Preliminary algebra. Polynomial equations. and three real roots altogether. Continue an investigation of its properties as follows.

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1 Student Solutions Manual for Mathematical Methods for Physics and Engineering: 1 Preliminary algebra Polynomial equations 1.1 It can be shown that the polynomial g(x) =4x 3 +3x 6x 1 has turning points at x = 1 and x = 1 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 and. Use it and the information given above to draw a graph and so determine the roots of g(x) =0as accurately as possible. (b) Find one accurate root of g(x) =0by inspection and hence determine precise values for the other two roots. (c) Show that f(x) =4x 3 +3x 6x k =0has 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 g(x) (b) It is apparent from the table alone that x = 1 is an exact root of g(x) =0and so g(x) can be factorised as g(x) =(x 1)h(x) =(x 1)(b x +b 1 x+b 0 ). Equating the coefficients of x 3, x, x and the constant term gives 4 = b, b 1 b =3, b 0 b 1 = 6 and b 0 = 1, respectively, which are consistent if b 1 = 7. To find the two remaining roots we set h(x) =0: 4x +7x +1=0. 1

2 Student Solutions Manual for Mathematical Methods for Physics and Engineering: The roots of this quadratic equation are given by the standard formula as α 1, = 7 ± (c) When k = 1 (i.e. the original equation) the values of g(x) at its turning points, x = 1 andx = 1,are4and 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 by proceeding as follows. f(x) =x 7 +5x 6 + x 4 x 3 + x =0, (a) By writing the fifth-degree polynomial appearing in the expression for f (x) in the form 7x 5 +30x 4 + a(x b) + 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) =x 7 +5x 6 + x 4 x 3 + x =0, f (x) =x(7x 5 +30x 4 +4x 3x +) = x[7x 5 +30x 4 +4(x 3 8 ) 4( 3 8 ) +] = x[7x 5 +30x 4 +4(x 3 8 ) ]. 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 inthesamerange.however,f(+ ) =+ and f(0) = < 0andsof(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) = andf( 1)=5,andsothereisatleastonerootineach of the ranges 0 <x<1and 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

3 Student Solutions Manual for Mathematical Methods for Physics and Engineering: 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: x 7 +5x 6 is positive for x> 5, x 4 x 3 is positive for x>1 and x<0, x is positive for x> and x<. Thus, all three terms are positive in the range(s) common to these, namely 5 <x< andx>1. It follows that f(x) is positive definite in these ranges andtherecanbenorootsoff(x) = 0 within them. However, since f(x) is negative for large negative x, there must be at least one root α with α< Construct the quadratic equations that have the following pairs of roots: (a) 6, 3; (b) 0, 4; (c), ; (d) 3 + i, 3 i, wherei = 1. Starting in each case from the product of factors form of the quadratic equation, (x α 1 )(x α ) = 0, we obtain: (a) (x +6)(x +3)=x +9x +18=0; (b) (x 0)(x 4) = x 4x =0; (c) (x )(x ) = x 4x +4=0; (d) (x 3 i)(x 3+i) =x + x( 3 i 3+i) +(9 6i +6i 4i ) = x 6x +13=0. Trigonometric identities 1.7 Prove that by considering cos π 1 = 3+1 (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 ( ) ( ) A + B A B sin A +sinb =sin cos, 3

4 Student Solutions Manual for Mathematical Methods for Physics and Engineering: we have (b) Using, successively, the identities sin π 3 +sinπ 6 =sinπ 4 cos π 1, = 1 cos π 1, cos π 1 = 3+1. sin(a + B) =sina cos B +cosa sin B, sin(π θ) =sinθ and cos( 1 π θ) =sinθ, we obtain ( π sin 3 + π ) 4 =sin π 3 cos π 4 +cosπ 3 sin π 4, sin 7π 3 1 = sin 5π 1 = , 3+1, cos π 1 = Find the real solutions of (a) 3 sin θ 4cosθ =, (b) 4 sin θ +3cosθ =6, (c) 1 sin θ 5cosθ = 6. We use the result that if then a sin θ + b cos θ = k θ =sin 1 ( k K ) φ, where K = a + b and φ =tan 1 b a. 4

5 Student Solutions Manual for Mathematical Methods for Physics and Engineering: 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 =,K = 3 +4 =5,φ =tan 1 ( 4/3) and so θ =sin 1 5 tan =1.339 or.66. (b) k =6,K = 4 +3 =5.Sincek>Kthere is no solution for a real angle θ. (c) k = 6, K = 1 +5 = 13, φ =tan 1 ( 5/1) and so θ =sin 13 tan 1 1 = or Find all the solutions of sin θ + sin 4θ = sin θ + sin 3θ that lie in the range π <θ π. What is the multiplicity of the solution θ =0? Using and ( ) ( ) A + B A B sin A +sinb =sin cos, ( ) ( ) A + B A B cos A cos B = sin sin, and recalling that cos( φ) =cos(φ), the equation can be written successively as sin 5θ ( cos 3θ ) =sin 5θ ( cos θ ), sin 5θ ( cos 3θ cos θ ) =0, sin 5θ sin θ sin θ =0. The first factor gives solutions for θ of 4π/5, π/5, 0, π/5 and4π/5. The second factor gives rise to solutions 0 and π, whilst the only value making the third factor zero is θ =0.Thesolutionθ = 0 appears in each of the above sets and so has multiplicity 3. 5

6 Student Solutions Manual for Mathematical Methods for Physics and Engineering: Coordinate geometry 1.13 Determine the forms of the conic sections described by the following equations: (a) x + y +6x +8y =0; (b) 9x 4y 54x 16y +9=0; (c) x +y +5xy 4x + y 6=0; (d) x + y +xy 8x +8y =0. (a) x + y +6x +8y =0. The coefficients of x and y 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) +(y +4) (3 +4 )=0. The equation is therefore that of a circle of radius 3 +4 =5centredon ( 3, 4). (b) 9x 4y 54x 16y + 9 = 0. This equation contains no xy term and so the centre of the curve will be at ( 54/( 9), 16/[ ( 4)] ) = (3, ), and in standardised form the equation is 9(x 3) 4(y +) = 0, or (x 3) (y +) = 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) = ±(y + ), i.e. lines of slope ± 3 passing through its centre at (3, ). (c) x +y +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 a i x+b i y+1 = 0, (i =1, ) and comparing the product (a 1 x+b 1 y+1)(a x+b y+1) with 1 6 (x +y +5xy 4x + y 6). The comparison produces five equations which the four constants a i, b i,(i =1, ) must satisfy: and a 1 a = 6, b 1b = 6, a 1 + a = 4 6, b 1 + b = 1 6 a 1 b + b 1 a =

7 Student Solutions Manual for Mathematical Methods for Physics and Engineering: Combining the first and third equations gives 3a 1 a 1 1 = 0 leading to a 1 and a having the values 1 and 1 3, in either order. Similarly, combining the second and fourth equations gives 6b 1 + b 1 = 0 leading to b 1 and b having the values 1 and 3, again in either order. Either of the two combinations (a 1 = 1 3, b 1 = 3, a =1,b = 1 )and(a 1 =1, b 1 = 1, a = 1 3, b = 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 +y 3=0and x + y +=0. (d) x + y +xy 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) =8(x y). The two lines given by x + y =0andx y = 0 are orthogonal and so the equation is of the form u =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) x +1 x +3x 10, (b) 4 x 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 ), the partial fraction expansion must have the form x +1 x +3x 10 = A x +5 + B x. 7

8 Student Solutions Manual for Mathematical Methods for Physics and Engineering: (i) A x +5 + B x x(a + B)+(5B A) =. (x +5)(x ) Solving A + B =and A +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 ; 3 6 = A 6 + B 1, 1 =A 5B; 3 =A 6B, (iii) with solution A = 9 7 and B = 5 7. A = ( 5) = 9 7 ; B = () = 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 x 3x = A x + B x 3. (i) A x + B x(a + B) 3A = x 3 (x 0)(x 3). Solving A + B =0and 3A = 4 gives A = 4 3 and B = 4 3. (ii) Setting x equal to 1 and, say, gives the pair of equations 4 = A 1 + B ; 4 = A + B 1, 4 =A B; 4 =A B, (iii) with solution A = 4 3 and B = 4 3. A = = 4 3 ; B = = 4 3. Again, all three methods give the same decomposition. 8

9 Student Solutions Manual for Mathematical Methods for Physics and Engineering: 1.17 Rearrange the following functions in partial fraction form: x 6 (a) x 3 x +4x 4, (b) x3 +3x + x +19 x 4 +10x. +9 (a) For the function f(x) = x 6 x 3 x +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 x 3 x +4x 4=(x 1)(x + b 1 x + b 0 ). Equating coefficients: 1 =b 1 1, 4 = b 1 + b 0 and 4 = b 0, giving b 1 =0 and b 0 = 4. Thus, x 6 f(x) = (x 1)(x +4). The factor x + 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 x +4 = Ax +4A + Bx + Cx Bx C (x 1)(x. +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 =1and 4A C = 6, which in turn yield A = 1, B =1andC =. Thus, (b) By inspection, the denominator of f(x) = 1 x 1 + x + x +4. x 3 +3x + x +19 x 4 +10x +9 factorises simply into (x +9)(x + 1), but neither factor can be broken down further. Thus, as in (a), we write f(x) = Ax + B x +9 + Cx + D x +1 = (A + C)x3 +(B + D)x +(A +9C)x +(B +9D) (x +9)(x. +1) 9

10 Student Solutions Manual for Mathematical Methods for Physics and Engineering: Equating coefficients gives A + C =1, B + D =3, A +9C =1, B +9D =19. From the first and third equations, A =1andC = 0. The second and fourth yield B =1andD =. Thus f(x) = x +1 x +9 + x +1. Binomial expansion 1.19 Evaluate those of the following that are defined: (a) 5 C 3, (b) 3 C 5, (c) 5 C 3, (d) 3 C 5. (a) 5 C 3 = 5! 3!! = 10. (b) 3 C 5. This is not defined as 5 > 3 > 0. For (c) and (d) we will need to use the identity m k m(m +1) (m + k 1) C k =( 1) =( 1) k m+k 1 C k. k! (c) 5 C 3 =( 1) C 3 = 7! 3! 4! = 35. (d) 3 C 5 =( 1) C 5 = 7! 5!! = 1. Proof by induction and contradiction 1.1 Prove by induction that n n r = 1 n(n +1) and r 3 = 1 4 n (n +1). r=1 r=1 To prove that n r=1 r = 1 n(n +1), 10

Calculus First Semester Review Name: Section: Evaluate the function: (g o f )( 2) f (x + h) f (x) h. m(x + h) m(x)

Calculus First Semester Review Name: Section: Evaluate the function: (g o f )( 2) f (x + h) f (x) h. m(x + h) m(x) Evaluate the function: c. (g o f )(x + 2) d. ( f ( f (x)) 1. f x = 4x! 2 a. f( 2) b. f(x 1) c. f (x + h) f (x) h 4. g x = 3x! + 1 Find g!! (x) 5. p x = 4x! + 2 Find p!! (x) 2. m x = 3x! + 2x 1 m(x + h)