Power Series

Lecture 23

Corollary 23.1.

If (23.1) diverges at some point z = z₁, then it diverges at all points zthat satisfy the inequality|z−z₀|>|z₁−z₀|.

Corollary 23.2.

LetRbe the radius of convergence of (23.1). For each 0< R₁< R,the series converges uniformly on the closed diskB(z₀, R₁).

Example 23.1.

From Example 21.3 and Theorem 23.1, it follows that the geometric series

∞ j=0

z^j = 1 +z+z²+· · ·+z^j+· · ·, (23.2) which is in fact a power series with z0 = 0, aj = 1, converges absolutely and uniformly on |z| ≤r <1 to the analytic function 1/(1−z).Since at z= 1 the series (23.2) diverges, from Corollary 23.1 we find that it diverges on |z|>1. Thus, the radius of convergence of (23.2) is R = 1. If|z| = 1, then the terms of (23.2) do not tend to zero, and hence it also diverges on

|z|= 1.

From Theorem 23.1 and Corollary 23.1, it is clear that the radius of convergenceRof (23.1) is the least upper bound of the distances|z−z₀|from the pointz₀ to the pointzat which the series (23.1) converges. However, for the computation ofR,one of the following methods is usually employed:

d’Alembert’s Ratio Test. R= (lim_j→∞|a_j+1/a_j|)⁻¹,provided the limit exists.

Cauchy’s Root Test. R=

limj→∞|aj|^1/j₋1

,provided the limit exists.

Cauchy-Hadamard Formula. R =

lim sup_j→∞|aj|^1/j₋1

; this limit always exists.

Example 23.2.

Since in (23.2), aj = 1, the ratio test confirms its radius of convergenceR= 1.Similarly, for the series$_∞

j=0z^j+2/(j+ 1),we have|a_j+1/aj|=|(j−1)/j| →1 asj → ∞.Thus, for this series also, the radius of convergenceR = 1. This series diverges at z = 1 and converges at z = −1. For the series $_∞

j=0(−1)^j(z−2−3i)^j+1/(j+ 1)!, we have

|a_j+1/a_j| = 1/(j+ 2) → 0 as j → ∞. Thus, the radius of convergence R=∞.

Example 23.3.

For the series $_∞

j=0(j+ 1)^j(z+i)^j,we have|aj|^1/j = (j+ 1)→ ∞asj→ ∞,and hence by the root test its radius of convergence R= 0; i.e., it converges only atz=−i.For the series$_∞

j=0(z−3−2i)^j/(j+ 1)^j,the root test givesR=∞, whereas for the series$_∞

j=0[(j+ 1)/(2j+ 3)]^j(z−2−i)^j, R= 2.

Example 23.4.

In the series$_∞

j=0z^2j/2^j, a2m= 1/2^manda2m+1= 0.

Thus, |a_j|^1/j = 0 ifj is odd and|a_j|^1/j = (1/2^m)^1/2m = 1/√

2 ifj = 2m.

Power Series 153 Hence, lim sup_j→∞|a_j|^1/j = 1/√

2. Therefore, by the Cauchy-Hadamard formula,R=√

2.

We shall now prove the following theorem.

Theorem 23.2.

LetR be the radius of convergence of (23.1).

(I). s(z) =$_∞

j=0aj(z−z0)^j is an analytic function onB(z0, R).

(II). (Term-by-Term Integration). Ifγ is a contour inB(z0, R) andg(z) is a continuous function onγ,then

γ

g(z)s(z)dz =

γ

g(z) ∞ j=0

aj(z−z0)^jdz = ∞ j=0

aj

γ

g(z)(z−z0)^jdz.

(23.3) In particular, if g(z)≡1,then

γ

∞ j=0

aj(z−z0)^jdz = ∞ j=0

aj

γ

(z−z0)^jdz. (23.4)

(III). (Term-by-Term Differentiation).

s(z) = d dz

∞ j=0

aj(z−z₀)^j = ∞ j=1

jaj(z−z₀)^j−1. (23.5)

The radius of convergence of the series$_∞

j=1ja_j(z−z₀)^j−1 is alsoR.

Proof.

(I). Letz1∈B(z0, R).Choose anyrsuch that|z1−z0|< r < R.

Then,sn(z) =$n

j=0aj(z−z0)^j→s(z) uniformly onB(z0, r) by Theorem 23.1. Now, since each term of the seriesaj(z−z₀)^j is entire, the function s(z) is analytic onB(z₀, r) by Corollary 22.3. Hence, in particular,s(z) is analytic atz₁.

(II). Let 0< r < Rbe such that{γ} ⊂B(z0, r). Then, by Theorem 23.1, sn(z)→s(z) uniformly onB(z₀, r).Since{γ}is a closed subset ofB(z₀, r), from Lemma 22.1 it follows that g(z)sn(z)→ g(z)s(z) uniformly on {γ}. Now, from Theorem 22.3, we have

nlim→∞

γ

g(z)sn(z)dz =

γ

nlim→∞g(z)sn(z)dz, and hence

∞ j=0

aj

γ

g(z)(z−z0)^jdz =

γ

g(z)s(z)dz.

(III). Letz ∈B(z₀, R), and let γ_r ={ξ :|ξ−z|=r} ⊂ B(z₀, R). Then, from part (II) withg(ξ) = 1/[2πi(ξ−z)²], it follows that

1 2πi

γ_r

s(ξ)

(ξ−z)²dξ = ∞ j=0

aj

1 2πi

γ_r

(ξ−z₀)^j (ξ−z)²dξ.

Now, from (18.3), we have s(z) =

∞ j=0

aj

d

dz(z−z₀)^j = ∞ j=1

jaj(z−z₀)^j−1. Finally, since

jlim→∞

(j+ 1)aj+1

jaj

= lim

j→∞

j+ 1 j

lim

j→∞

aj+1

aj

= lim

j→∞

aj+1

aj

, the radius of convergence of the series$_∞

j=1ja_j(z−z₀)^j−1 is alsoR.

Corollary 23.3.

LetRbe the radius of convergence of (23.1). Then, s(z) =$_∞

j=0a_j(z−z₀)^j is infinitely differentiable for all z∈B(z₀, R). In fact, for anyk,

s^(k)(z) = ∞ j=k

j(j−1)· · ·(j−k+ 1)a_j(z−z₀)^j−k. (23.6)

Remark 23.1.

From (23.6), it immediately follows thatak=s^(k)(z0)/k!, k= 0,1,· · ·.

Example 23.5.

From Example 23.1, we have$_∞

j=0z^j= 1/(1−z), z∈ B(0,1). Applying Theorem 23.2 (III) to differentiate the series term-by- term, it follows that

1 + 2z+ 3z²+· · · = ∞ j=0

(j+ 1)z^j = 1 (1−z)².

Example 23.6.

We shall show that onB(0,1), Log (1 +z) =

∞ j=0

(−1)^jz^j+1

j+ 1 = z−z² 2 +z³

3 +· · ·. (23.7) For this, lets(z) be the right-hand side of (23.7). Since limj→∞|a_j+1/aj|= limj→∞|j/(j+1)|= 1,the radius of convergence ofs(z) is 1.Using Theorem 23.2 (III), we find

s(z) = 1−z+z²−z³− · · · = (1 +z)⁻¹ = d

dzLog (1 +z) (23.8)

Power Series 155 on B(0,1); i.e., [s(z)−Log (1 +z)] = 0. The relation (23.7) now follows from Theorem 7.2 and the fact thats(0)−Log 1 = 0. Clearly, (23.7) also follows if we begin with (23.8) and use Theorem 23.2 (II) to integrate the series term-by-term along any path connecting 0 andz∈B(0,1).

We conclude this lecture by stating the following theorem.

Theorem 23.3.

Assume that the power seriesf(z) = ∞ j=0

aj(z−z0)^j and g(z) =

∞ j=0

bj(z −z0)^j have the radius of convergence R1 and R2, respectively. Then,

(i). f(z)±g(z) = ∞ j=0

(a_j±b_j)(z−z₀)^j andf(z)g(z) = ∞ j=0

c_j(z−z₀)^j,

where cj = j k=0

akbj−k = j k=0

aj−kbk,have the radius of convergenceR = min{R1, R2},and

(ii). ifg(z)= 0 in the diskB(z0, r) (necessarilyb0= 0), then the quotient f(z)/g(z) =

∞ j=0

cj(z−z0)^j,where the coefficients satisfy the equationaj= j

k=0

bj−kck,has the radius of convergenceR= min{r, R₁, R₂}.

Problems

23.1. Use the ratio test to compute the radius of convergence of the following series:

(a).

∞ j=0

1

(1 + 3i)^j+1(z−7i)^j, (b).

∞ j=0

(j!)²

(2j)!(z−3−2i)^j.

23.2. Use the root test to compute the radius of convergence of the following series:

(a).

∞ j=0

11j+ 9 2j+ 5

j

(z−2−5i)^j, (b).

∞ j=0

4j²

2j+ 1 − 6j² 3j+ 4

(z−3i)^j. 23.3. Use the Cauchy-Hadamard formula to compute the radius of convergence of the series$_∞

j=0a_jz^j,where

(a). a_j = _j

k=0

1 k!

^j

, (b). a_j = (8−(−3)^j)^j.

23.4. Suppose that (23.1) has radius of convergence R. Show that

$_∞

j=0a²_j(z−z0)^j has radius of convergenceR². 23.5. Show that

lim

n→∞

n!

nⁿ 1/n

= 1 e

and use it to find the radius of convergence of the series 1 +z+2²

2!z²+· · ·+j^j

j!z^j+· · ·. 23.6. TheBessel function of ordernis defined by

J_n(z) = ∞ j=0

(−1)^j j!(j+n)!

z 2

2j+n

.

(a). Show thatJ_n(z) is entire.

(b). Verify the identity [zⁿJn(z)]=zⁿJn−1(z).

23.7. Consider the function f(z) = 1 +

∞ j=1

a(a−1)(a−2)· · ·(a−j+ 1)

j! z^j.

(a). Show that its radius of convergence is 1.

(b). For all |z| < 1, f(z) = af(z)/(1 +z). Hence, deduce thebinomial expansion (1 +z)^a=f(z).

23.8. Use power seriesf(z) =$_∞

j=0ajz^j to solve thefunctional equa- tionf(z) =z+f(z²).

23.9. If the sum of two power series in a neighborhood of the point of expansionz₀ is the same, then show that the identical powers of (z−z₀) have identical coefficients; i.e., there is a unique power series that has a given sum in a neighborhood ofz₀.

23.10.For the power seriess(z) =$_∞

j=0a_jz^j,let the radius of conver- gence beR.Show that

(a). the coefficients of the odd powers of z vanish if s(z) is even; i.e., if s(−z) =s(z),and

Power Series 157 (b). the coefficients of the even powers of z vanish if s(z) is odd; i.e., if s(−z) =−s(z).

23.11.For the power seriess(z) =$_∞

j=0ajz^j,let the radius of conver- gence beR.

(a). Show that s(z) is a power series inz with the same radius of conver- genceR.

(b). Suppose R > 1 and|s(e^iθ)| ≤4 for 0 ≤ θ ≤π and |s(e^iθ)| ≤ 9 for π < θ≤2π.Show that|s(0)| ≤6.

23.12.Letf(z) =$_∞

j=0a_jz^j= 1 +z+ 2z²+ 3z³+ 5z⁴+ 8z⁵+ 13z⁶+ 21z⁷+· · ·,where the coefficientsaj are theFibonacci numbers defined by a0 = 1, a1 = 1, aj =aj−1+aj−2, j ≥2.Show that f(z) = 1/(1−z− z²), z ∈B(0, R),whereR= (√

5−1)/2.

23.13.Let f(z) = $_∞

j=0a_jz^j, where the coefficientsa_j are the Lucas numbers defined bya0 = 1, a1 = 3, aj =aj−1+aj−2, j ≥2. Show that f(z) = (1 + 2z)/(1−z−z²), z∈B(0, R),where R= (√

5−1)/2.

23.14 (Weierstrass Double Series Theorem). Suppose that for eachk = 0,1,2,· · · the power seriesf_k(z) =$_∞

j=0a^(k)_j (z−z₀)^j converges in the disk B(z₀, R), R ≤ ∞; i.e., each power series defines the function fk(z) in the disk B(z0, R). Suppose that F(z) = $_∞

k=0fk(z) converges uniformly for |z−z0| ≤ρfor everyρ < R.Show that F(z) is analytic on

|z−z0| ≤ ρand has a power series expansion,F(z) =$_∞

j=0Aj(z−z0)^j, that converges for all |z−z₀| < R; here, Aj = $_∞

k=0a^(k)_j . Furthermore, show that

(a).

∞ j=1

z^j

1 +z^2j converges uniformly for|z| ≤r <1,and (b).

∞ j=1

z^j 1 +z^2j =

∞ k=0

(−1)^k z^2k+1

1−z^2k+1 for|z|<1.

Answers or Hints

23.1. (a).√

10,(b). 4.

23.2. (a). 2/11,(b). 3/5.

23.3. (a). 1/e,(b). 1/11.

23.4.lim sup_j→∞|a²_j|^1/j =

lim sup_j→∞|aj|^1/j2

. 23.5.eⁿ = 1 +_1!ⁿ +· · ·+_(nⁿⁿ⁻¹_−1)!+ⁿ_n!ⁿ

1 +_n+1ⁿ +_(n+1)(n+2)ⁿ² +· · · implies

that ⁿ_n!ⁿ < eⁿ< nⁿ_n!ⁿ+ⁿ_n!ⁿ

1 +_n+1ⁿ + n

n+1

2

+· · ·

= (2n+ 1)ⁿ_n!ⁿ.Hence,

1

eⁿ <_n^n!_n < ²ⁿ⁺¹_en . R= 1/e.

23.6. (a).aj =

0 ifj is odd

(−1)^j/2/[2^j+n(j/2)! (j/2 +n)!] ifj is even. We will show that lim sup_j→∞|aj|^1/j = 0.Clearly,|aj|^1/j = 0 ifj is odd, and forj even we have|a_j|^1/j ≤1/[(j/2)!]^2/j.Now use [(k)!]^1/k → ∞ask→ ∞. (b). [zⁿJn(z)]=$_∞

j=0(−1)^j2ⁿ(j+n) j!(j+n)!

_z

2

2j+2n−1

=zⁿ$_∞

j=0 (−1)^j j!(j+n−1)!

×_z

2

2j+n−1

.

23.7. (a). limj→∞|a_j+1/aj|= limj→∞|(a−j)/(j+ 1)|= 1.(b). Compute (1 +z)f(z) directly and show that it is the same asaf(z).

23.8. f(z) =a₀+$_∞

j=0z^2j, |z|<1.

23.9. If$_∞

j=0aj(z−z0)^jand$_∞

j=0bj(z−z0)^j have the same sums(z) in a neighborhood ofz₀,then from Corollary 23.3 and Remark 23.1 it follows that aj=bj=s^(j)(z0)/j!, j= 0,1,· · ·.

23.10.(a).s^(2k+1)(0) = 0, k= 0,1,· · ·,(b).s^(2k)(0) = 0, k= 0,1,· · ·. 23.11.(a).s(z) =$_∞

j=0ajz^j,sos(z) =$_∞

j=0ajz^j,and since lim|aj|^1/j = lim|a_j|^1/j,they have the same radius of convergence. (b). Boths(z) and s(z) converge absolutely and uniformly on B(0, R), R > 1, so they are analytic on B(0,1). Hence, g(z) = s(z)s(z) is also analytic on B(0,1).

Since|s(e^iθ)| ≤4 for 0≤θ≤πand|s(e^iθ)| ≤9 forπ < θ≤2π, |g(e^iθ)|=

|s(e^iθ)s(e^−iθ)|=|s(e^iθ)||s(e^i(2π−θ))| ≤4×9 or 9×4 according to whether 0≤θ≤πorπ < θ≤2π.Thus,|g(e^iθ)| ≤36 for all 0≤θ≤2π.Now, by the Maximum Modulus Principle, |g(0)| ≤ |g(e^iθ)|for some θ,so |g(0)| ≤ 36;

i.e.,|s(0)|²≤36,and hence|s(0)| ≤6.

23.12. 1 +zf(z) +z²f(z) = 1 +$_∞

j=0ajz^j+1+$_∞

j=0ajz^j+2 = 1 +z+

$_∞

j=2(aj−1+aj−2)z^j = 1+z+$_∞

j=2ajz^j.Hence, 1+zf(z)+z²f(z) =f(z).

23.13.Show thatf(z) = 1 + 2z+zf(z) +z²f(z).

23.14.Use results of Lecture 22. For (a) and (b), expand each function in the power series.

Lecture 24