Asymptotic sequence and expansions
Lesson 6
Definition of asymptotic expansion as x → ∞
The expansion
f(x) = a0 + ax1 + xa22 + . . . + xaNN + RNis an asymptotic expansion if for any
NRN = O( 1
xN+1) x → ∞
In this case we write
f(x) ∼
∞
X
n=0
an
xn
as
x → ∞ (not
n → ∞)Finding the coefficients
•
How to find
a0, a1, a2, . . .? Since
f(x) = a0 + O1 x
, then
a0 = limx→∞
f(x)
•
From
f(x) = a0 + ax1 + O1 x2
we get
x(f(x) − a0) = a1 + O
x1
. So
a1 = lim
x→∞
x(f(x) − a0)
Finding the coefficients
•
How to find
a0, a1, a2, . . .? Since
f(x) = a0 + O1 x
, then
a0 = limx→∞
f(x)
•
From
f(x) = a0 + ax1 + O1 x2
we get
x(f(x) − a0) = a1 + O
x1
. So
a1 = lim
x→∞
x(f(x) − a0)
Finding the coefficients
•
Further, from
f(x) = a0 + ax1 + ax22 + Ox13
a2 = lim
x→∞
x2
f(x) − a0 − ax1
•
In general
aN = lim
x→∞
xN
f(x) −
N−1
X
n=0
an xn
• {1, x1, x12 , x13 , . . .}
is an asymptotic sequence
Finding the coefficients
•
Further, from
f(x) = a0 + ax1 + ax22 + Ox13
a2 = lim
x→∞
x2
f(x) − a0 − ax1
•
In general
aN = lim
x→∞
xN
f(x) −
N−1
X
n=0
an xn
• {1, x1, x12 , x13 , . . .}
is an asymptotic sequence
Finding the coefficients
•
Further, from
f(x) = a0 + ax1 + ax22 + Ox13
a2 = lim
x→∞
x2
f(x) − a0 − ax1
•
In general
aN = lim
x→∞
xN
f(x) −
N−1
X
n=0
an xn
• {1, x1, x12 , x13 , . . .}
is an asymptotic sequence
Asymptotic sequences
If we write
ε = x1for
x ∼ ∞, then
{1, ε, ε1, ε2, ε3, . . .}
{1, ε1/2, ε1, ε3/2, ε2, . . .}
en general if
δn(ε)is a term of asymptotic sequence then
{δ0(ε), δ1(ε), δ2(ε), . . .}
δn+1(ε) = o(δn(ε)) ε → 0
Examples of asymptotic sequences
{1, sin ε, sin2 ε, sin3 ε, . . . , sinn ε, . . .}
{1, ln(1 + ε), ln(1 + ε2), . . . ln(1 + εn), . . .}
{tan(ε1/2), tan(ε), tan(ε3/2), . . . , tan(εn/2), . . .}
{ln ε ε , 1
ε, ln ε, 1, ε ln ε, ε, ε2 ln ε, ε2, . . .}
Expansion in terms of δ
n(ε)
f(ε) ∼ P∞
n=0 anδn(ε)
that means
f(ε) =
N
X
n=0
anδn(ε) + RN, RN = O(δn+1(ε)) ε → 0 aN = lim
ε→0
f(ε) − PN−1
n=0 anδn(ε) δN(ε)
Expansion in terms of δ
n(ε)
f(ε) ∼ P∞
n=0 anδn(ε)
that means
f(ε) =
N
X
n=0
anδn(ε) + RN, RN = O(δn+1(ε)) ε → 0
aN = lim
ε→0
f(ε) − PN−1
n=0 anδn(ε) δN(ε)
Positive example
Let us decompose the function
eεusing
δn(ε) = ln(1 + εn)
, e. g.
eε ∼ a0 + a1 ln(1 + ε) + a2 ln(1 + ε2) + . . . a0 = 1
eε − 1 = a1 ln(1 + ε) + o(ln(1 + ε))
,
εlim→0
eε − 1 ln(1 + ε)
= a1 + lim
ε→0
o(ln(1 + ε)) ln(1 + ε)
Positive example
eε − 1 − ln(1 + ε) = a2 ln(1 + ε2) + o(ln(1 + ε2))
,
a2 = lim
ε→0
eε − 1 − ln(1 + ε) ln(1 + ε2)
= lim
ε→0
1 + ε + ε2/2 + O(ε3) − 1 − ε + ε2/2 ε2 + O(ε3)
= 1
eε = 1 + ln(1 + ε) + ln(1 + ε2) − 1
6 ln(1 + ε3) + o(ln(1 + ε3))
Negative example
Let us attempt to expand
cos(ε1/2 + ε)using the same asymptotic equation
δn(ε) = ln(1 + εn)cos(ε1/2 + ε) ∼ a0 + a1 ln(1 + ε) + a2 ln(1 + ε2) + . . . a0 = cos(ε1/2 + ε) = 1
a1 = lim
ε→0
cos(ε1/2 + ε) − 1 ln(1 + ε)
= lim 1 − (ε1/2 + ε)2/2 + O(ε2) − 1
= −1
Negative example
a2 = lim
ε→0
cos(ε1/2 + ε) − 1 ln(1 + ε)
= lim
ε→0
1 − (ε + 2ε3/2 + ε2)/2 + O(ε2) − 1 + ε/2 − ε2/4 ε2 + O(ε4)
= lim
ε→0
− ε3/2 + O(ε2) ε2 + O(ε4)
= −∞
What is the solution of the problem? Add the term
ln(1 + ε3/2)
and its subsequent powers.
Uniqueness of expansions
•
Any function
f(x)produce unique expansion.
•
Let us consider
f1 = sin εand
f2 = sin ε + exp(−1/ε2)
. If
sinε + exp(−1/ε2) = a0 + a1ε + a2ε2 + a3ε3 + . . .
then
a0 = 0,
a1 = lim
ε→0
sin ε + exp(−1/ε2) ε
= lim
ε→0
sinε ε
+ lim
ε→0
+ exp(−1/ε2) ε
= 1 + 0 = 1
Uniqueness of expansions
•
Any function
f(x)produce unique expansion.
•
Let us consider
f1 = sin εand
f2 = sin ε + exp(−1/ε2)
. If
sinε + exp(−1/ε2) = a0 + a1ε + a2ε2 + a3ε3 + . . .
then
a0 = 0,
a1 = lim
ε→0
sin ε + exp(−1/ε2) ε
= lim
ε→0
sinε ε
+ exp(−1/ε2)
Uniqueness of expansions
a2 = lim
ε→0
sin ε + exp(−1/ε2) − ε ε2
= 0
a3 = lim
ε→0
sin ε + exp(−1/ε2) − ε ε3
= −1 6 exp(−1/ε2) = 0 · 1 + 0 · ε + 0 · ε2 + 0 · ε3 + . . .
Taylor series is asymptotic expansion
•
Why the Taylor series at
0are asymptotic expansions?
• f(x) = PN
n=0 anxn + RN
with
RN = xN+1
(N + 1)!f(N+1)(z).
So
RN = O(xN+1) x → 0.
xlim→0
RN
xN+1 = 1
(N + 1)!f(N+1)(0) = 0
Taylor series is asymptotic expansion
•
Why the Taylor series at
0are asymptotic expansions?
• f(x) = PN
n=0 anxn + RN
with
RN = xN+1
(N + 1)!f(N+1)(z).
So
RN = O(xN+1) x → 0.
R 1