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[ Pobierz całość w formacie PDF ] r j z cj R for some point c the centre of the annulus. Then it will in general have an expansion in terms of integral powers some or all of which may be negative. This is called a Laurent Series for the function. More formally De nition 5.3.1 Laurent Series 5.3. LAURENT SERIES 139 For any w the integer power series 1 X ak z c k k 2 Z 1 is de ned to be 1 1 X X 1 ak z c k a k k z c 0 1 when both of these series converge. Theorem 5.3 Laurent s Theorem If f is analytic on the annulus r j z cj R for some point c then f is equal to the Laurent series 1 X f z ak z c k 1 where the coe cients ak can be computed from I 1 f z ak dz 2 i z c 1 k C if k is positive or zero and I 1 f z ak dz k 2 i z c 1 C when k is negative. C is any simple closed loop around the centre c which is contained in the annulus and goes in the positive sense. No Proof See Mathews and Howell or any standard text. 2 Exercise 5.3.1 Try showing a similar result for the Taylor series for an an alytic function i.e. try to get an expression for the Taylor series coe cients in terms of path integrals. It is the case that Laurent series about any point like Power series are unique when they converge. The following result is extremely useful 140 CHAPTER 5. TAYLOR AND LAURENT SERIES Theorem 5.4 Di erentiability of Laurent Series The Laurent series for a function analytic in an annulus if di erentiated termwise gives the derivative of the function. No Proof 2 Since the case where all the negative coe cients are zero reduces to the case of the Taylor series this is also true for Taylor Series. It is not generally true that if a function is given by a sequence of approximating functions the derivative is given by the sequence of derivatives. After all 1 sin nx n gets closer and closer to the zero function as n increases. But the derivatives cos nx certainly do not get closer to anything. This tells us yet again that the analytic functions are very special and that they behave in particularly pleasant ways all things considered. 5.4 Some Sums The series 1 X 1 1 z z2 z3 1 nzn 1 z converges if j zj 1. This is Well Known fact number 137 or thereabouts. We can get a Laurent Series for this as follows nd a series for 1 1 1 w by the usual trick of doing very long division to get 1 X 1 w w2 w3 w4 1 n 1wn 1 1 w 1 and then put z 1 w to get 1 X 1 1 1 1 1 1 1 n 1 1 z z z2 z3 z4 zn 1 5.4. SOME SUMS 141 This converges for j zj 1. It clearly goes bung when z 1 and equally clearly is a geometric series with ratio less than 1 provided j zj 1. 1 There are therefore two di erent Laurent series for one inside the unit 1 z disk one outside. One is actually a Taylor Series which is just a special case. Suppose we have a function like 1 1 1 2i 1 z z 2i z2 1 2i z 2i This has singularities at z 1 and z 2i where the modulus of the function goes through the roof. The function can be expanded about the origin to get i z z2 1 z z2 z3 2 4 8 which converges inside the unit disk. In the annulus given by 1 j zj 2 it can be written as 1 1 1 i z z2 z z2 z3 2 4 8 And in the annulus j zj 2 it can be written 1 1 1 1 2i 4 8i 16 z z2 z3 z z2 z3 z4 z5 Exercise 5.4.1 Con rm the above or nd my error. Substitutions for terms are valid providing care is taken about the radius of convergence of both series. Laurent series expansions about the origin have been produced by some sim ple division. The uniqueness of the Laurent expansions tells us that these have to be the right answers. Next we consider some simple tricks for getting expansions about other points 1 Example 5.4.1 Find a Laurent expansion of about i 1 z 142 CHAPTER 5. TAYLOR AND LAURENT SERIES We write 1 1 1 1 z i 1 z 1 i z i 1 i 1 1 i Then we have the Taylor expansion 1 z i z i 2 z i 3 1 z i 1 1 i 1 i 2 1 i 3 1 i p z i valid for j j 1 i.e. for j z ij 2. We also have 1 i 1 1 i 1 i 2 z i 1 z i z i 2 1 i p valid when j z ij 2. Example 5.4.2 Find a Laurent expansion for 1 z 3 z 2 about 1. 1 z 3 z 1 3 z 1 3 z 2 2 z 1 z 1 Putting w z 1 we get 1 1 1 1 w3 w3 1 w w w2 w3 To give the nal result 1 1 z 1 2 z 1 1 z 1 z 1 2 which is valid for j z 1j 1. A small amount of ingenuity may be required to beat the expressions into the correct shape practice does it. 5.5. POLES AND ZEROS 143 Exercise 5.4.2 Find the Laurent expansion for 1 z z 3 about 1 valid for j z 1j 2. Exercise 5.4.3 Make up a problem of this type and solve it. Exercise 5.4.4 Go through the exercises on page 230 232 of Mathews and Howell. 5.5 Poles and Zeros We have been looking at functions which are analytic at all points except some set of bad or singular points. In the case where every point in some neighbourhood of a singular point is analytic we say that we have an isolated singularity. Almost all examples have such singularities poles of the function places where the modulus goes through the roof no matter how high the roof is put more formally points w such that lim j f z j 1 z w We can distinguish di erent types of singularity there are those that look like 1 z those that look like 1 z2 those that look like Log at the origin and there are those that are just not de ned at some point w but could have been if we wanted to. The last are called removable singularities because we can remove them. For example if I give you the real function f x x2 x you might in a careless mood just cancel the x and assume that it is the same as the function f x x. This is so easily done you can do it by accident but strictly speaking you have a new function. It so happens that it agrees with the old function everywhere except at zero where the original function is not de ned. Moreover the new function is di erentiable everywhere while the old function has a singularity at the origin. But not the sort of singularity which should worry a reasonable man the gap can be plugged in only one way that will make the resulting function smooth and indeed in nitely di erentiable. And if I d made the x a z and said it was 144 CHAPTER 5. TAYLOR AND LAURENT SERIES a complex function exactly the same applies. Of course it may take a bit more e ort to see that a singularity of a function is removable. But if there is a new function which is analytic at w and which agrees with the old function in a neighbourhood of w the w is a removable singularity. To be more formal in our de nitions we can say that if w is an isolated singularity of a function f then f has a Laurent expansion about w and the cases are as follows 1 X f z ak z w k 1 1. If ak 0 for all negative k then f has a removable singularity. 2. If ak 0 for all negative k less than negative n and a 0 then we 6 n say that w is a pole of order n. Thus 1 z has a pole of order 1. Its Laurent expansion has every other coe cient zero 3. If there are in nitely many negative k non zero then we say that w is a pole of in nite order. The singularity is said to be essential. Exercise 5.5.1 Give examples of all types of poles. Exercise 5.5.2 Verify that sin z z 1 has a removable singularity at 0 and remove it i.e. de ne a value for the function at 0 . We can do the same kind of thing with zeros as we have done with poles. If w is a point such that f z is analytic at w then if f w 0 we say that w is a zero of f. It is an isolated zero if there is a neighbourhood of w such that [ Pobierz całość w formacie PDF ] |