1610912322-b551b095a53deaf3d3fbd1ed05ae9b84 (824701), страница 58

Просмтор этого файла доступен только зарегистрированным пользователям. Но у нас супер быстрая регистрация: достаточно только электронной почты!

Текст из файла (страница 58)

IfNow consider the ratio R(x) = Q(x)the degree of P (x) is larger or equal than that of Q(x), we apply the division algorithm and represent P (x) as p(x)Q(x) + r(x), where p(x) and r(x) are polynomials, the degree of r(x) being less than that of Q(x). Thus we obtain a representationr(x)r(x), where the fraction Q(x)is now a proper fractionof the form R(x) = p(x) + Q(x)in the sense that the degree of r(x) is less than that of Q(x).The corollary we are about to state involves the representation of a proper fractionas a sum of fractions called partial fractions.Corollary 4 a) If Q(z) = (z − z1 )k1 · · · (z − zp )kp andthere exists a unique representation of the fractionP (z)Q(z)is a proper fraction,in the form kjpP (z) aj k=.Q(z)(z − zj )kj =1 k=1P (z)Q(z)(5.134)2845Differential Calculusb) If P (x) and Q(x) are polynomials with real coefficients andmmQ(x) = (x − x1 )k1 · · · (x − xl )kl x 2 + p1 x + q1 1 · · · x 2 + pn x + qn n ,there exists a unique representation of the proper fractionP (x)Q(x)in the form kj mjlnaj kbj k x + cj kP (x) =+,Q(x)(x − xj )k(x 2 + pj x + qj )kj =1 k=1(5.135)j =1 k=1where aj k , bj k , and cj k are real numbers.We remark that there is a universal method of finding the expansions (5.134) and(5.135) known as the method of undetermined coefficients, although this method isnot always the shortest way.

It consists of putting all the terms on the right-hand sideof (5.134) or (5.135) over a common denominator, then equating the coefficients ofthe resulting numerator to the corresponding coefficients of P (x). The system oflinear equations that results always has a unique solution because of Corollary 4.Since we shall as a rule be interested in the expansion of a specific fraction, whichwe shall obtain by the method of undetermined coefficients, we require nothingmore from Corollary 4 than the assurance that it is always possible to do so. For thatreason, we shall not bother to go through the proof.

It is usually couched in algebraiclanguage in a course of modern algebra and in analytic language in a course in thetheory of functions of a complex variable.Let us consider a specially chosen example to illustrate what has just been explained.Example 17 LetP (x) = 2x 6 + 3x 5 + 6x 4 + 6x 3 + 10x 2 + 3x + 2,Q(x) = x 7 + 3x 6 + 5x 5 + 7x 4 + 7x 3 + 5x 2 + 3x + 1.P (x)Find the partial-fraction expansion (5.135) of the fraction Q(x).First of all, the problem is complicated by the fact that we do not know thefactors of the polynomial Q(x). Let us try to simplify the situation by eliminatingany multiple roots there may be of Q(x). We findQ (x) = 7x 6 + 18x 5 + 25x 4 + 28x 3 + 21x 2 + 10x + 3.By a rather fatiguing, but feasible computation using the Euclidean algorithm,we find the greatest common divisord(x) = x 4 + 2x 3 + 2x 2 + 2x + 1of Q(x) and Q (x).

We have written the greatest common divisor with leading coefficient 1.5.5 Complex Numbers and Elementary Functions285Dividing Q(x) by d(x), we obtain the polynomialq(x) = x 3 + x 2 + x + 1,which has the same roots as Q(x), but each with multiplicity 1. The root −1 iseasily guessed. After q(x) is divided by x + 1, we find a quotient of x 2 + 1. Thusq(x) = (x + 1) x 2 + 1 ,and then by successively dividing d(x) by x 2 + 1 and x + 1, we find the factorizationof d(x);d(x) = (x + 1)2 x 2 + 1 ,and then the factorization2Q(x) = (x + 1)3 x 2 + 1 .Thus, by Corollary 4b, we are seeking an expansion of the fractionP (x)Q(x)in the forma11a12P (x)a13b11 x + c11 b12 x + c12=++ 2++.23Q(x) x + 1 (x + 1)(x + 1)x2 + 1(x + 1)2Putting the right-hand side over a common denominator and equating the coefficients of the resulting numerator to those of P (x), we arrive at a system of sevenequations in seven unknowns, solving which, we finally obtain12x+11x −1P (x)=−+++.Q(x) x + 1 (x + 1)2 (x + 1)3 x 2 + 1 (x 2 + 1)25.5.6 Problems and Exercises1.

Using the geometric interpretation of complex numbersa) explain the inequalities |z1 + z2 | ≤ |z1 | + |z2 | and |z1 | + · · · + |zn | ≤ |z1 | +· · · + |zn |;b) exhibit the locus of points in the plane C satisfying the relation |z − 1| +|z + 1| ≤ 3;c) describe all the nth roots of unity and find their sum;d) explain the action of the transformation of the plane C defined by the formulaz → z̄.2. Find the following sums:a) 1 + q + · · · + q n ;b) 1 + q + · · · + q n + · · · for |q| < 1;c) 1 + eiϕ + · · · + einϕ ;2865d)e)f)g)h)i)Differential Calculus1 + reiϕ + · · · + r n einϕ ;1 + reiϕ + · · · + r n einϕ + · · · for |r| < 1;1 + r cos ϕ + · · · + r n cos nϕ;1 + r cos ϕ + · · · + r n cos nϕ + · · · for |r| < 1;1 + r sin ϕ + · · · + r n sin nϕ;1 + r sin ϕ + · · · + r n sin nϕ + · · · for |r| < 1.3.

Find the modulus and argument of the complex number limn→∞ (1 + nz )n andverify that this number is ez .4. a) Show that the equation ew = z in w has the solution w = ln |z| + i Arg z. It isnatural to regard w as the natural logarithm of z. Thus w = Ln z is not a functionalrelation, since Arg z is multi-valued.b) Find Ln 1 and Ln i.c) Set zα = eα Ln z .

Find 1π and i i .d) Using the representation w = sin z = 2i1 (eiz − e−iz ), obtain an expression forz = arcsin w.e) Are there points in C where | sin z| = 2?15. a) Investigate whether the function f (z) = 1+z2 is continuous at all points ofthe plane C.1b) Expand the function 1+z2 in a power series around z0 = 0 and find its radiusof convergence.c) Solve parts a) and b) for the function 1+λ12 z2 , where λ ∈ R is a parameter.Can you make a conjecture as to how the radius of convergence is determined bythe relative location of certain points in the plane C? Could this relation have beenunderstood on the basis of the real line alone, that is, by expanding the function1, where λ ∈ R and x ∈ R?1+λ2 x 26.

a) Investigate whether the Cauchy function!2e−1/z , z = 0,f (z) =0,z=0is continuous at z = 0.b) Is the restriction f |R of the function f in a) to the real line continuous?c) Does the Taylor series of the function f in a) exist at the point z0 = 0?d) Are there functions analytic at a point z0 ∈ C whose Taylor series convergeonly at the point z0 ?ne) Invent a power series ∞n=0 cn (z − z0 ) that converges only at the one pointz0 .in the power series7. a) Making the formal substitution z − a = (z − z0 ) + (z0 −a)∞∞n and gathering like terms, obtain a seriesnA(z−a)n=0 nn=0 Cn (z − z0 ) andkexpressions for its coefficients in terms of Ak and (z0 − a) , k = 0, 1, .

. . .b) Verify that if the original series converges in the disk |z − a| < R and |z0 −a| = r < R, then the series defining Cn , n = 0, 1, . . . , converge absolutely and thenseries ∞n=0 Cn (z − z0 ) converges for |z − z0 | < R − r.5.6 Examples of Differential Calculus in Natural Science287Fig. 5.33nc) Show that if f (z) = ∞n=0 An (z − a) in the disk |z − a| < R and |z0 − a| <R, then in the disk |z − z0 | < R − |z0 − a| the function f admits the representationnf (z) = ∞n=0 Cn (z − z0 ) .8. Verify thata) as the point z ∈ C traverses the circle |z| = r > 1 the point w = z + z−1traverses an ellipse with center at zero and foci at ±2;b) when a complex number is squared (more precisely, under the mapping w →w 2 ), such an ellipse maps to an ellipse with a focus at 0, traversed twice;c) under squaring of complex numbers, any ellipse with center at zero maps toan ellipse with a focus at 0.5.6 Some Examples of the Application of Differential Calculusin Problems of Natural ScienceIn this section we shall study some problems from natural science that are very different from one another in their statement, but which, as will be seen, have closelyrelated mathematical models.

That model is none other than a very simple differential equation for the function we are interested in. From the study of one suchexample – the two-body problem – we really began the construction of differentialcalculus. The study of the system of equations we obtained for this problem wasinaccessible at the time. Here we shall consider some problems that can be solvedcompletely at our present level of knowledge. In addition to the pleasure of seeingmathematical machinery in action in a specific case, from the series of examplesin this section we shall in particular acquire additional confidence in both the naturalness with which the exponential function exp x arises and in the usefulness ofextending it to the complex domain.5.6.1 Motion of a Body of Variable MassConsider a rocket moving in a straight line in outer space, far from gravitating bodies(Fig.

Характеристики

Список файлов книги