Chapter 6: Power Series

Question 1

CONSTRUCTING

Answer 1

condition that the terms of the series are non-negative real numbers can be relaxed.

Question 2

THM 6.1

increasing sequence of reals is….

Answer 2

An increasing sequence of real numbers is either
(i) bounded above and tends to a finite limit
or
(ii) not bounded above and tends to infinity.

Question 3

THM 6.2 COMPARISON TEST

Answer 3

Suppose that aₙ ≥ 0, bₙ ≥ 0 and that
0 ≤ bₙ ≤ aₙ for all n ≥ 1.

If Σn=1 to ∞ of 
aₙ 
converges, then 
Σn=1 to ∞ of 
bₙ
also converges.

Question 4

DEF 6.3: definition of convergence of a summation

Answer 4

Suppose that zₙ ∈ C for all n and that S_N = z1 +z2 +· · · +z_N . Then we
say that
Σn=1 to ∞ of zₙ
converges if there is a complex number L such that S_N → L as N → ∞
i.e. there is a complex number L such that |S_N − L| → 0 as N → ∞.
If no such complex number exists, then we say that the series diverges.

easy to show that S_N → L as N → ∞ if and only if Re (SN ) → Re L and
Im (S_N ) → Im L as N → ∞.

Question 5

THM 6.4: ‘‘divergence test”

Answer 5

If Σn=1 to ∞ of zₙ converges, then zₙ → 0 as n → ∞.

doesnt mean if tends to 0 converges NOT SUFFICIENT
——guarantees
If zₙ doesnt tend to 0 as n → ∞, then Σn=1 to ∞ of zₙ diverges.

Question 6

HARMONIC SERIES

Answer 6

Σn=1 to ∞ of 1/n
= 1 + (1/2) + (1/3) + (1/4) +…
is a DIVERGENT series for which zₙ tends to 0 as n tends to infinity

(divergence test passed)

Question 7

DEF 6.5: ABSOLUTE CONVERGENCE

Answer 7

A series Σn=1 to ∞ of zₙ of complex numbers is said to be absolutely convergent
if Σn=1 to ∞ of |zn| converges.

Question 8

THM 6.6: absolute convergence relationship with z_n convergence

Answer 8

An absolutely convergent series of real or complex numbers is also convergent.

For all complex numbers zn, |zn| is a non-negative real number and so any
result for series of non-negative real numbers is applicable to the series Σn=1 to ∞ of |zₙ|

Question 9

DEF 6.7: POWER SERIES

Answer 9

Suppose that aₙ ∈ C for all n and z₀ ∈ C. A series of the form
Σn=0 to ∞ of aₙ(z − z₀)ⁿ
is called a power series (centred on z₀).

note sum starts at 0!

z₀ is also commonly 0

Question 10

Σn=0 to ∞ of n!zⁿ

here, 0! ≡ 1

Answer 10

-converges only at 0

Question 11

Σn=0 to ∞ of zⁿ

Answer 11

• converges when |z| less thn 1

Question 12

Σn=0 to ∞ of zⁿ/n!

Answer 12

absolutely converges for all z in complex plane

Question 13

THM 6.8

absolute convergence

Answer 13

Suppose w≠ 0 and Σaₙwⁿ
is convergent. Then Σaₙzⁿ
is absolutely convergent for all z such that |z| < |w|.

So if a power series converges at some point, it is absolutely convergent at any point
nearer the origin

proof:
if Σaₙwⁿ is convergent then aₙwⁿ → 0 as n →∞.
ie there is some M st |aₙwⁿ| ≤ M for all n. Take z with |z| LESS THAN |w|
then
|aₙzⁿ | = |aₙwⁿ | |z/w|ⁿ ≤ M |z/w|ⁿ

M |z/w|ⁿ is a convergent geometric progression since |z/w| is less than 1 and by COMPARISON TEST Σ|aₙzⁿ| is convergent ie Σaₙzⁿ is absolutely convergent

bₙ→ 0 implies that |bₙ|≤ M for some M strictly bigger than 0 and all n. As bₙ→ 0 we can find some N∈ N such that |bn| ≤ 1 for all n ≥ N. Take M = max{ |b_1|,…, |b_N-1|, 1}.

Question 14

THM 6.9 Abel

radius of convergence existence

Answer 14

For the power series Σaₙzⁿ

one of the following is true:

1. the power series converges only at z = 0 (e.g., Σn!zⁿ)
2. the power series is absolutely convergent for all z ∈ C (e.g., Σzⁿ/n!)
3. we can find a real number R with 0 < R < ∞ such that the power series is absolutely
   convergent if |z| < R and divergent if |z| > R.

R = sup{|z| :Σ|aₙzⁿ| converges.}

Question 15

DEF 6.10 : Radius of Convergence, Disc of Convergence

Answer 15

The quantity R
in theorem 6.9 (case (3)) is called the radius of convergence of the power series.
case (1) R = 0
(2) R = ∞. DISC OF CONVERGENCE is C
(3) the disc D = {z ∈ C : |z| < R} is the DISC OF CONVERGENCE of Σaₙzⁿ

Question 16

THM 6.11: radius of convergence finding

Answer 16

If | aₙ / aₙ₊₁|→ R as n → ∞, then R is the radius of convergence of Σaₙzⁿ

Question 17

Corollary 6.12 infinite radius of convergence

Answer 17

if |a_n / a_n+1| tends to infinity as n tends to infinity then the power series Σa_n z^n has infinite radius of convergence
ie converges absolutely for all z

Question 18

radius of convergence for any power series?

Answer 18

A power series has a radius of convergence irrespective of whether |a_n / a_n+1 | tends to a limit as n to infinity

Question 19

FIND THE RADIUS OF CONVERGENCE FOR POWER SERIES:

Σ from 0 to ∞ of (sinh n) z^n

Answer 19

Σ from 0 to ∞ of (sinh n) z^n

aₙ =sinh n

= 0.5(e^n - e^-n) / 0.5(e^n+1 - e^n+1)

=(1- e^-2n) / (e -e^-{2n-1}) tends to 1-0/ e-0 = 1/e as n tends to infinity

power series has radius of convergence R=1/e

aₙ / aₙ₊₁| =| (sinh n) / (sinh n+1)| = (sinh n) / (sinh n+1) =

Question 20

FIND THE RADIUS OF CONVERGENCE FOR POWER SERIES:

Σ from 1 to ∞ of (2^n z^n)/(n^2)

Answer 20

Σ from 1 to ∞ of (2^n z^n)/(n^2)

aₙ = 2^n / n^2

p.s has r.o.c R=0.5 by theorem 6.11

aₙ / aₙ₊₁| = 0.5( 1 + 1/n)^2 tends to 0.5 as n to infinity

Question 21

FIND THE RADIUS OF CONVERGENCE FOR POWER SERIES:

Σ from 1 to ∞ of (2^n z^4n)/(n^2)

Answer 21

Σ from 1 to ∞ of (2^n z^4n)/(n^2)

we cant directly apply thm 6.11 as coeff of z^n is 0 unless n is a multiple of 4.

let w=z^4

p.s= sum of a_n w^n

which has radius 0f convergence 0.5, thus p.s is absolutely convergent if |w| LESS than 0.5 and divergent if |w| bigger than 0.5

Absolutely convergent if |z^4| less than 0.5 and divergent if |z^4| larger than 0.5

hence absolutely convergent if |z| less than 2^-1/4 and divergent if |z| bigger than 2^-1/4
R= 2^-1/4

Question 22

FIND THE RADIUS OF CONVERGENCE FOR POWER SERIES:

Σ from 1 to ∞ of ((3n)!n!)/(4n)! * z^n

Answer 22

Σ from 1 to ∞ of ((3n)!n!)/(4n)! * z^n
aₙ =((3n)!n!)/(4n)!

each term dividing by 4n as 1/4n tends to 0 as n to infinity

aₙ / aₙ₊₁| tends to 4^4 /3^3 = R

Question 23

THM 6.13
Σ from 0 to ∞ of aₙ z^n

Σ from 1 to ∞ of naₙ z^{n-1}

Answer 23

The power series
Σ from 0 to ∞ of aₙ z^n
and

Σ from 1 to ∞ of naₙ z^{n-1}

both have the same radius of convergence

Question 24

THM 6.14

power series diffferentiated term by term inside the disc of convergence

Answer 24

Suppose that p.s has radius of cinvergence R. Define f by

f(z) =
Σ from 0 to ∞ of aₙ z^n (|z| less than R).

Then the function is differentiable on the disc of convergence D = {z∈C : |z| less thn R}

and

f’(z) =

Σ from 1 to ∞ of naₙ z^{n-1} (|z| less thn R)

Question 25

DISC OF CONVERGENCE

Answer 25

D is open set on which f is differentiable, analytic on D.
eg infinite R then sum funct f is analytic on C=

the sum funct of a power series is analytic on the disc of convergence

handled similarly to polynomials in the disc

Question 26

power series for exp

Answer 26

Σ from 0 to ∞ of z^n /n!

used to derive trig

Question 27

power series for sinz

Answer 27

sinz=

Σ from k=0 to ∞ of (-1)^k z^{2k+1} / (2k+1)!

valid for C

Question 28

power series for cosz

Answer 28

cosz=

Σ from k=0 to ∞ of (-1)^k z^2k / (2k)!

as i^n + (-i)^n =0 for odd n and (-1)n/2 x2 if even

valid for C

Question 29

SWITCHING ORDERS :

summation and differentiation

Answer 29

if each function
g_n (n = 1, 2, . . .) is analytic on a region D, and if for each z ∈ D the sum \
Σ_n g_n(z) is convergent, then
Σ_n g_n is analytic on D and (Σ_n g_n) ‘ = Σ_n g’_n
on D.

Question 30

SWITCHING ORDERS :

summation and integration

Answer 30

You can switch orders of summation and integration:

Σ(∫_γ g_n) = ∫_γ ( Σ_n g_n)

Question 31

SWITCHING ORDERS :

differentiation and integration

Answer 31

A3. You can switch the order of differentiation and integration:

d/dz( ∫_γ g(z,w) .dw ) = ∫_γ ∂/∂z g (z/w) .dw