Chapter 6: Power Series Flashcards
CONSTRUCTING
condition that the terms of the series are non-negative real numbers can be relaxed.
THM 6.1
increasing sequence of reals is….
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.
THM 6.2 COMPARISON TEST
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.
DEF 6.3: definition of convergence of a summation
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 → ∞.
THM 6.4: ‘‘divergence test”
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.
HARMONIC SERIES
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)
DEF 6.5: ABSOLUTE CONVERGENCE
A series Σn=1 to ∞ of zₙ of complex numbers is said to be absolutely convergent
if Σn=1 to ∞ of |zn| converges.
THM 6.6: absolute convergence relationship with z_n convergence
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|
DEF 6.7: POWER SERIES
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
Σn=0 to ∞ of n!zⁿ
here, 0! ≡ 1
-converges only at 0
Σn=0 to ∞ of zⁿ
- converges when |z| less thn 1
Σn=0 to ∞ of zⁿ/n!
absolutely converges for all z in complex plane
THM 6.8
absolute convergence
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}.
THM 6.9 Abel
radius of convergence existence
For the power series Σaₙzⁿ
one of the following is true:
- the power series converges only at z = 0 (e.g., Σn!zⁿ)
- the power series is absolutely convergent for all z ∈ C (e.g., Σzⁿ/n!)
- 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.}
DEF 6.10 : Radius of Convergence, Disc of Convergence
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
THM 6.11: radius of convergence finding
If | aₙ / aₙ₊₁|→ R as n → ∞, then R is the radius of convergence of Σaₙzⁿ
Corollary 6.12 infinite radius of convergence
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
radius of convergence for any power series?
A power series has a radius of convergence irrespective of whether |a_n / a_n+1 | tends to a limit as n to infinity
FIND THE RADIUS OF CONVERGENCE FOR POWER SERIES:
Σ from 0 to ∞ of (sinh n) z^n
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) =
FIND THE RADIUS OF CONVERGENCE FOR POWER SERIES:
Σ from 1 to ∞ of (2^n z^n)/(n^2)
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
FIND THE RADIUS OF CONVERGENCE FOR POWER SERIES:
Σ from 1 to ∞ of (2^n z^4n)/(n^2)
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
FIND THE RADIUS OF CONVERGENCE FOR POWER SERIES:
Σ from 1 to ∞ of ((3n)!n!)/(4n)! * z^n
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
THM 6.13
Σ from 0 to ∞ of aₙ z^n
Σ from 1 to ∞ of naₙ z^{n-1}
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
THM 6.14
power series diffferentiated term by term inside the disc of convergence
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)
