Solutions to Tutorial 6 (Week 7)

The University of Sydney School of Mathematics and Statistics Solutions to Tutorial 6 (Week 7)

MATH2962: Real and Complex Analysis (Advanced) Semester 1, 2017 Web Page: http://www.maths.usyd.edu.au/u/UG/IM/MATH2962/ Lecturer: Florica C. Cˆırstea Questions marked with * are more diﬃcult questions.

Material covered (1) power series and radius of convergence. (2) double series and Cauchy products.

Outcomes This tutorial helps you to (1) know and be able to apply convergence tests for sequences and series; (2) have a working knowledge of (real and complex) power series and their properties;

Summary of essential material

N N Let (an) be a sequence in R or C and z ∈ R or C.A power series is a series of the form ∞ X n anz . (PS) n=0

Cauchy-Hadamard Theorem There exists a number % ∈ [0, ∞], called the radius of convergence, such that the power series (PS)

• converges absolutely if |z| < % (makes only sense if % > 0);

• diverges if |z| > % (makes only sense if % < ∞).

The radius of convergence can be found as follows:

(CH1) Derived from root test: 1 (formula is always valid, but the limit superior may % = pn be impossible to compute) lim sup kank n→∞ (CH2) Derived from ratio test: 1 ka k % = = lim n (only valid if the limit exists!!) ka k n→∞ ka k lim n+1 n+1 n→∞ kank (CH3) Find other means of getting convergence for |z| < % and divergence for |z| > %, for instance by substituting certain values of z, or by using known convergence properties of series, for instance, the geometric series.

• Formula (CH1) does not work well if the coeﬃcients an of the series contain factorials, n! √ for instance a = because we cannot easily handle n n!. It is better to use (CH2) in n nn such cases because we can cancel many factors before computing the limit.

pn • If we use (CH1), then the n in kank must match the power of z. ∞ X Example. To ﬁnd the radius of convergence for the series 4nz2n, we can proceed in √ n=0 √ two ways. To use (CH1), we need to look at 2n 4n = 2, and not n 4n = 4. Indeed, we P∞ k write the series as k=0 akz , where the coeﬃcients ak for k ∈ N are given by ( 4n if k = 2n for some n ∈ N, ak = 0 if k = 2n + 1 for some n ∈ N.

pk Here, the sequence ( |ak|)k≥1 has two convergent subsequences: The one with odd indices 2n+1p being identically zero (i.e., limn→∞ |a2n+1√| = 0), and the one with even indices having 2pn 2n n a positive limit: limn→∞ |a2n| = limn→∞ 4 = 2. Then

pk lim sup |ak| = max{0, 2} = 2 k→∞ (the largest between the two accumulation points). Hence, the radius of convergence for 1 1 the given series is ρ = = . Alternatively, we can denote z2 = y and pk lim supk→∞ |ak| 2 P∞ n n ﬁnd that the radius of convergence for n=0 4 y is 1/4 (using either (CH1) or (CH2)), that is the series converges absolutely for |y| < 1/4 and diverges for |y| > 1/4. This shows P∞ n 2n that n=0 4 z converges absolutely for |z| < 1/2 and diverges for |z| > 1/2 (use that 2 P∞ n 2n |z| = |y|). Thus the series n=0 4 z has radius of convergence 1/2.

Double series theorem N N Suppose that xjk ∈ R or C such that

m n XX sup kxjkk < ∞, m,n∈ N j=0 k=0

and that σ : N → N × N is a bijection. Then the series

∞ ∞ ∞ ∞ ∞ XX XX X xjk , xjk , xσ(i) j=0 k=0 k=0 j=0 i=0

converge absolutely, and they all have the same limit.

Cauchy products ∞ ∞ X X Suppose that aj and bk are absolutely convergent series in R or C. Then, j=0 k=0

∞ ∞ ∞ n X X X X aj bk = cn, where cn := akbn−k. j=0 k=0 n=0 k=0

2 Questions to complete during the tutorial

P∞ n 1. It is given that the power series n=0 anz converges for z = 1 + i and diverges for z = 3 + 4i. Find bounds for its radius of convergence %.

P∞ n Solution: By the Cauchy–Hadamard theorem a power series n=0 anz converges for |z| < %, and diverges for |z| > %. Since, by assumption, the power series converges for z = 1 + i√, it follows that 1 + i is either√ inside the ball B(0,%) or on its boundary. As |1 + i| = 2, we conclude that % ≥ 2. We are told that the power series diverges for z = 3 + 4i. This means that 3 + 4i cannot belong to B(0,%). Hence, % ≤ |3 + 4i| = 5. Consequently, we have √ 2 ≤ % ≤ 5.

2. Determine the radius of convergence % of the following power series. ∞ X zn (a) , p > 0; np n=1 p Solution: In the given series an = 1/n for n ≥ 1. We have that

|a | n p α = lim n+1 = lim = 1. n→∞ |an| n→∞ n + 1 Hence, by formula (CH2) the radius of convergence for the given series is 1/1 = 1. ∞ X √ (b) n(1 + i 3)nzn; n=0 Solution: We use formula (CH1) to compute %. We have q √ √ √ √ √ √ n |n(1 + i 3)n| = n n|1 + i 3| = n n 1 + 3 = 2 n n → 2

as n → ∞. Hence from (CH1) we get 1 % = . 2 If the coeﬃcients in the series are sign changing or complex as in this example, remember to take absolute values! ∞ X n3 (c) zn; (1 + i)n n=0 Solution: We use formula (CH1) to compute %. We have s √ √ n3 n n)3 n n)3 1 n = = √ → √ (1 + i)n |1 + i| 2 2

as n → ∞. Hence from (CH1) we get 1 √ % = √ = 2. 1/ 2 If the coeﬃcients in the series are sign changing or complex as in this example, remember to take absolute values!

3 ∞ X nn (d) zn; n! n=1 nn Solution: We denote a = for n ≥ 1. Because of the n! in a we compute n n! n the radius of convergence using formula (CH2). We have

n+1 n n |an+1| (n + 1) n! n + 1 1 lim = lim · n = lim = lim 1 + = e. n→∞ |an| n→∞ (n + 1)! n n→∞ n n→∞ n Hence by formula (CH2), the radius of convergence is % = 1/e. ∞ X (e) (log n)zn; n=2 Solution: For n ≥ 2 we have √ pn 1 ≤ log n ≤ n n. √ √ As n n → 1 as n → ∞, the squeeze law implies that n log n → 1 as n → ∞. Hence formula (CH1) gives % = 1/1 = 1. ∞ X (f) n1/nzn; n=1 Solution: We compute the radius of convergence for the power series using formula . For n ≥ 1 we have 1 ≤ n1/n ≤ n. Hence, for all n ≥ 1 √ √ 1 ≤ n n1/n ≤ n n √ √ Because n n → 1 as n → ∞, the squeeze law implies that n n1/n → 1 as n → ∞. Hence, by formula (CH1) the radius of convergence is % = 1/1 = 1. ∞ X n2 (g) z3n; 2n n=0 k Solution: This is a power series where the coeﬃcient ak of z is given by n2  n if k = 3n and n ∈ N, ak := 2 0 otherwise.

We ﬁnd the radius of convergence using formula (CH1). We cannot directly use (CH2) because we would need to divide by zero when taking the ratio of successive terms. pk When k ∈ N is not a multiple of 3, then |ak| = 0. If k ∈ N is a multiple of 3, that is k = 3n, then r √ n2 ( n n)2/3 1 pk |a | = 3pn |a | = 3n = → k 3n 2n 21/3 21/3

pk 1 as k = 3n → ∞. Hence, lim sup |ak| = 1/3 . Then, by formula (CH1), the k→∞ 2 1 radius of convergence is % = = 21/3. pk lim supk→∞ |ak|

4 An alternative approach is to consider the series as a power series in the new variable y = z3. Then, ∞ ∞ X n2 X n2 z3n = yn. 2n 2n n=0 n=0 For the latter series we can ﬁnd the radius of convergence using (CH1) or (CH2). Using either formula we get a radius of 2, that is, the series for y converges if |y| < 2 and diverges if |y| > 2. Since |z|3 = |y|, we see that the original series converges when |z| < 21/3 and diverges when |z| > 21/3. Again we see that the ∞ X n2 radius of convergence for z3n is 21/3. 2n n=0 z z2 z3 z4 z5 *(h)1 − + − + − + ··· . 2 32 23 34 25 ∞ X 1 Solution: This is a power series a zn, where a is given by if n ∈ is n n 3n N n=0 1 even, and − if n ∈ is odd. It follows that 2n N r 1 1  n = if n ≥ 0 is even  n pn |a | = 3 3 n r 1 1  n = if n ≥ 1 is odd.  2n 2

pn Hence, lim supn→∞ |an| = 1/2 since the limit superior is the largest accumulation point. Hence, by formula (CH1) the radius of convergence is 1 % = = 2. 1/2

3. Find the radius of convergence of the power series

∞ X 3n zn n n=1 3n in 2. Here the coeﬃcient a = is a vector. R n n Solution: We use (CH2) to compute the radius of convergence: r ka k p(n + 1)2 + 32(n+1) (n + 1)23−2n + 32 n+1 √ = = 2 −2n → 3 kank n2 + 32n n 3 + 1 as n → ∞. Here we used the elementary limit that nkan → 0 if |a| < 1 and k ≥ 0 is ﬁxed. Hence the radius of convergence is % = 1/3.

Extra questions for further practice 4. Find power series expansions in z and their radius of convergence for the following maps. Hint: Use the formula for the geometric series

∞ 1 1 1 1 Xz k = = , (GS) a − z a 1 − z/a a a k=0

5 which converges if and only if |z/a| < 1, that is, |z| < |a|. Also make use of Cauchy products. 1 (a) ; 3 − z Solution: By formula (GS)

∞ ∞ 1 1 Xz k X zk = = 3 − z 3 3 3k+1 k=0 k=0 if and only if |z| < 3. The radius of convergence is % = 3. 1 (b) ; (3 − z)2 ∞ 1 X 1 Solution: Since the series = a zk with a = is absolutely 3 − z k k 3k+1 k=0 convergent for |z| < 3, we can write 1/(3 − z)2 as a Cauchy product. So, for |z| < 3 we have

∞ ! ∞ ! ∞ 1 X X X = a zj a zk = c zn, (3 − z)2 j k n j=0 k=0 n=0

where cn is given by n n n X X 1 1 X 1 n + 1 c = a a = = = . n k n−k 3k+1 3n−k+1 3n+2 3n+2 k=0 k=0 k=0 Thus, for |z| < 3 we ﬁnd that

∞ ∞ 1 X X n + 1 = c zn = zn. (3 − z)2 n 3n+2 n=0 n=0

|cn+1| 1 Since limn→∞ = , it follows that the radius of convergence for the latter |cn| 3 series is 3. 1 *(c) ; z2 − 5z + 6 Solution: We ﬁrst factorize the denominator and then use partial fractions: 1 1 1 1 = = − . z2 − 5z + 6 (z − 2)(z − 3) 2 − z 3 − z Then by applying (GS) with a = 2 and a = 3, we obtain that for |z| < 2

∞ ∞ ∞ 1 1 1 X zk X zk X 1 1 = − = − = − zk. z2 − 5z + 6 2 − z 3 − z 2k+1 3k+1 2k+1 3k+1 k=0 k=0 k=0

P∞ k k+1 Since k=0 z /2 diverges for |z| > 2 the radius of convergence is % = 2. z + 1 *(d) . z − 1 Solution: Observe that z + 1 (z − 1) + 2 2 2 = = 1 + = 1 − . z − 1 z − 1 z − 1 1 − z 6 Hence, by using (GS), we ﬁnd that

∞ z + 1 X = 1 − 2 zk = −1 − 2z − 2z2 − 2z3 − ... z − 1 k=0 if and only if |z| < 1. Hence, the radius of convergence for (z + 1)/(z − 1) is % = 1.

n 5. (a) Let n ∈ , n 6= 0. Show that pk(n − k) ≤ for all k = 0, . . . , n. N 2 Solution: The points k(n − k), k = 0, . . . , n, lie on the parabola x(n − x). The parabola has x-intercepts x = 0 and x = n, and attains a maximum at x = n/2. Hence, r rn n n2 n pk(n − k) ≤ n − = = 2 2 2 2 for every k = 0, . . . , n as claimed. ∞ X k+1 1 (b) Consider the series ak with ak = (−1) √ . We deﬁne a0 = 0. The series k=1 k converges due to the Leibniz test. The Cauchy product of the series with itself is ∞ X given by cn, where n=0 n X cn := akan−k. k=0

Use part (a) to show that cn 6→ 0, and hence the Cauchy product diverges.

Solution: Since a0 = 0, we have c0 = c1 = 0. By deﬁnition of cn, we ﬁnd that for n ≥ 2,

n−1 n−1 n−1 X X k+1 n−k+1 1 1 n X 1 cn = akan−k = (−1) (−1) √ √ = (−1) . n − k p k=1 k=1 k k=1 k(n − k) Using part (a) we conclude that

n−1 n−1 X 1 X 2 2(n − 1) 2 |c | = ≥ = = 2 − . n p n n n k=1 k(n − k) k=1 P∞ for all n ≥ 2, so cn 6→ 0 as n → ∞. Hence, the Cauchy product n=0 cn diverges.

*6. Try to construct an array xjk with j, k ∈ N, such that the row and the column series ∞ X converge absolutely, but xσ(i) diverges for some bijection σ : N → N × N. i=0 Solution: Consider the array

2 −1 0 0 0 0 0 ... −1 2 −1 0 0 0 0 ... 0 −1 2 −1 0 0 0 ... 0 0 −1 2 −1 0 0 ... 0 0 0 −1 2 −1 0 ......

7 Then ∞ ∞ ∞ ∞ X X X X xjk = xjk = 1 + 0 + 0 + ··· = 1, j=0 k=0 k=0 j=0 showing that the row and column series converge absolutely. Now we sum a diﬀerent way over the array. We go along the diagonals with indices j + k = n with n ≥ 0. This gives the series

2 − 1 − 1 + 2 − 1 − 1 + 2 − 1 − 1 + 2 − 1 − 1 + ....

It is evident that the latter series diverges because the terms do not converge to zero.

Challenge questions (optional)

*7. Let B ∈ RN×N be a matrix. Deﬁne kBk be as in Tutorial 1, Question 9. (a) Show that the series ∞ X Bk k=0 converges in RN×N if s(B) := lim sup pk kBkk < 1, k→∞ and in particular if kBk < 1. (The series is called the Neumann series for the matrix B.) ∞ X Solution: By the root test, the series Bk converges if s(B) < 1. Since k=0 kBkk ≤ kBkk for all k ∈ N, we have s(B) ≤ kBk. Hence, if kBk < 1, then the ∞ X series Bk converges. k=0 (b) If the Neumann series converges, show that I − B is invertible, and that

∞ X (I − B)−1 = Bk. k=0 (Note the similarity to the geometric series.) Pn k Solution: Denote by Sn := k=0 B the n-th partial sum and set

∞ X k S := lim Sn = B . n→∞ k=0 We do a proof similar to the proof for the geometric series, where B is just a real or complex number. Then

n n n X k X k X k+1 n+1 (I − B)Sn = (I − B) B = B − B = I − B k=0 k=0 k=0 and similarly

n n n X k X k X k+1 n+1 Sn(I − B) = B (I − B) = B − B = I − B . k=0 k=0 k=0

8 We conclude that n+1 (I − B)Sn = Sn(I − B) = I − B (1) P∞ k n+1 for all n ∈ N. Because k=0 B converges we know that kB k → 0. Hence the right hand side converges to the identity matrix I as n → ∞. We also have that

k(I − B)(Sn − S)k ≤ kI − BkkSn − Sk → 0

k(Sn − S)(I − B)k ≤ kSn − SkkI − Bk → 0

as n → ∞. We therefore conclude from (1) that

(I − B)S = lim (I − B)Sn = lim Sn(I − B) = Sn(I − B) = I, n→∞ n→∞ or equivalently ∞ ∞ X X Bk (I − B) = (I − B) Bk = I. k=0 k=0 P∞ k −1 This ﬁnally implies that k=0 B = (I − B) . (c) Show that |λ| ≤ s(B) for all eigenvalues λ of B, where s(B) is as in (a). Give an example where equality holds. Solution: If λ 6= 0 we write λI − B = λ(I − λ−1B). Hence λI − B is invertible if and only if I − λ−1B is invertible. By part (a) the latter is invertible if

s(λ−1B) = lim sup kλ−kBkk1/k = |λ|−1 lim sup kBkk1/k = |λ|−1 s(B) < 1. k→∞ k→∞ Hence λI − B is invertible if |λ| > s(B), which means that if λ is an eigenvalue then |λ| ≤ s(B). Note that equality is possible. For example, if B = I then 1 is an eigenvalue. Also √ √ 1/n kIk = N, so s(I) = limn→∞ N = 1. For a general matrix one can even show that there is always an eigenvalue λ such that |λ| = s(B), so s(B) is the maximal modulus of the eigenvalues of B. (d) Show that the above series may converge, no matter how large kBk is. 0 a Solution: Consider the matrix B = . Then kBk = |a| can be as large 0 0 as we like, but since B2 = 0 we have s(B) = 0 for all a ∈ K. Hence there is no connection between the magnitude of kBk and the convergence of the series under consideration in general.