Chapter 1, Exercise 22

Transcription

1 Chapter, Exercise 22 Let N = {,2,3,...} denote the set of positive integers. A subset S N is said to be in arithmetic progression if S = {a,a+d,a+2d,a+3d,...} where a,d N. Here d is called the step of S. Show that N cannot be partitioned into a finite number of subsets that are in arithmetic progressionwith distinct steps (except for the trivial case a = d = ). [Hint: write n N zn as a sum of terms of the type z a z d.] SupposethatwecanpartitionNintodisjointarithmeticprogressionsS,...,S n, where S j has smallest element a j and step d j. Then we have for all z <. We also have j= j S j z j = z j = z z k= = z aj z aj+kdj k= = zaj z dj z kdj Since the S j form a partition of N, we get, for z <, that z n z = z aj z dj. j= Without loss of generality, assume d is the largest of {d,...,d n }, Because the d j are distinct, it follows that d j < d for j. Consider the root of unity ζ = e /d. For this root of unity we have z d =, but z dj for any j d j. Thus the right hand side of the above equation is unbounded as z ζ from inside the unit disk. But ζ so the left hand side remains bounded as z ζ from inside the unit disk. Therefore, the left hand side is not equal to the right hand side and no such partition exists.

2 Chapter, Exercise 25 Part a Evaluate the integrals z n γ for all integers n. Here γ is any circle centered at the origin with the positive (counterclockwise) orientation. We can parameterize γ via the parameterization z(t) = e t, where t goes from to. Under this parameterization, the integral becomes: Which is This integral can be written as e nt e t dt. e (n+)t.dt cos(2π(n+)t)+isin(2π(n+)t)dt Both trigonometric functions will run through an integer number of periods, and therefore the integral will vanish unless n+ =, in which case the integrand is and the integral evaluates to. So for integers n the integral is zero; for n = the integral is. Part b Same question as before, but with γ any circle not enclosing the origin. Let γ be any such circle. We can parameterize γ with the parameterization γ(t) = z +re t where z > r. In the region enclosed by γ, the function z n has the explicit primitive n+ zn+ for n. Therefore, the integral along γ is zero by Corollary 3.3. If n =, then we have, after parametrizing, (z +re t ) e t. This is z (+r/z e t ) e t. 2

3 Since r/z <, this has a power series expansion z ( r/z ) j e (j+)t j= and by absolute convergence of the sum, we can interchange the sum and integral. Because j+ is a positive integer for any nonnegative integer j, each term in the sum integrates to and the integral is. Part c Show that if a < r < b, then dz = (z a)(z b) a b, γ where γ denotes the counterclockwise circle centered at the origin, of radius r, with the positive orientation. For this problem, it is best to perform a partial fraction decomposition on (z a)(z b). We rewrite (z a)(z b) = ( a b z a ). z b We then evaluate a b γ z a dz a b γ z b dz. The second integral vanishes, as we computed in part b. For the first integral, we parameterize γ by γ = re θ : r a b (re t a) e t dt Now, r > a, so we can pull out re t to get which gives r a b r a b r re t( a r e t ) e t dt a j r j e t(j+) dt j= As before this integral vanishes unless j =, so we get r r(a b) = a b. 3

4 Chapter 2, Problem Here are some examples of analytic functions on the unit disc that cannot be extended analytically past the unit circle. The following definition is needed. Let f be a function defined in the unit disc D, with boundary circle C. A point w on C is said to be regular for f if there is an open neighborhood U of w and an analytic function g on U, such that f = g on D U. A function f defined on D cannot be continued analytically past the unit circle if no point of C is regular for f. Part a Let f(z) = z 2n for z <. n= Noticethattheradiusofconvergenceoftheaboveseriesis. Showthatf cannot be continued analytically past the unit disc. [Hint: Suppose θ = 2πp/2 k, where p and k are positive integers. Let z = re iθ ; then f(re iθ ) as r.] Let θ = 2πp 2 k, and let z = re iθ. Then we have k f(z) = r 2n e 2n p 2 k + n= r 2n. Where we used the fact that the exponent was an integer multiple of for n k. There are only k summands in the first sum, so the first sum is bounded by k in absolute value for r <. As r from the left, the sum is therefore bounded from below by n r2n k, which is unbounded as r. Thus f(z) cannot be analytically extended to any ball containing a point of the form e p/2k. Since points of this form aredense in the unit circle, f(z) is not regular at any point of the unit circle. Part b n=k Fix < α <. Show that the analytic function f defined by f(z) = 2 nα z 2n for z < n= can be continuously extended to the unit circle, but cannot be analytically continued past the unit circle. [Hint: There is a nowhere differentiable function lurking in the background. See Chapter 4 in Book I.] 4

5 The function f(z) = n= 2 nα z 2n converges absolutely and uniformly in the disc z < ; by Abel s theorem, the function f can be extended continuously to the unit circle and is given by f(z) = n= 2 nα z 2n on the unit circle z. For < α <, The function f(z), restricted to the unit circle z = e θ, is given by n= cos(2n 2πθ)+isin(2 n 2πθ). This is a nowhere differentiable Weierstrass function. Thus the function f(z) is nowhere differentiable on the unit circle and therefore cannot be analytically extended past the circle. For non-integer α >, we can define the operator T by Tf(z) = zf (z). Letting k < α < k +, we take the function T k f, which an be expanded for z < as a power series: Tf(z) = 2 n(α k) z 2n. n= By the previous argument, this cannot be analytically continued past the unit circle. Since Tf(z) is analytic wherever f(z) is analytic, it follows that f(z) cannot be extended past the unit circle either. For integers α = k, and for z <, T k f is given by n= which cannot be analytically extended past the unit circle by part a. z 2n 5