Exam (elaborations) TEST BANK FOR Mathematical Methods for Physics and

Exam (elaborations) TEST BANK FOR Mathematical Methods for Physics and a) Straightforward evaluation of g(x) at integer values of x gives the following table: x −2 −1 01 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 i 2 = −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 − 4i 2 ) = 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 = 2 sin
A + B 2  cos
A − B 2  , 3 PRELIMINARY ALGEBRA we have sin π 3 + sin π 6 = 2 sin 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 + cosAsinB, sin(π − θ) = sin θ and cos( 1 2π − θ) = sin θ, 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 = √ 3+1 2 √ 2 . 1.9 Find the real solutions of (a) 3 sin θ − 4 cos θ = 2, (b) 4 sin θ + 3 cos θ = 6, (c) 12 sin θ − 5 cos θ = −6. We use the result that if a sin θ + b cos θ = k then θ = sin−1
k K  − φ, where K2 = a2 + b2 and φ = tan−1 b a . 4 PRELIMINARY ALGEBRA Recalling that the inverse sine yields two values and that the individual signs of a and b have to be taken into account, we have (a) k = 2, K = √ 32 + 42 = 5, φ = tan−1(−4/3) and so θ = sin−1 2 5 − tan−1 −4 3 = 1.339 or − 2.626. (b) k = 6, K = √ 42 + 32 = 5. Since kK there is no solution for a real angle θ. (c) k = −6, K = √ 122 + 52 = 13, φ = tan−1(−5/12) and so θ = sin−1 −6 13 − tan−1 −5 12 = −0.0849 or − 2.267. 1.11 Find all the solutions of sin θ + sin 4θ = sin 2θ + sin 3θ that lie in the range −πθ ≤ π. What is the multiplicity of the solution θ = 0? Using sin(A + B) = sinAcosB + cosAsinB, and cosA − cosB = −2 sin
A + B 2  sin
A − B 2  , and recalling that cos(−φ) = cos(φ), the equation can be written successively as 2 sin 5θ 2 cos
−3θ 2  = 2 sin 5θ 2 cos
−θ 2  , sin 5θ 2
cos 3θ 2 − cos θ 2  = 0, −2 sin 5θ 2 sin θ sin θ 2 = 0. The first factor gives solutions for θ of −4π/5, −2π/5, 0, 2π/5 and 4π/5. The second factor gives rise to solutions 0 and π, whilst the only value making the third factor zero is θ = 0. The solution θ = 0 appears in each of the above sets and so has multiplicity 3. 5 PRELIMINARY ALGEBRA Coordinate geometry 1.13 Determine the forms of the conic sections described by the following equations: (a) x2 + y2 + 6x + 8y = 0; (b) 9x2 − 4y2 − 54x − 16y + 29 = 0; (c) 2x2 + 2y2 + 5xy − 4x + y − 6 = 0; (d) x2 + y2 + 2xy − 8x + 8y = 0. (a) x2 + y2 + 6x + 8y = 0. The coefficients of x2 and y2 are equal and there is no xy term; it follows that this must represent a circle. Rewriting the equation in standard circle form by ‘completing the squares’ in the terms that involve x and y, each variable treated separately, we obtain (x + 3)2 + (y + 4)2 − (32 + 42 )=0. The equation is therefore that of a circle of radius √ 32 + 42 = 5 centred on (−3, −4). (b) 9x2 − 4y2 − 54x − 16y + 29 = 0. This equation contains no xy term and so the centre of the curve will be at ( 54/(2 × 9), 16/[2 × (−4)] ) = (3, −2), and in standardised form the equation is 9(x − 3)2 − 4(y + 2)2 + 29 − 81 + 16 = 0, or (x − 3)2 4 − (y + 2)2 9 = 1. The minus sign between the terms on the LHS implies that this conic section is a hyperbola with asymptotes (the form for large x and y and obtained by ignoring the constant on the RHS) given by 3(x − 3) = ±2(y + 2), i.e. lines of slope ±3 2 passing through its ‘centre’ at (3, −2). (c) 2x2 + 2y2 + 5xy − 4x + y − 6=0. As an xy term is present the equation cannot represent an ellipse or hyperbola in standard form. Whether it represents two straight lines can be most easily investigated by taking the lines in the form aix+biy+1 = 0, (i = 1, 2) and comparing the product (a1x+b1y+1)(a2x+b2y+1) with −1 6 (2x2 + 2y2 + 5xy − 4x + y − 6). The comparison produces five equations which the four constants ai, bi, (i = 1, 2) must satisfy: a1a2 = 2 −6 , b1b2 = 2 −6 , a1 + a2 = −4 −6 , b1 + b2 = 1 −6 and a1b2 + b1a2 = 5 −6 . 6 PRELIMINARY ALGEBRA Combining the first and third equations gives 3a2 1 − 2a1 − 1 = 0 leading to a1 and a2 having the values 1 and −1 3 , in either order. Similarly, combining the second and fourth equations gives 6b2 1 +b1 −2 = 0 leading to b1 and b2 having the values 1 2 and −2 3 , again in either order. Either of the two combinations (a1 = −1 3 , b1 = −2 3 , a2 = 1, b2 = 1 2 ) and (a1 = 1, b1 = 1 2 , a2 = −1 3 , b2 = −2 3 ) also satisfies the fifth equation [note that the two alternative pairings do not do so]. That a consistent set can be found shows that the equation does indeed represent a pair of straight lines, x + 2y − 3 = 0 and 2x + y + 2 = 0. (d) x2 + y2 + 2xy − 8x + 8y = 0. We note that the first three terms can be written as a perfect square and so the equation can be rewritten as (x + y) 2 = 8(x − y). The two lines given by x + y = 0 and x − y = 0 are orthogonal and so the equation is of the form u2 = 4av, which, for Cartesian coordinates u, v, represents a parabola passing through the origin, symmetric about the v-axis (u = 0) and defined for v ≥ 0. Thus the original equation is that of a parabola, symmetric about the line x + y = 0, passing through the origin and defined in the region x ≥ y. Partial fractions 1.15 Resolve (a) 2x + 1 x2 + 3x − 10, (b) 4 x2 − 3x into partial fractions using each of the following three methods: (i) Expressing the supposed expansion in a form in which all terms have the same denominator and then equating coefficients of the various powers of x. (ii) Substituting specific numerical values for x and solving the resulting simultaneous equations. (iii) Evaluation of the fraction at each of the roots of its denominator, imagining a factored denominator with the factor corresponding to the root omitted – often known as the ‘cover-up’ method. Verify that the decomposition obtained is independent of the method used. (a) As the denominator factorises as (x+ 5)(x−2), the partial fraction expansion must have the form 2x + 1 x2 + 3x − 10 = A x + 5 + B x − 2 . 7 PRELIMINARY ALGEBRA (i) A x + 5 + B x − 2 = x(A + B) + (5B − 2A) (x + 5)(x − 2) . Solving A + B = 2 and −2A + 5B = 1 gives A = 9 7 and B = 5 7 . (ii) Setting x equal to 0 and 1, say, gives the pair of equations 1 −10 = A 5 + B −2 ; 3 −6 = A 6 + B −1 , −1=2A − 5B; −3 = A − 6B, with solution A = 9 7 and B = 5 7 . (iii) A = 2(−5) + 1 −5 − 2 = 9 7 ; B = 2(2) + 1 2+5 = 5 7 . All three methods give the same decomposition. (b) Here the factorisation of the denominator is simply x(x−3) or, more formally, (x − 0)(x − 3), and the expansion takes the form 4 x2 − 3x = A x + B x − 3 . (i) A x + B x − 3 = x(A + B) − 3A (x − 0)(x − 3) . Solving A + B = 0 and −3A = 4 gives A = −4 3 and B = 4 3 . (ii) Setting x equal to 1 and 2, say, gives the pair of equations 4 −2 = A 1 + B −2 ; 4 −2 = A 2 + B −1 , −4=2A − B; −4 = A − 2B, with solution A = −4 3 and B = 4 3 . (iii) A = 4 0 − 3 = −4 3 ; B = 4 3 − 0 = 4 3 . Again, all three methods give the same decomposition. 8 PRELIMINARY ALGEBRA 1.17 Rearrange the following functions in partial fraction form: (a) x − 6 x3 − x2 + 4x − 4 , (b) x3 + 3x2 + x + 19 x4 + 10x2 + 9 . (a) For the function f(x) = x − 6 x3 − x2 + 4x − 4 = g(x) h(x) the first task is to factorise the denominator. By inspection, h(1) = 0 and so x − 1 is a factor of the denominator. Write x3 − x2 + 4x − 4=(x − 1)(x2 + b1x + b0). Equating coefficients: −1 = b1 − 1, 4 = −b1 + b0 and −4 = −b0, giving b1 = 0 and b0 = 4. Thus, f(x) = x − 6 (x − 1)(x2 + 4). The factor x2 + 4 cannot be factorised further without using complex numbers and so we include a term with this factor as the denominator, but ‘at the price of’ having a linear term, and not just a number, in the numerator. f(x) = A x − 1 + Bx + C x2 + 4 = Ax2 + 4A + Bx2 + Cx − Bx − C (x − 1)(x2 + 4) . Comparing the coefficients of the various powers of x in this numerator with those in the numerator of the original expression gives A+ B = 0, C − B = 1 and 4A − C = −6, which in turn yield A = −1, B = 1 and C = 2. Thus, f(x) = − 1 x − 1 + x + 2 x2 + 4. (b) By inspection, the denominator of x3 + 3x2 + x + 19 x4 + 10x2 + 9 factorises simply into (x2 + 9)(x2 + 1), but neither factor can be broken down further. Thus, as in (a), we write f(x) = Ax + B x2 + 9 + Cx + D x2 + 1 = (A + C)x3 + (B + D)x2 + (A + 9C)x + (B + 9D) (x2 + 9)(x2 + 1) . 9 PRELIMINARY ALGEBRA Equating coefficients gives A + C = 1, B + D = 3, A + 9C = 1, B + 9D = 19. From the first and third equations, A = 1 and C = 0. The second and fourth yield B = 1 and D = 2. Thus f(x) = x + 1 x2 + 9 + 2 x2 + 1. Binomial expansion 1.19 Evaluate those of the following that are defined: (a) 5C3, (b) 3C5, (c) −5C3, (d) −3C5. (a) 5C3 = 5! 3! 2! = 10. (b) 3C5. This is not defined as 5 3 0. For (c) and (d) we will need to use the identity −mCk = (−1)k m(m + 1) ··· (m + k − 1) k! = (−1)k m+k−1 Ck. (c) −5C3 = (−1)3 5+3−1C3 = − 7! 3! 4! = −35. (d) −3C5 = (−1)5 5+3−1C5 = − 7! 5! 2! = −21. Proof by induction and contradiction 1.21 Prove by induction that n r=1 r = 1 2 n(n + 1) and n r=1 r 3 = 1 4 n2 (n + 1)2 . To prove that n r=1 r = 1 2 n(n + 1), 10 PRELIMINARY ALGEBRA assume that the result is valid for n = N and consider N  +1 r=1 r =  N r=1 r + (N + 1) = 1 2N(N + 1) + (N + 1), using the assumption, = (N + 1)( 1 2N + 1) = 1 2 (N + 1)(N + 2). This is the same form as in the assumption except that N has been replaced by N + 1; this shows that the result is valid for n = N + 1 if it is valid for n = N. But the assumed result is trivially valid for n = 1 and is therefore valid for all n. To prove that n r=1 r3 = 1 4 n2 (n + 1)2 , assume that the result is valid for n = N and consider N  +1 r=1 r 3 =  N r=1 r 3 + (N + 1)3 = 1 4N2 (N + 1)2 + (N + 1)3 , using the assumption, = 1 4 (N + 1)2 [ N2 + 4(N + 1) ] = 1 4 (N + 1)2 (N + 2)2 . This is the same form as in the assumption except that N has been replaced by N + 1 and shows that the result is valid for n = N + 1 if it is valid for n = N. But the assumed result is trivially valid for n = 1 and is therefore valid for all n. 1.23 Prove that 32n + 7, where n is a non-negative integer, is divisible by 8. As usual, we assume that the result is valid for n = N and consider the expression with N replaced by N + 1: 32(N+1) +7=32N+2 +7+32N − 32N = (32N + 7) + 32N(9 − 1). By the assumption, the first term on the RHS is divisible by 8; the second is clearly so. Thus 32(N+1) + 7 is divisible by 8. This shows that the result is valid for n = N + 1 if it is valid for n = N. But the assumed result is trivially valid for n = 0 and is therefore valid for all n. 11 PRELIMINARY ALGEBRA 1.25 Prove by induction that n r=1 1 2r tan
θ 2r  = 1 2n cot
θ 2n  − cot θ. (∗) Assume that the result is valid for n = N and consider N  +1 r=1 1 2r tan
θ 2r  = 1 2N cot
θ 2N  − cot θ + 1 2N+1 tan
θ 2N+1  . Using the half-angle formula tan φ = 2r 1 − r2 , where r = tan 1 2φ, to write cot(θ/2N) in terms of t = tan(θ/2N+1), we have that the RHS is 1 2N
1 − t 2 2t  − cot θ + 1 2N+1 t = 1 2N+1
1 − t 2 + t 2 t  − cot θ = 1 2N+1 cot
θ 2N+1  − cot θ. This is the same form as in the assumption except that N has been replaced by N + 1 and shows that the result is valid for n = N + 1 if it is valid for n = N. But, for n = 1, the LHS of (∗) is 1 2 tan(θ/2). The RHS can be written in terms of s = tan(θ/2): 1 2 cot
θ 2  − cot θ = 1 2s − 1 − s2 2s = s 2 , i.e. the same as the LHS. Thus the result is valid f