Math 220A Fall 2007 Homework #7. Will Garner A

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1 Math A Fall 7 Homework #7 Will Garner

2 Pg 74 #13: Find the power series expansion of about ero and determine its radius of convergence Consider f( ) = and let ak f ( ) = k! k = k be its power series expansion about ero What is the radius of convergence? Show that n+ 1 n+ 1 = a + a1+ + a 1 n Using the fact that f() + (1/) is an even function, show that a k = for k odd and k > 1 The numbers B n = (-1) n 1 a n are called the Bernoulli numbers for n 1 Calculate B, B 4,, B 1 n Let g ( ) = and consider its power series expansion n 3 n e = 1 = = = n! 1!! 3!! 3! ( n+ The radius of convergence is given by: an 1( n+ ( n+ )! R= lim = lim = lim = lim n+ = a 1( n+ )! ( n+ n n n n n+ 1 ak k Let f ( ) = be the power series expansion of f( ) = about ero The k = k! radius of convergence is the farthest distance from (in this case) ero, we can go before we encounter a non-removable singularity Notice that the denominator is ero, hence we have a singularity, whenever e 1 = That is, = pik, k an integer At =, despite the fact that the denominator is undefined, there is no singularity (It is removable, as is evident in the construction of the first power series) Hence, the only non-removable singularities occur at the other places where pik, k So, the singularities closest to are given by pi and -pi The distance from to both of these points is given by p, so we have that R = p 1

3 n+ 1 n+ 1 To show that = a + a1+ + an, we make the following observation: 1 n l ak k 1= 1 = e 1 l= ( l+ k= k! on both sides On the left, we have that it will be equal to On the right, we have that it will be precisely: a + a1+ a + + a ( n+ n!1! ( n! 1! n! n Consider the coefficient of n (n > ) Equating both sides and multiplying by (n +, we have: ( n+ ( n+ ( n+ n+ 1 n+ 1 = a + a1+ + an, ie = a + a1+ + an ( n+ n!1! 1! n! 1 n To find the actual power series expansion of f(), notice that f() = 1/g() 1 n n n f( ) = = = 1 n + + e 1 1 ( n+ 1 ( n+ 1 ( n+ 1+ ( n+ The terms involving m occur only in the first m+1 summands We can compute the coefficients of the power series expansion by collecting coefficients of m in each of the first m+1 summands and discarding the remaining summands This is justified by the uniform convergence of the geometric series for near Collecting coefficients, we have: c = 1, c 1 =, c = +, (1 + ( + ( c 3 = + (3 + (1 + ( + (1 +, c 4 = (4+ (3+ (1 + (+ ( + (+ (1 + (1 + (1+ Simplified, we have: c = 1, c 1 = -1/, c = -1/6 + 1/4 = 1/1, c 3 = -1/4 + 1/6 1/8 =, c 4 = -1/1 + 1/4 + 1/36 1/8 + 1/16 = -1/7 In terms of the a k (where we have to multiply c k by k!), we have: a = 1, a 1 = -1/, a = 1/6, a 3 =, a 4 = -1/3 4

4 Continuing this process out to a 1, we have: a 5 =, a 6 = 1/4, a 7 =, a 8 = -1/3, a 9 =, a 1 = 5/66 Since we are told that f() + (1/) is an even function, that implies that the power series expansion of f() + (1/) consists only of even powers of, ie there are no odd powers of so a k = for k odd and k > 1 So, we have that Bernoulli numbers are given by: B = (-1) 1 1 a = 1/6 B 4 = (-1) 1 a 4 = 1/3 B 6 = (-1) 3 1 a 6 = 1/4 B 8 = (-1) 4 1 a 8 = 1/3 B 1 = (-1) 5 1 a 1 = 5/66 Pg 8 #1: Let f be an entire function and suppose there is a constant M, an R >, and an integer n 1 such that f() M n for > R Show that f is a polynomial of degree n Suppose f is an entire function By Proposition 33, f can be expressed as a power series, which is essentially a polynomial of infinite degree It may be that infinitely many of the coefficients in the expansion are, though So, suppose also that f is a polynomial of degree > n Let g() = f()/ n n α j Doing this division, g ( ) = h ( ) +, for some constants a j j We are told that g() M for all such that > R However, h() is unbounded, as it consists of positive powers of Hence, we have reached a contradiction to the boundedness of g() and thus f must be a polynomial of degree n Intuitively, the n provides a maximum degree to the polynomial f In the case where n is ero, f must be a polynomial of degree (at most), which makes it a constant That is Liouville s Theorem This is just a generaliation of that result j= #3: Find all entire functions f such that f(x) = e x for x in R Let g() = f() e is analytic such that f restricted to R is e x Since f(x) = e x for x in R, we have that { œ C: g() = } has a limit point in R Œ C Letting G = C and applying Theorem 37, we have that g() =, ie f() = e So, we have that the only entire functions of f is e 3

5 #8: Let G be a region and let f and g be analytic functions on G such that f()g() = for all in G Show that either f() = or g() = Let f = { œ G: f() = } and g = { œ G: g() = } There are three cases to consider Case (i): f has countably many eros This necessarily implies that g has uncountably many eros, since f()g() = for all œ G and G consists of uncountably many points Hence, around any point in G, there exists infinitely many eros of g(), ie g has a limit point in G By Theorem 37, we have that g() = Case (ii): g has countably many eros By an identical argument as Case (i), we have that f() = Case (iii): Both f and g have uncountably many eros By the same reasoning as in Case (i), we have that both f() and g() will be equal to Regardless, in all three cases, either f() = or g() =, as desired #9: Let U: C ö R be a harmonic function such that U() for all in C; prove that U is constant U() harmonic implies that there exists V() which is the harmonic conjugate of U() This implies the existence of f() = U() + iv(), where f() is entire We are told that U() for all in C, hence we have that image(f) Œ G, where G = {: Re() } We claim that f is constant, as that will force U() to be constant Consider a Möbius transformation T which maps G to the unit closed unit disk, D We have that this composition, (T é f)() is a bounded, analytic function Hence, by Louiville s Theorem, it must be constant, call it K Mobius transformations are bijective, we have that f() = T -1 (K) We have that this is not equal to infinity, since f is analytic Hence, we have that f is constant, completing the proof 4

6 Additional Problems to Look At and Know Pg #1: Evaluate d, where g(t) = re it, t p, for all possible values of r, γ ( + 4) < r < and < r < For < r <, there is only one singularity at =, and we have that ( + 1)/( + 4) is analytic in a ball of radius r centered at Hence, by Proposition 6, we have that π d = πi = i γ ( + 4) + 4 For < r <, there are three singularities, at =, i and -i Using partial fractions, we have: d = d + d + d γ ( + 4) γ γ + i γ i Each component has only one singularity and the remaining functions 1/4, 3/8, 3/8, respectively are analytic Hence, d = ( πi) + ( πi) + ( πi) = πi γ ( + 4) #14: Find the power series expansion of tan() about =, expressing the coefficients in terms of Bernoulli numbers (Hint: use Exercise 13 and the formula cot() = (1/)cot() (1/)tan()) ak k By Exercise 13, we have that =, where a k = -1/ if k = 1, for k > 1 k = k! and odd and (-1) n 1 B k for k even ak k So, we have that + =, where a k = if k is odd and a k = (-1) k 1 B k k = k! B n n 1 n Hence, we have that + = ( 1) ( n)! e + 1 e + e + = + 1 = = The left-hand side is given by e e πi πi e + e B n n 1 n Now, replacing with pi, we have: πi ( 1) ( πi) πi πi = e e ( n)! = πcot( π) 5

7 B n n n Cleaning up the right-hand side, we have: πcot( π) = ( π) ( n)! B n n n Replacing p with, we have: cot( ) = ( n)! Since cot() = (1/)cot() (1/)tan(), we see that tan() = cot() cot() Multiplying both sides by, we have: tan() = cot() cot() Hence, tan( ) = cot( ) cot( ) B B = ( ) ( n)! ( n)! = n n n n n n n n ( 1) B n n ( n)! Will Garner Notice that the constant term is, thus we can divide through by and get our final n n ( 1) Bn n 1 expression: tan( ) = ( n)! Pg 8 #5: Prove that cos(a + b) = cos(a)cos(b) sin(a)sin(b) by applying Corollary 38 Let f = cos(a + b) and g = cos(a)cos(b) sin(a)sin(b) We have that {: f() = g()} has a limit point in R Œ C Letting G = C and applying Corollary 38, we have that f = g, ie cos(a + b) = cos(a)cos(b) sin(a)sin(b) #6: Let G be a region and suppose that f: G ö C is analytic and a œ G such that f(a) f() for all in G Show that either f(a) = or f is constant Clearly the result holds if f(a) = Now, suppose that f(a) We want to show that f(a) f() for all in G implies that f is constant Since f(a), we have that f(a) >, which implies that f() for all in G Hence, we can consider the reciprocal of f(a) f(), ie 1 f ( ) 1 f( a) Since f(a) is non-ero, the quantity on the right-hand side is finite and since f() is non-ero, we have that 1/f() is analytic in G By the Maximum Modulus Theorem, we have that 1/f() must be constant, ie f() is constant 6

8 #1: Show that if f and g are analytic functions on a region G such that fg is analytic then either f is constant or g = Clearly if f and g are analytic and g =, then fg is analytic So, suppose that g Then there exists an a such that g(a) This implies that g() for all œ B(a, e) Hence, we have that in B(a, e) Thus, we have that both f and f are analytic in B(a, e) f fg = is analytic g The Cauchy-Riemann equations imply that f () = for all œ B(a, e) This follows from the fact that if f() = u(x, y) + iv(x, y), then f ( ) = u( x, y) iv( x, y) We know that they both need to satisfy the Cauchy-Riemann equations since they are analytic The first function gives u x = v y and u y = -v x while the second function gives u x = -v y and u y = v x Adding these two sets of equations together, we have that u x = and u y = Hence, both u x and u y are ero From that, it follows that v x and v y are ero Hence, f () = for all œ B(a, e) By Theorem 37, though, we have that since f () = on an open set, then we have that f () = for all œ G But if f () =, it must be that f is constant, which completes the proof 7