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Unformatted text preview: CHAPTER 0 Preliminaries 1. Introduction In this preliminary chapter we consider briefly some important concepts from calculus and algebra which we shall require for our study of differen- tial equations. Many of these concepts may be familiar to the student, in which case this chapter can serve as a review. First the elementary proper— ties of complex numbers are outlined. This is followed by a discussion of functions which assume complex values, in particular polynomials and power series. Some consequences of the Fundamental Theorem of Algebra are given. The exponential function is defined using power series; it is of central importance for linear differential equations with constant coefli- cients. The role that determinants play in the solution of systems of linear equations is discussed. Lastly we make a few remarks concerning principles of discovery, and methods of proof, of mathematical results. 2. Complex numbers It is a fundamental fact about real numbers that the square of any such number is never negative. Thus there is no real x which satisfies the equa- tion 902 + 1 = 0. We shall use the real numbers to define new numbers which include numbers which satisfy such equations. A complex number 2 is an ordered pair of real numbers (cc, y), and we write 2 = (96,21). If 21 = ($11111), 22 = (x2) y2)) 4 Preliminaries Chap. 0 agrees with our earlier identifications 0 = (0, 0) , 1 = (1, 0) .] In this sense, the complex numbers contain the real numbers. The properties (i)—(ix), which hold for complex numbers, are also valid for real numbers, and thus we see that we have succeeded in enlarging the set of real numbers without losing any of these algebraic properties. We have gained something also, since there are complex numbers 2 which satisfy the equation z2+1=0. One such number the imaginary unit 1' = (0, 1), as can be easily checked, and this provides one justification for our definition of multiplication. If 2 = (x, y) is a complex number, the real number x is called the real part of z, and we write Re 2 = x; whereas y is called the imaginary part of z, and we write Im z = y. Thus 2 = (x,y) =x(1,0) +y(0,1) =x+iy = Rez+i(Imz). Hereafter it will be convenient to denote a complex number (x, y) as :c + z'y. It is clear that the complex numbers are in a one-to-one correspondence with the points of the (re, y)-plane, the complex number 2 = x + iy corre- sponding to the point with coordinates (x, y). Then thought of in this way the x—axis is of ten called the real axis, the y—axis is called the imaginary axis, and the plane is called the complex plane. If 2 = x + iy, its mirror image in the real axis is the point x — iy. This number is called the complex conjugate of z, and is denoted by 2. Thus 2 = x — iy if z = x + iy. We see immediately that Z = Z, 21 + 22 = 21 + 52, 2122 = 5152, 2-1 = (2)_l, for any complex numbers 2, 21, 22. Introducing polar coordinates (r, 0) in the complex plane via x=rcos(9, y=rsin6, (r;0,0§0<21r), we see that we may write 2 = x+iy = r(cos0+isin0). The magnitude of z = x + iy, denoted by l2 l , is defined to be 7‘. Thus lzl = (x2 + ye)”2 = (25)“, where the positive square root is understood. Clearly = Suppose z is real (that is, Inn 2 = 0). Then 2 = x + £0, for some real 2:, and Izl = (222)1’2, Sec. 2 Preliminaries ' 5 which is the magnitude of :1: considered. as a real number. In addition the magnitude of a complex number obeys the same rules as the magnitude of a real number, namely: lzl 20: [2| =0 ifandonlyif 2:0, lzl, |21+22l § W + [22L 1 33. ll lzlllz2l- lzlzzl We show that I21 +22] g lzll + I22], for example. First we note that Rez g [2| for any complex number 2. Then I21+22|2 (21+22)m = |21|2+ I22|2+zlé2+2122 I21|2+ |22|2+2Re (2152) l21|2+ lz2|2+2|2152l \ |21l2+|22|2+2l21l l22l (l21l+lzzl)2, from which it follows that |zl + all g [21] + [22 I. From the above rules one can deduce further that II "A ll ll llzll “IZle §l31+z2| é lzll‘l‘lzzb fig =lzll 22 lzzl Geometrically we see that [21 — 22I represents the distance between the two points 21 and 22 in the complex plane. EXERCISES 1. Compute the following complex numbers, and express in the form x + iy, Where x, y are real: (a) (2 — i3) + (-1+ 756) . (b) (4 i2) — (6 — 2'3) (c) (6— ifi)(2+i4) (d) 1+5 ' _ 7: (e) I4—i5l (f) Re(4—i5) (g) Im (6 + 2'2) 6 Preliminaries Chap. 0 2. Express the following complex numbers in the form r (cos 0 + t sin 0) with r20and0§0<21rz _ ,. (a) 1+ N37 0)) (1+ 02 (c) I“ (d) (1+i)<1— 2') l—t ,3. Indicate graphically the set of all complex numbers 2 satisfying: (a)|z-—2l=1 .(b)[z+2|<2 (c)|Rez[§3 ‘ (d)[Imz[>1 .(e) |z—1[+Iz+2l=8. .4. Prove that: (a)z+é=2Rez (b)z—2=2iImz (c)lRezI§|z| ‘(d)|z|§|Rez!+IImz| 5. If r is a real number, and 2 complex, show that Re (72) = 7‘ (Re 2), Im (r2) = 7* (Im z). 6. Provethat l|21l*lz2ll§l21+22|. (Hint: 21 = 21 + 22 + (-Z2), and 22 = 21 + 22 + (—-21).) 7. Provethat I21+22l2+ I21—22l2=2l21l2+2l22l2, for all complex 21, Z2. _ ISI? 8. If I a I < 1, what complex 2 satisfy _ _ l 1 — azl , 9. If n is any positive integer, prove that 7"" (cos n0 + i sin n0) = [r (cos 0 + i sin (9)1". (H int: Use induction.) 10. Use the result of Ex. 9 to find (a) two complex numbers satisfying 22 = 2, (b) three complex numbers satisfying 23 = 1. 3. Functions Suppose D is a set whose elements are denoted by P, Q, - - - , which are called the points of the set. Let R be another set. A function on D to R is a law f which associates with each point P in D exactly one point in R, which we denote by f(P). The set D is called the domain of f. The point f(P) is called the value of f at P. We can Visualize the concept of a function as See. 3 Preliminaries 7 Figure l in Fig. 1, where each P in D is connected to a unique f ( P) in R by a string according to some rule. This rule, or what amounts to the same thing, the collection of all these strings, is the function f on D to R. We say that two functions f and g are equal, f = 9, if they have the same domain D, and f(P) = g(P) for all P in D. The idea of a function is very general, and is a fundamental one in mathematics. We shall consider some examples which are of importance for our study of differential equations. (a) Complex-valued functions. If the set R which contains the values of f is the set of all complex numbers, we say that f is a complex-valued func— tion. If f and g are two complex-valued functions with the same domain D, we can define their sum f + g and product fg by (f+g)(P) =f(P) +9(P), (fg)(P) =f(P)g(P), for each P in D. Thus f + g and fg are also functions with domain D. If a is any complex number the function which assigns to each P in a domain D the number a is called a constant function, and is also denoted by (2. Thus if f is any complex—valued function on D we have (af)(P) = 04(1)) for all P in D. A real—valued function f defined on D is one whose values are real num- bers. Such a function is a special case of a complex-valued function. Clearly the sum and product of two real—valued functions on D are real—valued functions. Real-valued functions are usually the principal object of study in first courses in calculus. Every complex—valued function f defined on a domain D gives rise to two real—valued functions Re f, Im f defined by (Ref)(P) = Re [f(P)], (Imf)(P) = ImEf(P)l 8 Preliminaries Chap. 0 for all P in D. Re f and Im f are called the real and imaginary parts of f respectively and we have f=Ref+iImf. Thus the study of complex-valued functions can be reduced to the study of pairs of real-valued functions. To obtain examples of complex—valued func- tions we must specify their domains. (b) Complex—valued functions with real domains. Many of the functions we consider in this book have a domain D which is an interval I of the real axis. Recall that an interval is a set of real as satisfying one of the nine inequalities "A II/\ a a- b, a§x<b, a<x§b, a<x<b, IIA /\ a x 00, —OO<1’§I), a<x<°°, —°°<x<b) —oo<x<00, where a, b are distinct real numbers. The calculus of complex-valued func- tions defined on real intervals is entirely analogous to the calculus of real- Valued functions defined on intervals. We sketch the main ideas. Suppose f is a complex-valued function defined on a real interval I. Then f is said to have the complex number L as a limit at so in I, and we write limf(x) = L, or flat) —> L, (cc—>960), 192:0 if |f(x)—-L|—>0, as 0<Ix—xol—-)0. This means that given any 6 > 0 there is a 6 > 0 such that |f(:v) -— LI < e, whenever 0 <13: — 230! < 6, xinI. Note that here we are using the magnitude of complex numbers. Formally our definition is the same as that for real limits of real-valued functions. Because of this the usual rules for limits, and their proofs, are valid. In particular, if f and g are cOmplex-valued functions defined on I such that for some so in I (x6530): then ' ' (f+g)(x)—->L+M, (f9)(x)-+LM, (ac—>960)- Sec. 3 Preliminaries 9 Suppose I has a limit L since ll L; + iLg at so, where L1, L2 are real. Then [(Rech) - Lll lReU(x) - LJI é |f(x) — Li, and l(Imf)(x) — Lzl = IImEflx) - L]! é We) — Ll, it follows that (Ref)($)—*L1, (Imf)(x)->L2, (as—>330). Conversely, if Re f and Im f have limits L1, L2 respectively at me, then f will have the limit L = L1 + iLz at me. We say that a complex-valued function f defined on an interval I is continuous at are in I if f has the limit f (are) at so, that is, |f(a:) —f(xo)I——>0, as 0<Ix—:co[—>0. Equivalently, f is continuous at $0 if both Re f and Im f are continuous at are. We say f is continuous on I if it is continuous at each point of I. The sum and product of two functions which are continuous at so are continuous there. The complex-valued function f defined on an interval I is said to be difierentiable at we in I if the ratio f(:v) - f(xo) x__x0 1 (x;£x0)7 has a limit at so. If f is differentiable at so we define its derivative at 220, f’ (we), to be this limit. Thus, iff’(xo) exists, f(z) - f(xo) x_xo —f’(xo) —>0, as 0<fx—x0[—+0. An equivalent definition is: f is differentiable at are if both Ref and Im f are differentiable at :50. The derivative of f at me is given by f’(xo) = (Ref)’(xo)+i(1mf)’(xo)- Using these definitions one can show that the usual rules for differentiating real—valued functions are valid for complex—valued functions. For example, if f, g are differentiable at so in I , then so are f + g and fg, and (f + g)’(xo) = f’(xo) + g’(xo), (fg)’(xo) =f’(xo)g(xo) +f(xo)g’(xo). If f is differentiable at every x in an interval I , then f gives rise to a new function f’ on I whose value at each x on I is f’ (x) . 10 Preliminaries Chap. 0 A complex—valued function f with domain'the interval a g :1: § b is said to be integrable there if both Re f and Im f are, and in this case we define its integral by b b b / f(x) d3; =/ (Re f) (x) dx +if (Im f) (x) dx. Every function f which is continuous on a § 00 g b is integrable there. This definition implies the usual integration rules. In particular, if f and g are integrable on a g x g b, and a, B are two complex numbers, b b b / W + new dz = a [ f(x) dx + a / gm) dz. An important inequality connected with the integral of a continuous complex—valued function f defined on (1 § 5c é b is free) dx s flies) Idrc-* This inequality is valid if f is real-valued, and the proof for the case when f is complex—valued can be based on this fact. Let F = [7 f(x) dx. If F = 0 the inequality is obvious. If F 75 0, let F=|F|u, u=cos0+isin0, (0§0<27r). Then ml = 1, and we have [fee dx =iI/abf(x) dx = Re[a£bf(x) dx] = beeEfiflxfldx g [him ldx, *By b f lf(x)ldx is meant the integral of the function If | given by If I (x) = lf(x)| for a g :3 § b. Thus a more appropriate notation would be b / |f|(x)dx. We shall use the former notation since it is commonly used, and there will be no chance of confusion, Sec. 3 Preliminaries 11 since Re [mm s We)! = We) I. As particular examples of complex-valued functions let f(x) = x + (1 — 17762, 9(x) = (1 +0962, for all real 95. Then (Re f) (x) = a: + 962, (1m f) (x) = -x2, (f+g)(x) =x+2x2, (EH18) = (1+i)$3+2$‘, f'($) =1+ (2*2i)$y [0f(x)dx=Axdx+(1—i)£x2dx=g_g (c) Complex—valued functions with complex domains. We shall need to know a little about complex-valued functions whose domains consist of complex numbers. An example is the function f given by f(3) = z": for all complex z, where n is a positive integer. Let f be a complex—valued function which is defined on some disk D: Iz—al<r with center at the complex number a and radius r > 0. Much of the calculus for such functions can be patterned directly after the calculus of complex- valued functions defined on a real interval I. We say that f has the com- plex number L as a limit at 20 in D if |f(z) — LI —>0, as 0 < Iz—zol -—)0, and we write limf(z) =L, or f(z)—>L, (2—920). 2—»: 0 If f and g are two complex-valued functions defined on D such that for some 20 in D I‘M—>11, 9(2)—>M, (z-Mo), 12 Preliminaries Chap. 0 then (f+ g)(2) —>L + M, (fg)(z) —>LM, (2—wa- The proofs are identical to those for functions defined on real intervals. The function f, defined on the disk D, is said to be continuous at 20 in D if |f(z) —f(zo) l——>0, as 0 < [2 — all—>0. It is said to be continuous on D if it is continuous at each point of D. The sum and product of two functions which are continuous at 20 are continuous there. Examples of continuous functions on the whole complex plane are f0?) = lzt 9(2) = 23. Let g be defined on some disk D1 containing 20, and let its values be in some disk D2, where a function f is defined. If g is continuous at 20, and f is continuous at g(zo), then “the function of a function” F given by 17(2) = (z in D1); is continuous at 20. The proof follows the same lines as in calculus for real- valued functions defined for real 2:. If f is defined on a disk D containing .20 we say that f is dififerentiable at 20 f(z) — f(zo) 2’20 ’ (z ¢ 20)! has a limit at 20. If f is differentiable at 20 its derivative at 20, f’ (zo) , is defined to be this limit. Thus 2"‘20 ——f’(zo) ——>0, asO<|z-—zo|—->0. Formally our definition is the same as that for the derivative of a complex- valued function defined on a real interval. For this reason if f and g are functions which have derivatives at 20 in D then f + g, fg have derivatives there, and (f + g)'(Zo) = f’(Zu) + g'(zo), (3.2) (fg)’(zo) = f’(Zo)g(Zo) + f(20)g'(20)~ Also, suppose f and g are two functions as given in (3.1), and that g is differentiable at 20, whereas f is differentiable at g(20). Then F is differen- tiable at zo, with 'F' (20) = f'(9(zo) )9' (20}- Sec. 3 Preliminaries 13 It is clear from the definition of a derivative that the function (1 defined by (1(2) = c, where c is a complex constant, has a derivative which is zero everywhere, that is, q’(z) = 0. Also, if 121(2) = z for all 2, then p{(z) = 1. Combining these results with the rules (3.2) we obtain the fact that every polynomial has a derivative for all z. A polynomial is a function 1) whose domain is the set of all complex numbers and which has the form 72(2) = aoz" + an“ + -- - + (In—12 + an, where a0, a1, - ~ -, a" are complex constants. The rules (3.2) imply that for such a p p’(z) = %7zz”“1 + a,1(n —— Dan—2 + - + and. Thus 12’ is also a polynomial. It is a rather strong restriction on a function defined on a disk D to demand that it be differentiable at a point 20 in D. To illustrate this we note that the real—valued function f given by f(x) = [$1, for all real .1, is differentiable at all cc 75 0. Indeed f ’( x) is +1 or ——1 accord- ing as a: is positive or negative. However the continuous complex-valued function g given by 9(2) = Izl, for all complex 2, is not differentiable for any 2. Suppose 20 = x0 + 1101' sf 0, for example, and let 2 = x + yi. Then for z ¢ 20 V! - leol = (962 + 7J2)”2 - (x3 + 2/3)“2 2—20 (x—xo)+i(y-yo) = (x2 + W - (xi + 2/3) [(96 — me) + My * yoljflwz + 3/2)“2 + (x3 + 313W]- If we let Iz — col 90 using 2 of the form 2 = $0 + yz‘ (that is y—> go) we see that izl_izoi ) 1/0 z—a' un+nww (3.3) Whereas if we let I2 — 20[ —>0 using 2 of the form 2 = x + W (that is x —+ me) we obtain l2! - IZOI we a . .4. 2—a "Wa+aw2 (3) 14 Preliminaries Chap. 0 The two limits (3.3) and (3.4) are different. However, in order that g be differentiable at 20 we must obtain the same limit no matter how I2 —— zol —> 0. This shows that g is not differentiable at 20. (d) Other functions. Other types of functions which are important for our study of differential equations are usually combinations of the types discussed in (b), (c) above. Typical is a complex—valued function f which is defined for real as on some interval Ix — xol g a (mo real, (1 > 0), and for complex 2 on some disk lz —— zol § b (20 complex, b > 0). Thus the do- main D of f is given by D: lx~xol ga, Iz—zol éb, and the value of f at (x, z) is denoted by f (x, 2). Such a function f is said to be continuous at (E, n) in D if lf(x,z) “f(£,n)l—*0, as 0 < Ix— 21 + 12— nl ->0- There are two important facts which we shall need in Chap. 5 concern- ing such continuous functions. The first is that a continuous f on the D given above (with the equality signs included) is bounded, that is, there is a positive constant M such that lf(x,z)l é M, for all (x, z) in D. This result is usually proved in advanced calculus courses. The second result relates to “plugging in” a complex—valued func- tion 4) into f. Suppose 45 is a complex—valued function defined on lit—550' gay which is continuous there, and has values in la — zol g b. Then if f is continuous on D, the function F given by FCC) =f(x,¢(x)), for all x such that [as —— :50] § (1, is continuous for such as. A slightly more complicated type of complex—valued function f is one which is defined for real 2' and complex 21, - - - , 2,. on a domain D: [1— fuel é a, In —' zlol + + lzn— znol é b. Here are is real, 2m, - - -, 2,0 are complex, and a, b are positive. The value of fat cc, 21, - ~ - , 2,. is denoted byf(x, zl, - - - , 2,.) . Continuity offis defined just Sec. 3 as in the case of one 2. Thqu is continuous at E, m, - - o, 17,. in D if 0< lat-El + IZi—ml +-~+ lzn_nn|—)0. Such an f is bounded on D, and if (351, ~ - valued functions defined on Ix — xol _S_ (1, having the property that Preliminaries “(37,211 "'5zn) -fQNTI: "3771!” *0; 15 -, qsn are n continuous complex- |¢1(x) "Ziol + "' + |¢n(x) — Znol é b for all such 25, then the function F given by Ft”) =f(xi¢l(x)) ""¢n(x)) for |x —- x0] é a is continuous there. EXERCISES ,l. Leta: 2+i3,b= 1— i. Ifforallreala: f(2) = ax+ (bx)? compute : (a) (Ref)(rc) (c) f’(x) ‘ 2. If for all real 1: compute : f(rv) = x+ ix”. (b) (Imoe) l d dz < > [0 f(x) x2 9(90) = —, (a) The function F given by F (x) = f (g(:c)) 2 (b) F’(x) 3. If a is a real—valued function defined on an interval I, and f is a complex- valued function defined there, show that Re (W) = a(Re D, 4. Let f(z) = 22 for all complex 2, and let 24(33, y) = (313 fXx + W); (a) Compute u(x, y) and 0(23, y). (b) Show that du 82) 62: w fly, Im (af) = a(Im 0(1', y) = (1m f)(x + iy). 16 Preliminaries Chap. 0 (0) Show that 6% 62a 62v 620 6x2 + 6y2 ’ 6x2 fly2 5. Let f be a complex-valued function defined on a disk D: Iz|<r (r>0), which is differentiable there. Let "(95, y) = (Re f)($ + 1'31), 9(3), y) = (1m no + W)- Show that 6a 80 Ba _ 6v ace—6y, 53—; 6:15, forallz = x+ iyinD. (Hintlfzo = x0+ iyoisinD,let0 < [2 — eel—>0, in the definition of f’(2o), through 2 of the form 2 = as + iyo, and then of the form 2 = x0 + lg, to obtain 6 6 rec) = is... 2/0) + rice, yo) =—( )—-“( l x ’t x, . 0,110 a 03/0 The equations (*) are called the Cauchy-Riemann equations.) . 6. Let f be the complexwalued function defined on DI lxlél, HMS—2, (a: real, 2 complex) by f(x, 2) = 32;2 + x2 + 22, and let d) be the function defined on I a: | g l by We) = x + i. (a) Compute the function F given by F03) = f(x, ¢(x)), (l xl é 1). (b) Compute F'(:v). (0) Compute [01 F(x) dz. — -— (*)4 Sec. 4 Preliminaries 17 , 7. If r is a complex number, and 29(2) = (2 — r)", Where n is a positive integer, show that W) = p’(r) = = p‘"“’(r) = 0, p‘"’(r) = n!- 4. Polynomials We have defined a polynomial as a complex-valued function 20 whose domain is the set of all complex numbers and which has the form 17(2) = 04:2" + (112"‘1 + - - - + an_1z + an, where n is a non-negative integer, and a0, a1 - - - , an are complex constants. The highest power of z with non-zero coefficient which appears in the expression defining a polynomial p is called the degree of p, and written deg 17. A root of a polynomial p is a complex number r such that p(r) = 0. A root of p is sometimes called a zero of 12. We shall require, and assume, the following important result.* Flmdamental theorem of algebra. If p is a polynomial such that deg p g 1, then p has at least one root. This is a rather remarkable result, and justifies our introduction of the complex numbers. We have seen that not every polynomial with real coefficients (for example 22 + 1) has a real root, but polynomials of degree greater than zero with complex coefficients always have a complex root. The remarkable fact is that we do not need to invent new numbers, which include the complex numbers, to guarantee a complex root. We derive some consequences of this fundamental theorem. Corollary 1. Let p be a polynomial of degree n g 1, with leading coefi‘i— cient 1 (the coefilcient of z"), and let r be a root of p. Then p(z) = (z — r)q(2) where q is a polynomial of degree n — 1, with leading coefiicient 1. Proof. Let 12(2) have the form 22(2) = z" + aiz'”l + - - - + tin—12 + an, * A proof can be found in G. Birkhofl' and S. MacLane, A survey of modern algebra, New York, rev. ed., 1953, p. 107, and also in K. Knopp, Theory of functions, New York, 1945, p. 114. 18 Preliminaries ‘ Chap. 0 and let 6 be any complex number. Then 22(2) — 10(6) = (2" -— c") + (1102"—1 — c“) + + an-1(z — c) = (z - c)q(z), where q is the polynomial given by q(z) = 2"‘1 + cam—2 + czz"‘3 + - ' ' + 0"”1 + a1(z"‘2 + 62"‘3 + - - - + c"") + - - - :I- am. Clearly deg q = n —- 1 and g has leading coefficient 1. In particular if c = r, a root of p, then we have p(z) = (z - 104(2), as desired. If n —— 1 _Z_ 1, the polynomial q has a root, and this root is also a root of p by Corollary 1. Thus applying the Fundamental Theorem of Algebra n times, together with Corollary 1, we obtain Corollary 2. If p is a polynomial, deg p = n g 1, with leading coefi‘l— cz'ent a0 ;é 0, then p has exactly 11 roots. If r1, r2, - - ~ , rn are these roots, then pt?) = ao(z — 1'1) (z — r2) (z — r"). (4.1) Note that aglp is a polynomial which has leading coefficient 1. We re— mark that the roots need not all be distinct. If r is a root of p; the number of times 2 — r appears as a factor in (4.1) is called the multiplicity of 7'. Theorem 1. If r is, a root of multiplicity In of a polynomial p, deg p g 1, then W”) = IN“) = = p‘“‘”(r) = 0, and p('"’(r) # 0. Proof. Let p have leading coefficient on ¢ 0, and degree n g m. It follows from Corollary 2 that 22(2) = aoe — We, (42) where q is a polynomial of degree n —— m, and q(r) ;é 0. Clearly p0“) = 0 by the definition of a root. Also p’Cz) = agm(z -— r)'"“q(z) + (10(3 — r)"‘g’(z), Sec. 4 Preliminaries 19 and this implies that, if m — 1 > 0, p’(r) = 0. If m = 1 we have P’(z) = aoq(2) + ao(z - r)q’(z), and thus p’(r) = a0q(r) 75 0. The general argument can be based on (4.2) and the formula Ic(k —— 1) (fg)"°’ = My + (CW-“9’ + 2. far—mg” + + fgoc) (4.3) I for the k-th derivative of the product fg of two functions having 19 deriva- tlves. Formula (4.3) can be established by induction. Applying (4.3) to the funct10nsf(z) = (z — 1')“, 9(2) = q(z) in (4.2), we obtain p‘Wz) = ao[m(m — 1) (m — k + 1) (z — r)"'”°q(z) + (terms with higher powers of (z — r) as a factor) ]. It is now clear that p0.) ___ pi“) = ... = p(m—1)(r) = 0’ and Mo) = aom! go) :75 0, which is the desired result. EXERCISES 1. Compute the roots, with multiplicities, of the following polynomials: (a)zz+z—6 (b)22+2+,1 (c) 23 — 322+ 4 (d) 23 — (2+ 7322+ (1+ 72).? -i (8)24—3 ’ 2. If r is such that r3 = 1, and 1' ¢ 1, prove that 1+ 7' + r2 = 0. 3. Let p be the polynomial given by '. M2) = aoz"+ W“ + + am with a0, a1, - - -, an all real. Show that 1-7?) = W). As a consequence show that if r is a root of p, then so is 7". . 4. Prove that every polynomial of degree 3 with real coefficients has at least one real root. ' 20 Preliminaries Chap. 0 5. Prove that if p is a polynomial, deg 12 > 1, and r is a complex number such that 11(r) = 11’0") = = p‘m‘1)(r) = 0. p‘“’(r) 75 0, then r is a root of p with multiplicity m. This is the converse of Theorem 1. 6. (a) Use the result of Ex. 5 to show that i is a root of the polynomial 1) given by 73(2) = 25+ (2 — 3024+ (fl — 6023+ (-—6 -- 51322 + (-6 + 2i)z + 2i, and compute the multiplicity of i. (b) Find the other roots of the polynomial p in (a). 7. Prove the formula (4.3). This can be written in the form k k _ ~ I (lg—2) II (for) — flk’g + (1)“ “g + (2)f g + + (Dr's—“gm + + fg‘k’, (k) _ ls! l _ l!(k— l)! is a binomial coefficient. H int: Use induction, and show that k + 1) - ( k ) (k) < l ' z— 1 + z ' 5. Complex series and the exponential function where If a: is a real number, and e is the base for the natural logarithms, the number 6" exists, and m xiv vr = —, 0! = 1 , ° a. k! ( ) where theseries converges for all real Indeed, this series may be taken as the definition of e”. We shall need to know what (22 is for complex 2. One way is to define e3 by a) 2k 6‘ = -. (5.1) k=0 kl Now we have to prove that this series converges for all complex 2, and in fact there is the problem of defining what we mean by a convergent series with complex terms. The method is the sameas that used to define con— vergcnt series with real terms. ,V “sageswwhw. . , u. “WWWWWY‘. inflnmn, . > «vm kwrr" «may»? mmm “ml”... was“ New“, We See. 5 Preliminaries 21 A series I; Ck, V (5.2) where all ck are complex numbers, is said to be convergent if the sequence of partial sums sn=zck1 (nz0r192yn')! k—O tends to a limit 3, as n -—> 00. That is, s is a complex number such that lsn - SI —>0, (n—+ 00), Where the magnitude is the magnitude for complex numbers. If the series (5.2) is convergent, and 3,. —> s, we call 8 the sum of the series, and write an S = Z 6),. kfil If the series is not convergent we say that it is divergent. The series (5.2) with 00mplex terms or gives rise to two series with real terms, namely ‘ f: Re ck, k=0 go: Im ca (5.3) and it is not difficult to see that the series (5.2) is convergent with sum 8 = Re 3 + iIm 8 if, and only if, the two real series in (5.3) are con- vergent with sums Re s and Im s respectively. In principle, therefore, the study of series with complex terms is the study of pairs of real series. The series (5.2) is said to be absolutely convergent if the series k; | a. | (5.4) is convergent. It can be shown that every absolutely convergent series is convergent. Since the series (5.4) has terms which are real and non-negative, any condition which implies the convergence of such series can be applied to guarantee the convergence of the series (5.2). One of the most important testsfor convergence is the ratio test. One version of this is the following. Ratio test. Consider the series a: Zlckl, 22 Preliminaries Chap. 0 where the 0,. are complex. If Ickl > 0 for all k beyond a certain positive integer, and ick+1i I Ch I —> L. (k —* w), (5-5) then the series is convergent if L < 1, and divergent for L > 1. Thus the series (5.2) is convergent if (5.5) is valid for an L < 1. An immediate application of this result is to the series co 3k En Here ck = z"/Ic!and |6k+1| 2"“ k! lzl = ———-— = k . Ickl (Ic+1)!z'= Ic+l—)0’ (Tm) Thus this series converges for every z such that < 00, that is, for all complex 2. Hence our definition (5.1) of e’ as the sum of this series makes sense. The function which associates with each 2 the complex number 6’ is called the exponential function. The series defining e‘ is an example of a power series 0:) Z ak(z '— 20),“ k-O about some point 20, the a]. being complex. Many of the properties of a power series of the type (D Z ak(x _ 1:0)"; i=0 where the ca, x, :30 are real, remain true for series of the form (5.6), and the proofs are identical. In particular, if a series (5.6) is convergent on a disk Dzlz — 201 < r (r > 0), then the functionf defined by on f(2) = Z M2 -— Zo)’°, k—O (z in D), has all derivatives in D, and these may be computed by differentiating term by term. Thus f’(z) = 5: me — 20),”! = 5': me — sow, lu-O i=1 ? . ~’ .. g, if, t} X5" é‘, ' «i i i ., Sec. 5 Preliminaries 23 where the last series converges in D. Applying this result to (5.1) we find that zk—l a, zk—l a, 2k (6‘) =Ekfi=§(k—1)!=.§m Another important property of the exponential function is that = e‘. ez1+22 = 621622 for every complex 21, 22. This can be proved by justifying the following steps I: k €21,622 = _ __ = . (2s, this, k!) 2 c" k Ic—O Here Thus formally we have the product of the series defining e'1 and e'2 is the series defining e'h‘”, and these steps can be justified to give a proof of the equality (5.17) . A consequence of (5.7) is that (ez)n = en: for every integer n. In particular l/e' = e“. Another property of the exponential function is that for all real 0, e“ = cos 0 + isin 0, (5.8) and the proof results from adding the series involved. Indeed, i2 = —1, i3 = —i, i4 = 1, etc., and thus 62 94 0089:]. "‘ (M2 (179)“ ._ 1 + 2' 4! _ 03 65 sm0—0—§+fi- "-, ‘ . 3 ' 5 ismfl = w+(1)+(w) 3! 5! 24 Preliminaries Chap. 0 Hence . . I “(9)2 (my cosfi+zsm0=1+w+ + +--- 2! 3! = 20.: (my: = e“. k=0 k! A consequence of (5.8) is that r” = cos 0 — isin 0, (5.9) since cos (—0) = cos 0, and sin (—0) = — sin 0. Using (5.8) and (5.9) we can solve for cos 0 and sin 6, obtaining eiO + 6—129 cos 0 = ———2— , eiB _ e—i9 sin 0 = —————— 22‘ If 2 is a complex number with polar coordinates (r, 0) , then z=r(cos0+isin0), (r;0,0§0<27r), and we have, using (5.8) , z = re”. (5.10) Note that [2] = r, Ieial = 1 for every real 6. The relation (5.10) can be employed to find the roots of polynomials p of the form 10(2) = z" — c, (5.11) where c is a complex constant. Suppose c = [cl 6"“, where a is real, 0 g a < 21r, and re“ is a root. Then rneino = iclem’ and taking magnitudes of both sides we see that r" = Icl, or r =|cl1/", where the positive n—th root is understood. Further em = eia, or ei(n9—a) = 1, Sec. 5 Preliminaries 25 There are exactly n distinct values of 0 satisfying this relation and 0 g 0 < 21r, namely, those for which n0 — a = 21rk, or 2k o=9~+7£, (lo=0,1,u-,n—-l). Thus the roots zl, - - -, 2,. of the polynomial p in (5.11) are given by zk+l = I c ll/n6i(a+21rlc)/n = lcll’" [cos(%12—dc)+ isin(c—¥—j—;L~2Lk)], (k =0,1, n-,n — 1). Geometrically we can describe the roots of p as follows. All roots lie on a circle about the origin with radius Icl 1’". One root has an angle 01/11 with the real axis, if c has angle a with the real axis. The remainder of the roots are located by cutting the circle into n even parts, with the first cut being at the root at angle a/n. As a particular example let us find the three cube roots of 41'. Thus we want the roots of 23 — 41'. Here 0 = 41', and hence the cube roots will all have a magnitude of [41' [1’3 = 4‘“. If we write 6 = Icle‘“, we see that a = 7r/2 in this case. Thus the three cube roots of 42’ are given by 21 = 41/361316, 22 = 41/361151”, as = 41/3ei9rI6’ Imaginary axis Figure 2. Three cube roots of 413 26 Preliminaries Chap. 0 or since 7r/6 represents 30°, 2.», = — 41/321 These roots are sketched in Fig. 2. E X E R C I S E S 1. Find the three cube roots of l. 2. Find the two square roots of i. 3. Find all roots of the polynomials: (a) 23 + 24 (h) z4 + 2'64 (c)z‘+4z2+4 x(d)2‘°°—1 : 4. If 2 = x + iy, where x, y are real, show that I e‘ I = e”. As a consequence / show that there is no complex 2 such that e‘ = 0. _ 5. If a, b, a; are real show that: (a) Re [e(“+"’)’] = e” cos b2: (b) Im [e(“+"’)’] = 9‘” sin bx 6. (a) If r = a + 2'6 75 0, where a, b are real, show that (e”)’ = re". (b) Using (a) compute: (i) [01 e" da: 1 (ii) / e” cos b2; d2: 0 1 (iii) / c“ sin bx d1: 0 , 7. (a) If ¢(:v) = e”, where r is a complex constant, and :1: is real, show that ¢’(x) — WW) = 0- (b) If (Mac) = em, where a is a real constant, show that: (i) ¢'(x) — ia¢(x) = 0 (ii) 45"(35) + a2¢(x) = 0 8. For What values of the constant r will the function 4) given by ¢(x) = e” satisfy ¢"(x) + 3¢’(x) — 24:05) = 0 for all real 1;? regime 2-2:, Sec. 6 Preliminaries 27 9. Let as = k! + (i/kl). For what real 1:, are the following series convergent? (a) f, (Re ak)x" (b) 2“: (Im mack k-o k—O (0) 2 am" [5-0 I}, 10. Consider the series . E 2k, (*) where z is complex. (a) Show that the partial sum 8,.(2) = Z": 2" = tan 1 _ zn+l l—z ’ ifz¢ l. (b) Show that the series (*) converges absolutely for I z I < l. (0) Compute the sum 8(2) of the series (*) for I zI < l. 6. Determinants We shall need to know the connection between determinants and the solution of systems of linear equations. Suppose we have such a system of n equations (11121 + (11232 + ' ' ' + alnzn 01 02 (6.1) can + (12222 + - - - + aznzn «1.121 + c.222 + ' - - + annzn = on, Where the a5,- and c; are given complex constants. The problem is to find complex numbers 21, ---, 2,. satisfying these equations. Such a set of n numbers is called a solution of (6.1). We say that two solutions 21, - - o, 2,. andzi, ---,z,:of (6.1) areequal ile = 21’, ---,z,. = 2;. Ifcl = 02 = = 6,. = O we say that the system is a homogeneous system of n linear equations, otherwise we say (6.1) is a non-homogeneous system. The determinant A of the coefficients in (6.1) is denoted by an (112 "' aln 1121 (122 "‘ Gen 28 Preliminaries Chap. 0 and is shorthand for the number A given by A = 2(i)ali1a2i2 ' '. ail-in) where the sum is over all indices i1, - - -, in such that i1, - - -, in is a permuta- tion of 1, - - -, n and each term occurs with a + or — sign according as i1, - - -, in is an even or odd permutation of 1, - - -, n. Thus an 012 =a114122 " (1121121, €121 (122 and ’ an 012 (113 = auazzaaa ‘ (11111231132 + (112023031 “21 €122 023 “ alzamaaz + (1130211132 — a13a22aal- 031 (132 (133 The principal results we require concerning determinants are contained in the following theorems. They are usually proved in elementary texts on linear algebra. Theorem 2. If the determinant A of the coefiicients in (6.1) is not zero there is a unique solution of the system for 21, ~ - -, 2". It is given by 216:2" (k=1)"’7n)a where Ah is the determinant obtained from A by replacing its kth column a”, ---,a,.kbycl, ---,c,.. Proof for the case n = 2. In this case suppose 21, 22 satisfy (11121 + (11252 = 01 (6.2) 02131 + 02222 = 02- Multiply the first equation by (122, the second equation by —-a12, and add. There results 01 612 21A 2 (12261 —' (21262 = = A1- 02 (122 Sec. 6 Preliminaries 29 Multiply the first equation by —a21, and the second by an, and add, ob- taining an 01 22A = -—a2101 + (11162 = = A2. 021 62 Thus if A ¢ 0, zk must be Ak/ A (k = 1, 2), and it is readily verified that these values satisfy (6.2). We note that for a homogeneous system (cl = 02 = -- - = 6,. = 0 in (6.1)) there is always the solution zl=z2=luu=zn=0. This solution is called the trivial solution. Theorem 3. If cl = (32 = = 0,. = 0 in (6.1), and the determinant of the coefiicients A = 0, there is a solution of (6.1) such that not all the zk are 0. Proof for the case n = 2. We are dealing with the case 01121 + (11222 = 0 azlzi + a2222 = 0, where (111022 — 021012 = 0. If an 75 0, _a12 21 = , 2'2 = 1, an is a solution. If an = 0, and a2; 75 0, _a22 is a solution. If an = 0, and a2; = 0, is a solution. ‘ Combining Theorem 3 with Theorem 2 we obtain Theorem 4. The system of equations (6.1) has a unique solution if, and only if, the determinant A of the coefiicients is not zero. Proof. If A ¢ 0 Theorem 2 says that there is a unique solution. Con- versely, suppose there is a unique solution 21, - - -, 2,. of (6.1). If A = 0, by 30 Preliminaries Chap. 0 Theorem 3 there is a solution {1, - - -, {n of the corresponding homogeneous system, which is not the trivial solution. Then it is easy to check that z; + {1, - - -, 2.. + £3. is a solution of (6.1) distinct from 21, ---, z”, and forces us to conclude that A ;é 0. EXERCISES 1. Consider the system of equations izl + 22 = 1 + 'I; 221 + (2 — 1:)22 = 1. (a) Compute the determinant of the coefficients. (b) Solve the system for 21 and 22. 2. Solve the following system for 21, 22 and 23: 321 + 22 — 23 = 0 221 — 23 = 1 32+2Z3=2 3. Does the following system of equations have any solution other than 21 = 22 = 23 = 0? If so find one. 421+222+223=0 321+ 722+ 223: 0 221+ 22+ 23:0 4. Consider the homogeneous system corresponding to (6.1) (the case 01 —— 62 = - ° - = cn = O). Show that if the determinant of the coefficients A = 0, there are an infinite number of solutions. (Hint: If 21, - - -, 2,. is a non-trivial solution, show that 0121, - - - , azn is also a solution for any complex number a.) 5. Prove that if the determinant A of the coefficients in (6.1) is zero then either there is no solution of (6.1), or there are an infinite number of solutions. (Hint: Use Ex. 4.) 7. Remarks on methods of discovery and proof Often a student studying mathematics has difficulty in understanding why or how a particular result, or method of proof, was ever conceived in the first place. Sometimes ideas seem to appear from nowhere. Now it is true that mathematical geniuses do invent radically new results, and meth- ods for proving old results, which often appear quite strange. The most that ordinary people can do is to accept these brilliant ideas for what m-mwm —. We..,...m...w,m..w mwsmwmwww... , Sec. 7 Preliminaries 31 they are, try to understand their consequences, and build on them to obtain further information. However, there are a few general principles which, if followed, can lead to a better understanding of mathematical dis- covery and proof. Concerning discovery, we mention two principles: (a) use simple examples as a basis for conjecturing general results, (b) argue in reverse. Both of these principles are illustrated in the proof we gave of Theorem 2 for the case n = 2. We were faced with trying to find out whether the system (6.1) of n linear equations has a solution or not, and what condition, or conditions, would guarantee a unique solution. We looked at the simplest example, which occurs for n = 2 (using ((1)). Then we assumed that we had a solution (principle (b)), and found out what must be true for a solution, namely, that 21A = A1, 22A = A2. We immediately saw that if A ¢ 0, then A1 A2- Zl=‘_ Zz=~.i A . A (7.1) Note that at this point we have not yet shown that there is a solution. All we have shown is that if 21, 22 is a solution, and A 75 0, it must be given by (7.1). We can now guess that if A 75 0, then 21, 22 given by (7.1) is a solu- tion. This can be readily verified by substituting ( 7 .1) into the given equations. An alternate procedure is to check that the steps leading to (7 .1) can be reversed, if A # 0. Once we have discovered the right condi- tion for the case n = 2, it is natural to conjecture that a similar condition will work for a general n. Three important methods of proving mathematical results are: (i) a constructive method, (ii) method of contradiction, (iii) method of induction. A typical example of aconstructive method appears in the proof of Theorem 3 for the case n = 2. We wanted to show that nontrivial solutions of the two homogeneous equations exist if A = 0. To do this we constructed solutions explicitly. An example of the method of contradiction appears in the proof of Theorem 4. We supposed that the system (6.1) had a unique solution. We assumed that A = 0, and, using logical arguments, we arrived at the fact that ( 6.1) does not have a unique solution. This is a contradic- tion, and the only thing that can be wrong is our assumption that A — 0. 32 Preliminaries Chap. 0 The only other alternative is that A 75 0, which is the conclusion we de— sired. ' The method of induction is concerned with proving an infinite number of statements 31, 82, - - -, one for each positive integer n. If 81 is true, and if for any positive integer k the statement 3,, implies the statement 8H1, then all the statements 81, 82, - - -, are true. An example of a result which can be proved using induction is the formula k Ic k k! (k) = (In—z) a) = “— (fg) Half ‘7 ’ (l) l!(Ic—l)!’ for the k-th derivative of the product of two complex-valued functions f, g which have It derivatives; see (4.3). The proof is the same as the induc- tion used to prove the binomial formula I: (a + w = 120 ak-lbl, (Is =1,2,---), for the poweis of the sum of two complex numbers a, b. The method of induction is equivalent to a property of the positive integers, and conse- quently we assume that this method is a valid method of proof. The principles of discovery (a), (b), and the methods of proof (i), (ii), (iii), will be used many times throughout this book. It will be instruc— tive for the student to identify which principles and methods are being used in any particular situation. CHAPTER 1 Introduction—Linear Equations of the First Order 1. Introduction In Sec. 2 we discuss what is meant by an ordinary differential equation and its solutions. Various problems which arise in connection with differ- ential equations are considered in Sec. 3, notably initial value problems, boundary value problems, and the qualitative behavior of solutions. In a succession of easy steps we solve the linear equation of the first order in Secs. 4—7. 2. Difl'erential equations Suppose f is a complex-valued function defined for all real :6 in an interval I , and for complex y in some set S. The value of f at (x, y) is denoted by f (x, 1/). An important problem associated with f is to find a (complex-valued) function d; on I , which is differentiable there, such that for all x on I, (i) ¢>(x) is in S, (ii) ¢’(x) =f(x, ¢(x)). This problem is called an ordinary difierential equation of the first order, and is denoted by y’ = f(x, fl)- (2-1) The ordinary refers to the fact that only ordinary derivatives enter into the problem, and not partial derivatives. If such a function ¢ exists on I satisfying (i) and (ii) there, then qs is called a solution of (2.1) on I. 33 ...
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