3. Asymptotic Expansions

Question 1

Function in terms of General Truncated Sum

Answer 1

f(ε) = Sn(ε) + Rn(ε)
-where:
Sn = Σ an*ε^n
-sum from n=0 to n=N
-here Sn is a partial sum involving expansion coefficients a0,a1,…,aN and Rn(ε) is a remainder term
-note that this is an exact expression for f(ε) so if Sn starts to diverge then Rn must diverge (in the opposite direction) so that the equality holds

Question 2

Convergent Series

Definition

Answer 2

-convergence asks about the expression:
f(ε) = Sn(ε) + Rn(ε)
-as N->∞ at fixed ε we say that Sn->f if:
lim Rn(ε) = 0 at fixed ε
-limit as N->∞
-then:
f(ε) = Σ an*ε^n
-sum from n=0 to n=∞

Question 3

Factorial vs Power Growth

Answer 3

-factorial growth always beats power growth

Question 4

Asymptotic Series

Definition

Answer 4

-we can also ask what happens if Rn decreases rapidly to 0 as ε->0
-in particular, we demand:
|Rn| &laquo_space;ε^N as ε->0
<=>
lim |Rn|/ε^n -> 0 as ε->0
-limit as ε->0
-since the final term of Sn is aNε^N this means that the remainder will be much smaller than Sn as ε->0
-if this limit applies for all N, then:
f(ε) ~ Σ anε^n as ε->0
-sum from n=0 to n=∞
-we call this the asymptotic expansion of f (as ε-> 0)

Question 5

Asymptotic vs. Convergent Series

Answer 5

-in practice the conditions for asymptotic and convergent series mean very different things for approximating f(ε)

Question 6

Convergent Series

Tolerance

Answer 6

-given a tolerance δ, a convergent series can achieve this tolerance with sufficiently large N, for some range of ε, i.e. |R| < radius of convergence

Question 7

Asymptotic Series

Tolerance

Answer 7

-for tolerance δ, an asymptotic series when truncated to any N can meet the tolerance but only for sufficiently small ε

Question 8

Evaluating Asymptotic Series

Answer 8

• for divergent series, there comes a point where adding more terms to the sequence makes the estimate worse and not better
• and even if a series is convergent, simply summing many terms form the infinite series may note be efficient, especially if convergence is slow

Question 9

Severe Truncation

Answer 9

-just take the first few terms of a series to make an estimate, e.g.
S1 = a0 + a1ε
S2 = a0 + a1ε + a2*ε²
-provided the series is asymptotic, Rn &laquo_space;ε^n as ε->0, so |R1|<

Question 10

Optimal Truncation

Answer 10

• here we seek the specific value of N=N* where the terms of the series are minimised in magnitude, i.e. minimise aN*ε^N
• for a divergent series, as a good working rule N* is found by taking N* to be the point where the terms of the series stop decreasing, i.e. when |aN*ε^N| is minimised
• for a convergent series there may well never be such a point since the terms typically keep decreasing
• this value, N*, depends on the chosen value of ε

Question 11

Pade Approximant

Outline

Answer 11

-a superior technique for accelerating the convergence of a series given only its first few terms
-the idea is to rewrite the asymptotic expression:
f(ε) ~ Σ ak*ε^k
-sum from k=0 to k=∞
-as a rational function in which the numerator and denominator are polynomials of degree M and N respectively

Question 12

Pade Approximant

Equation

Answer 12

-the [M/N] Pade approximant of f(ε):
PM/N = Σnkε^k / Σdkε^k
= [n0+n1ε+n2ε²+…+nMε^M] / [d0+d1ε+d2ε²+…+dNε^N]
-without loss of generality we can set d0=1 with M+N+1 remaining unknown constants

Question 13

Pade Approximant

Determining Coefficients

Answer 13

• the M+N+1 unknown constants are determined from the first M+N+1 coefficients in the asymptotic expansion for f(ε)
• specifically, we require that the Pade approximant is the same as f(ε) ~ Σ ak*ε^k for the first M+N+1 terms (i.e. up to and including O(ε^M+N)

Question 14

The Shanks Transform

Outline

Answer 14

• we consider a series f(ε) ~ Σ akε^k (sum k=0 to k=∞) along with its partial sums SN(ε)=Σ akε^k (sum k=0 to k=N)
• the Shanks transform is a technique for estimating f(ε) given a sequence of three consecutive partial sums
• can be used on a divergent series or to accelerate convergence of a convergent series

Question 15

The Shanks Transform

Equation

Answer 15

-given a sequence of three consecutive partial sums SN-1, SN, SN+1
-the prediction for f(ε) is:
σ(SN) = [SN-1*SN+1 - SN²] / [SN-1 + SN+1 - 2SN]

Question 16

The Shanks Transform

Reapplication

Answer 16

-if the Shanks transform is applied three times to produce σ(SN-1), σ(SN), σ(SN+1), the Shanks transform can then be reapplied on these values:

σ(σ(SN)) = [σ(SN-1)*σ(SN+1) - σ(SN)²] / [σ(SN-1) + σ(SN+1) - 2σ(SN)]

Question 17

The Shanks Transform

Basis

Answer 17

-the basis for the Shanks transform is to suppose that the partial sums are in a geometric progression:
SN-1 = b + cr^(N-1)
SN = b + cr^N
SN-1 = b + r^(N-1)
-we can then eliminate c and r to find b:
b = [SN-1SN+1 - SN²] / [SN-1 + SN+1 - 2SN]

Question 18

The Shanks Transform

ε

Answer 18

-we can also use the Shanks transform with general ε
e.g.
f(ε) = 1 + ε + ε² + …
=>
S0 = 1
S1 = 1+ε
S2 = 1+ε+ε²
-then sub these into the Shanks transform equation

Question 19

General Asymptotic Expansion

Form

Answer 19

f(ε) = Σ an δn(ε) + RN(ε)
-sum from n=0 to n=N
-where {δn(ε)} is an asymptotic sequence of functions with:
δ0(ε) >> δ1(ε) >> δ2(ε) >> ...
-as ε->0
-more precisely:
lim δn+1(ε)/δn(ε) = 0 
-limit as ε->0 for all n≥0

Question 20

General Asymptotic Expansion

Asymptotic Expansion

Answer 20

-if, for all N:
lim RN/δN(ε) = 0
-limit as ε->0
-i.e. RN(ε) << δN(ε)
-we say that f(ε) has an asymptotic expansion, write:
f(ε) ~ Σ an δn(ε) as ε->0

Question 21

o Notation

Answer 21

-if lim f(ε)/g(ε) = 0
-limit as ε->0
-then we say that:
f(ε) = o(g(ε)) as ε->0
-i.e. f(ε) &laquo_space;g(ε)
-can actually be any limit, doesn’t have to be ε->0

Question 22

Os Notation

Answer 22

-if lim f(ε)/g(ε) = C for C∈R{0}
-limit taken as ε->0
-if C is finite and non-zero we can write:
f(ε) = Os(g(ε)) as ε->0
-and we say that f(ε) is strictly of order g(ε) as ε->0
-can actually be any limit, doesn’t have to be ε->0

Question 23

O Notation

Answer 23

-if lim f(ε)/g(ε) = C for C∈R then:
f(ε) = O(g(ε)) as ε->0
-f(ε) is no larger than g(ε)
-can actually be any limit, doesn’t have to be ε->0

Question 24

Expansion Coefficients

Formula for aN

Answer 24

aN = lim [f(ε) - Σan*δn(ε)]/δN(ε)

• for N≥1
• limit as ε->0 and sum from n=0 to n=N-1

Question 25

Expansion Coefficients

Generating δn(ε)

Answer 25

• the formula for aN can also be used to generate δn(ε)
• sub into the formula as normal leaving δN(ε) as a general expression
• test possible function for δN(ε) searching for a function which gives a finite, non-zero limit

Question 26

Uniqueness of Asymptotic Expansions

Answer 26

• for a given function has different asymptotic expansions in terms of different asymptotic sequences {δn(ε)} and {δn^(ε)}
• for a given sequence {δn(ε)}, f(ε) has a unique asymptotic expansion
• for a given sequence {δn(ε)}, different functions may have the same asymptotic expansion

Question 27

Exponentially/Transcendentally Small Terms and Uniqueness of Asymptotic Expansions

Answer 27

-any two functions which differ by an exponentially small term, e.g. e^(-1/ε²), have the same asymptotic expansion in δn(ε)=ε^n

Question 28

Operations on Asymptotic Expansions

Addition / Subtraction / Multiplication / Division

Answer 28

-asymptotic expansions can be added, subtracted multiplied and divided in the obvious way

Question 29

Operations on Asymptotic Expansions

Integration

Answer 29

-asymptotic expansions can also be integrated term by term, although you may need to be careful with terms in logε or inverse powers of ε as ε->0

Question 30

Operations on Asymptotic Expansions

Differentiation

Answer 30

-asymptotic expansions cannot IN GENERAL be differentiated term by term

Question 31

Uniform Asymptotic Expansion

Definition

Answer 31

-consider f(x;ε) where ε«1 with {δn(ε)}:
f(x;ε) ~ Σ an(x) δn(ε), ε->0
-we then need an+1(x)δn+1(ε) &laquo_space;an(x)δn(ε) as ε->0 to have an asymptotic expansion
-if the above holds for all x for which f is defined then the expansion is a uniform asymptotic expansion

Question 32

Non-Uniform Asymptotic Expansion

Definition

Answer 32

ε«1 with {δn(ε)}:
f(x;ε) ~ Σ an(x) δn(ε), ε->0
-we then need an+1(x)δn+1(ε) &laquo_space;an(x)δn(ε) as ε->0 to have an asymptotic expansion
-if the inequality does not hold then the expansion is non-uniform