Mathematical Methods for Physics and Engineering - Matematica.NET

PRELIMINARY ALGEBRA1.11 Show that the equation is equivalent to sin(5θ/2) sin(θ)sin(θ/2) = 0.Solutions are −4π/5, −2π/5, 0, 2π/5, 4π/5,π.Thesolutionθ = 0 has multiplicity3.1.13 (a) A circle of radius 5 centred on (−3, −4).(b) A hyperbola with ‘centre’ (3, −2) and ‘semi-axes’ 2 and 3.(c) The expression factorises into two lines, x +2y − 3=0and2x + y +2=0.(d) Write the expression as (x+y) 2 =8(x−y) to see that it represents a parabolapassing through the origin, with the line x + y = 0 as its axis of symmetry.51.15 (a)7(x − 2) + 97(x +5) , (b) − 43x + 43(x − 3) .1.17 (a) x +2x 2 +4 − 1x +1, (b)x − 1 x 2 +9 + 2x 2 +1 .1.19 (a) 10, (b) not defined, (c) −35, (d) −21.1.21 Look for factors common to the n = N sum and the additional n = N +1term,so as to reduce the sum for n = N +1toasingleterm.1.23 Write 3 2n as 8m − 7.1.25 Use the half-angle formulae of equations (1.32) to (1.34) to relate functions ofθ/2 k to those of θ/2 k+1 .1.27 Divisible for k =1, 2,...,p− 1. Expand (n +1) p as n p + ∑ p−11p C k n k + 1. Applythe stated result for p = 5. Note that n 5 − n = n(n − 1)(n +1)(n 2 + 1); the productof any three consecutive integers must divide by both 2 and 3.1.29 By assuming x = p/q with q ≠ 1, show that a fraction −p n /q isequaltoaninteger a n−1 p n−1 + ···+ a 1 pq n−2 + a 0 q n−1 . This is a contradiction, and is onlyresolved if q = 1 and the root is an integer.(a) The only possible candidates are ±1, ±2, ±4. None is a root.(b) The only possible candidates are ±1, ±2, ±3, ±6. Only −3 isaroot.1.31 f(x) can be written as x(x +1)(x +2)+x(x +1)(x − 1). Each term consists ofthe product of three consecutive integers, of which one must therefore divide by2 and (a different) one by 3. Thus each term separately divides by 6, and sotherefore does f(x). Note that if x is the root of 2x 3 +3x 2 + x − 24 = 0 that liesnear the non-integer value x =1.826, then x(x + 1)(2x + 1) = 24 and thereforedivides by 6.1.33 Note that, e.g., the condition for 6a 4 + a 3 to be divisible by 4 is the same as thecondition for 2a 4 + a 3 to be divisible by 4.For the necessary (only if) part of the proof set n =1, 2, 3 and take integercombinations of the resulting equations.For the sufficient (if) part of the proof use the stated conditions to prove theproposition by induction. Note that n 3 − n is divisible by 6 and that n 2 + n iseven.40