 5910 0.73 7673 1.23 8907 1.73 9582 2.23 9871 2.73 9968 0.24 5948 0.74 7704 1.24 8925 1.74 9591 2.24 9875 2.74 9969 0.25 5987 0.75 7734 1.25 8944 1.75 9599 2.25 9878 2.75 9970 0.26 6026 0.76 7764 1.26 8962 1.76 9608 2.26 9881 2.76 9971 0.27 6064 0.77 7794 1.27 8980 1.77 9616 2.27 9884 2.77 9972 0.28 6103 0.78 7823 1.28 8997 1.78 9625 2.28 9887 2.78 9973 0.29 6141 0.79 7852 1.29 9015 1.79 9633 2.29 9890 2.79 9974 0.30 6179 0.80 7881 1.30 9032 1.80 9641 2.30 9893 2.80 9974 0.31 6217 0.81 7910 1.31 9049 1.81 9649 2.31 9896 2.81 9975 0.32 6255 0.82 7939 1.32 9066 1.82 9656 2.32 9898 2.82 9976 0.33 6293 0.83 7967 1.33 9082 1.83 9664 2.33 9901 2.83 9977 0.34 6331 0.84 7995 1.34 9099 1.84 9671 2.34 9904 2.84 9977 0.35 6368 0.85 8023 1.35 9115 1.85 9678 2.35 9906 2.85 9978 0.36 6406 0.86 8051 1.36 9131 1.86 9686 2.36 9909 2.86 9979 0.37 6443 0.87 8078 1.37 9147 1.87 9693 2.37 9911 2.87 9979 0.38 6480 0.88 8106 1.38 9162 1.88 9699 2.38 9913 2.88 9980 0.39 6517 0.89 8133 1.39 9177 1.89 9706 2.39 9916 2.89 9981 0.40 6554 0.90 8159 1.40 9192 1.90 9713 2.40 9918 2.90 9981 0.41 6591 0.91 8186 1.41 9207 1.91 9719 2.41 9920 2.91 9982 0.42 6628 0.92 8212 1.42 9222 1.92 9726 2.42 9922 2.92 9982 0.43 6664 0.93 8238 1.43 9236 1.93 9732 2.43 9925 2.93 9983 0.44 6700 0.94 8264 1.44 9251 1.94 9738 2.44 9927 2.94 9984 0.45 6736 0.95 8289 1.45 9265 1.95 9744 2.45 9929 2.95 9984 0.46 6772 0.96 8315 1.46 9279 1.96 9750 2.46 9931 2.96 9985 0.47 6808 0.97 8340 1.47 9292 1.97 9756 2.47 9932 2.97 9985 0.48 6844 0.98 8365 1.48 9306 1.98 9761 2.48 9934 2.98 9986 0.49 6879 0.99 8389 1.49 9319 1.99 9767 2.49 9936 2.99 9986 0.50 6915 1.00 8413 1.50 9332 2.00 9772 2.50 9938 3.00 9987 APP. 5 Tables A99 Table A8 Normal Distribution Values of z for given values or Q>(z) [see (3), Sec. 24.8] and D(z) = 0%) - <5(-z) Example: z = 0.279 if O(z) = 61%; z = 0.860 if D(z) = 61%. % em Z(O) % z(») % Z(D) 1 -2.326 0.013 41 -0.228 0.539 81 0.878 1.311 2 -2.054 0.025 42 -0.202 0.553 82 0.915 1.341 3 -1.881 0.038 43 -0.176 0.568 83 0.954 1.372 4 -1.751 0.050 44 -0.151 0.583 84 0.994 1.405 5 -1.645 0.063 45 -0.126 0.598 85 1.036 1.440 6 -1.555 0.075 46 -0.100 0.613 86 1.080 1.476 7 -1.476 0.088 47 0.075 0.628 87 1.126 1.514 8 -1.405 0.100 48 -0.050 0.643 88 1.175 1.555 9 -1.341 0.113 49 -0.025 0.659 89 1.227 1.598 10 -1.282 0.126 50 0.000 0.674 90 1.282 1.645 11 -1.227 0.138 51 0.025 0.690 91 1.341 1.695 12 -1.175 0.151 52 0.050 0.706 92 1.405 1.751 13 -1.126 0.164 53 0.075 0.722 93 1.476 1.812 14 -1.080 0.176 54 0.100 0.739 94 1.555 1.881 15 -1.036 0.189 55 0.126 0.755 95 1.645 1.960 16 -0.994 0.202 56 0.151 0.772 96 1.751 2.054 17 -0.954 0.215 57 0.176 0.789 97 1.881 2.170 18 -0.915 0.228 58 0.202 0.806 97.5 1.960 2.241 19 -0.878 0.240 59 0.228 0.824 98 2.054 2.326 20 -0.842 0.253 60 0.253 0.842 99 2.326 2.576 21 -0.806 0.266 61 0.279 0.860 99.1 2.366 2.612 22 -0.772 0.279 62 0.305 0.878 99.2 2.409 2.652 23 -0.739 0.292 63 0.332 0.896 99.3 2.457 2.697 24 -0.706 0.305 64 0.358 0.915 99.4 2.512 2.748 25 -0.674 0.319 65 0.385 0.935 99.5 2.576 2.807 26 -0.643 0.332 66 0.412 0.954 99.6 2.652 2.878 27 -0.613 0.345 67 0.440 0.974 99.7 2.748 2.968 28 -0.583 0.358 68 0.468 0.994 99.8 2.878 3.090 29 -0.553 0.372 69 0.496 1.015 99.9 3.090 3.291 30 -0.524 0.385 70 0.524 1.036 31 -0.496 0.399 71 0.553 1.058 99.91 3.121 3.320 32 -0.468 0.412 72 0.583 1.080 99.92 3.156 3.353 33 -0.440 0.426 73 0.613 1.103 99.93 3.195 3.390 34 -0.412 0.440 74 0.643 1.126 99.94 3.239 3.432 35 -0.385 0.454 75 0.674 1.150 99.95 3.291 3.481 36 -0.358 0.468 76 0.706 1.175 99.96 3.353 3.540 37 -0.332 0.482 77 0.739 1.200 99.97 3.432 3.615 38 -0.305 0.496 78 0.772 1.227 99.98 3.540 3.719 39 -0.279 0.510 79 0.806 1.254 99.99 3.719 3.891 40 -0.253 0.524 80 0.842 1.282 A100 APP. 5 Tables Table A9 t-Distribution Values of z for given values of the distribution function F(z) (see (8) in Sec. 25.3) Example: For 9 degrees of freedom, z = 1-83 when F(z) = 0.95. Number of Degrees of Freedom 1 2 3 4 5 6 7 8 9 10 0.5 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.6 0.32 0.29 0.28 0.27 0.27 0.26 0.26 0.26 0.26 0.26 0.7 0.73 0.62 0.58 0.57 0.56 0.55 0.55 0.55 0.54 0.54 0.8 1.38 1.06 0.98 0.94 0.92 0.91 0.90 0.89 0.88 0.88 0.9 3.08 1.89 1.64 1.53 1.48 1.44 1.41 1.40 1.38 1.37 0.95 6.31 2.92 2.35 2.13 2.02 1.94 1.89 1.86 1.83 1.81 0.975 12.7 4.30 3.18 2.78 2.57 2.45 2.36 2.31 2.26 2.23 0.99 31.8 6.96 4.54 3.75 3.36 3.14 3.00 2.90 2.82 2.76 0.995 63.7 9.92 5.84 4.60 4.03 3.71 3.50 3.36 3.25 3.17 0.999 318.3 22.3 10.2 7.17 5.89 5.21 4.79 4.50 4.30 4.14 Number of Degrees of Freedom F(z) 11 12 13 14 15 16 17 18 19 20 0.5 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.6 0.26 0.26 0.26 0.26 0.26 0.26 0.26 0.26 0.26 0.26 0.7 0.54 0.54 0.54 0.54 0.54 0.54 0.53 0.53 0.53 0.53 0.8 0.88 0.87 0.87 0.87 0.87 0.86 0.86 0.86 0.86 0.86 0.9 1.36 1.36 1.35 1.35 1.34 1.34 1.33 1.33 1.33 1.33 0.95 1.80 1.78 1.77 1.76 1.75 1.75 1.74 1.73 1.73 1.72 0.975 2.20 2.18 2.16 2.14 2.13 2.12 2.11 2.10 2.09 2.09 0.99 2.72 2.68 2.65 2.62 2.60 2.58 2.57 2.55 2.54 2.53 0.995 3.11 3.05 3.01 2.98 2.95 2.92 2.90 2.88 2.86 2.85 0.999 4.02 3.93 3.85 .3.79 3.73 3.69 3.65 3.61 3.58 3.55 Number of Degrees of Freedom F(z) 22 24 26 28 30 40 50 100 200 00 0.5 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.6 0.26 0.26 0.26 0.26 0.26 0.26 0.25 0.25 0.25 0.25 0.7 0.53 0.53 0.53 0.53 0.53 0.53 0.53 0.53 0.53 0.52 0.8 0.86 0.86 0.86 0.85 0.85 0.85 0.85 0.85 0.84 0.84 0.9 1.32 1.32 1.31 1.31 1.31 1.30 1.30 1.29 1.29 1.28 0.95 1.72 1.71 1.71 1.70 1.70 1.68 1.68 1.66 1.65 1.65 0.975 2.07 2.06 2.06 2.05 2.04 2.02 2.01 1.98 1.97 1.96 0.99 2.51 2.49 2.48 2.47 2.46 2.42 2.40 2.36 2.35 2.33 0.995 2.82 2.80 2.78 2.76 2.75 2.70 2.68 2.63 2.60 2.58 0.999 3.50 3.47 3.43 3.41 3.39 3.31 3.26 3.17 3.13 3.09 APP. 5 Tables A101 Table A10 Chi-square Distribution Values of x for given values of the distribution function F(z) (see Sec. 25.3 before (17)). Example: For 3 degrees of freedom, z = 11.34 when F(z) = 0.99. Number of Decrees of Freedom m 1 2 3 4 5 6 7 8 9 10 0.005 0.00 0.01 0.07 0.21 0.41 0.68 0.99 1.34 1.73 2.16 0.01 0.00 0.02 0.11 0.30 0.55 0.87 1.24 1.65 2.09 2.56 0.025 0.00 0.05 0.22 0.48 0.83 1.24 1.69 2.18 2.70 3.25 0.05 0.00 0.10 0.35 0.71 1.15 1.64 2.17 2.73 3.33 3.94 0.95 3.84 5.99 7.81 9.49 1 1.07 12.59 14.07 15.51 16.92 18.31 0.975 5.02 7.38 9.35 11.14 12.83 14.45 16.01 17.53 19.02 20.48 0.99 6.63 9.21 11.34 13.28 15.09 16.81 18.48 20.09 21.67 23.21 0.995 7.88 10.60 12.84 14.86 16.75 18.55 20.28 21.95 23.59 25.19 Number of Degrees of Freedom F(z) 11 12 13 14 15 16 17 18 19 20 0.005 2.60 3.07 .3.57 4.07 4.60 5.14 5.70 6.26 6.84 7.43 0.01 3.05 3.57 4.11 4.66 5.23 5.81 6.41 7.01 7.63 8.26 0.025 3.82 4.40 5.01 5.63 6.26 6.91 7.56 8.23 8.91 9.59 0.05 4.57 5.23 5.89 6.57 7.26 7.96 8.67 9.39 10.12 10.85 0.95 19.68 21.03 22.36 23.68 25.00 26.30 27.59 28.87 30.14 31.41 0.975 21.92 23.34 24.74 26.12 27.49 28.85 30.19 31.53 32.85 34.17 0.99 24.72 26.22 27.69 29.14 30.58 32.00 33.41 34.81 36.19 37.57 0.995 26.76 28.30 29.82 31.32 32.80 34.27 35.72 37.16 38.58 40.00 Number of Degrees of Freedom F(z) 21 22 23 24 25 26 27 28 29 30 0.005 8.0 8.6 9.3 9.9 10.5 11.2 11.8 12.5 13.1 13.8 0.01 8.9 9.5 10.2 10.9 11.5 12.2 12.9 13.6 14.3 15.0 0.025 10.3 11.0 11.7 12.4 13.1 13.8 14.6 15.3 16.0 16.8 0.05 11.6 12.3 13.1 13.8 14.6 15.4 16.2 16.9 17.7 18.5 0.95 32.7 33.9 35.2 36.4 37.7 38.9 40.1 41.3 42.6 43.8 0.975 35.5 36.8 38.1 39.4 40.6 41.9 43.2 44.5 45.7 47.0 0.99 38.9 40.3 41.6 43.0 44.3 45.6 47.0 48.3 49.6 50.9 0.995 41.4 42.8 44.2 45.6 46.9 48.3 49.6 51.0 52.3 53.7 F(z) Number of Degrees of Freedom 40 50 60 70 80 90 100 100 (Approximation) \(h - 2.58)2 \(h - 2.33)2 - 1.96)2 0.005 0.01 0.025 0.05 0.95 0.975 0.99 0.995 20.7 22.2 24.4 26.5 55.8 59.3 63.7 66.8 28.0 29.7 32.4 34.8 67.5 71.4 76.2 79.5 35.5 37.5 40.5 43.2 79.1 83.3 88.4 92.0 43.3 45.4 48.8 51.7 90.5 95.0 100.4 104.2 51.2 53.5 57.2 60.4 101.9 106.6 112.3 116.3 59.2 61.8 65.6 69.1 113.1 118.1 124.1 128.3 67.3 70.1 74.2 77.9 124.3 129.6 135.8 140.2 I.64)2 5in t 1.64)2 \, y, Phi K Kappa y Chi A, A Lambda i/y, Ý Psi M Mu w, fl Omega i j k "i bi b-2 b3 Vectors a*b = axbx + a2b2 + a3b3 a x b = ŕ) f d f d f grad/= Vf = ^-[ + ~f} + ~k ó X riV d z dt>i dv2 dv3 div v = V • v =--1---1-- dx dy dz curi v = V x v = i j k d d d dx dy dz v2 V3