Exam (elaborations) TEST BANK FOR Mathematical methods for physics and engineering 3rd Edition By Riley, Kenneth Franklin_ Hobson, Michael Paul (Solution Manual)-Converted

It can be shown that the polynomial g(x) = 4x3 + 3x2 − 6x − 1 has turning points at x = −1 and x = 1 2 and three real roots altogether. Continue an investigation of its properties as follows. (a) Make a table of values of g(x) for integer values of x between −2 and 2. Use it and the information given above to draw a graph and so determine the roots of g(x) = 0 as accurately as possible. (b) Find one accurate root of g(x) = 0 by inspection and hence determine precise values for the other two roots. (c) Show that f(x) = 4x3 + 3x2 − 6x − k = 0 has only one real root unless −5 ≤ k ≤ 7 4 . (a) Straightforward evaluation of g(x) at integer values of x gives the following table: x −2 −1 0 1 2 g(x) −9 4 −1 0 31 (b) It is apparent from the table alone that x = 1 is an exact root of g(x) = 0 and so g(x) can be factorised as g(x) = (x−1)h(x) = (x−1)(b2x2+b1x+b0). Equating the coefficients of x3, x2, x and the constant term gives 4 = b2, b1 − b2 = 3, b0 − b1 = −6 and −b0 = −1, respectively, which are consistent if b1 = 7. To find the two remaining roots we set h(x) = 0: 4x2 + 7x + 1 = 0. 1 PRELIMINARY ALGEBRA The roots of this quadratic equation are given by the standard formula as α1,2 = −7 ± √ 49 − 16 8 . (c) When k = 1 (i.e. the original equation) the values of g(x) at its turning points, x = −1 and x = 1 2, are 4 and −11 4 , respectively. Thus g(x) can have up to 4 subtracted from it or up to 11 4 added to it and still satisfy the condition for three (or, at the limit, two) distinct roots of g(x) = 0. It follows that for k outside the range −5 ≤ k ≤ 7 4 , f(x) [= g(x) + 1 − k] has only one real root. 1.3 Investigate the properties of the polynomial equation f(x) = x7 + 5x6 + x4 − x3 + x2 − 2 = 0, by proceeding as follows. (a) By writing the fifth-degree polynomial appearing in the expression for f(x) in the form 7x5 + 30x4 + a(x − b)2 + c, show that there is in fact only one positive root of f(x) = 0. (b) By evaluating f(1), f(0) and f(−1), and by inspecting the form of f(x) for negative values of x, determine what you can about the positions of the real roots of f(x) = 0. (a) We start by finding the derivative of f(x) and note that, because f contains no linear term, f can be written as the product of x and a fifth-degree polynomial: f(x) = x7 + 5x6 + x4 − x3 + x2 − 2 = 0, f  (x) = x(7x5 + 30x4 + 4x2 − 3x + 2) = x[ 7x5 + 30x4 + 4(x − 3 8 )2 − 4( 3 8 )2 + 2] = x[ 7x5 + 30x4 + 4(x − 3 8 )2 + 23 16 ]. Since, for positive x, every term in this last expression is necessarily positive, it follows that f(x) can have no zeros in the range 0 < x < ∞. Consequently, f(x) can have no turning points in that range and f(x) = 0 can have at most one root in the same range. However, f(+∞) = +∞ and f(0) = −2 < 0 and so f(x) = 0 has at least one root in 0 < x < ∞. Consequently it has exactly one root in the range. (b) f(1) = 5, f(0) = −2 and f(−1) = 5, and so there is at least one root in each of the ranges 0 < x < 1 and −1 < x < 0. There is no simple systematic way to examine the form of a general polynomial function for the purpose of determining where its zeros lie, but it is sometimes 2 PRELIMINARY ALGEBRA helpful to group terms in the polynomial and determine how the sign of each group depends upon the range in which x lies. Here grouping successive pairs of terms yields some information as follows: x7 + 5x6 is positive for x > −5, x4 − x3 is positive for x >1 and x < 0, x2 − 2 is positive for x > √ 2 and x < − √ 2. Thus, all three terms are positive in the range(s) common to these, namely −5 < x < − √ 2 and x > 1. It follows that f(x) is positive definite in these ranges and there can be no roots of f(x) = 0 within them. However, since f(x) is negative for large negative x, there must be at least one root α with α < −5. 1.5 Construct the quadratic equations that have the following pairs of roots: (a) −6,−3; (b) 0, 4; (c) 2, 2; (d) 3 + 2i, 3 − 2i, where i2 = −1. Starting in each case from the ‘product of factors’ form of the quadratic equation, (x − α1)(x − α2) = 0, we obtain: (a) (x + 6)(x + 3) = x2 + 9x + 18 = 0; (b) (x − 0)(x − 4) = x2 − 4x = 0; (c) (x − 2)(x − 2) = x2 − 4x + 4 = 0; (d) (x − 3 − 2i)(x − 3 + 2i) = x2 + x(−3 − 2i − 3 + 2i) + (9 − 6i + 6i − 4i2) = x2 − 6x + 13 = 0. Trigonometric identities 1.7 Prove that cos π 12 = √ 3 + 1 2 √ 2 by considering (a) the sum of the sines of π/3 and π/6, (b) the sine of the sum of π/3 and π/4. (a) Using sinA + sinB = 2sin
A + B 2  cos
A − B 2  , 3 PRELIMINARY ALGEBRA we have sin π 3 + sin π 6 = 2sin π 4 cos π 12 , √ 3 2 + 1 2 = 2 √1 2 cos π 12 , cos π 12 = √ 3 + 1 2 √ 2 . (b) Using, successively, the identities sin(A + B) = sinAcosB + cosAsin B, sin(π − θ) = sinθ and cos( 1 2π − θ) = sinθ, we obtain sin π 3 + π 4  = sin π 3 cos π 4 + cos π 3 sin π 4 , sin 7π 12 = √ 3 2 √1 2 + 1 2 √1 2 , sin 5π 12 = √ 3 + 1 2 √ 2 , cos π 12