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Asymptotic sequence and expansions

Lesson 6

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Definition of asymptotic expansion as x → ∞

The expansion

f(x) = a0 + ax1 + xa22 + . . . + xaNN + RN

is an asymptotic expansion if for any

N

RN = O( 1

xN+1) x → ∞

In this case we write

f(x) ∼

X

n=0

an

xn

as

x → ∞ (

not

n → ∞)

(3)

Finding the coefficients

How to find

a0, a1, a2, . . .

? Since

f(x) = a0 + O

1 x

, then

a0 = lim

x→∞

f(x)

From

f(x) = a0 + ax1 + O

1 x2

we get

x(f(x) − a0) = a1 + O

x1

. So

a1 = lim

x→∞

x(f(x) − a0)

(4)

Finding the coefficients

How to find

a0, a1, a2, . . .

? Since

f(x) = a0 + O

1 x

, then

a0 = lim

x→∞

f(x)

From

f(x) = a0 + ax1 + O

1 x2

we get

x(f(x) − a0) = a1 + O

x1

. So

a1 = lim

x→∞

x(f(x) − a0)

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Finding the coefficients

Further, from

f(x) = a0 + ax1 + ax22 + O

x13

a2 = lim

x→∞

x2

f(x) − a0ax1

In general

aN = lim

x→∞

xN

f(x) −

N1

X

n=0

an xn

{1, x1, x12 , x13 , . . .}

is an asymptotic sequence

(6)

Finding the coefficients

Further, from

f(x) = a0 + ax1 + ax22 + O

x13

a2 = lim

x→∞

x2

f(x) − a0ax1

In general

aN = lim

x→∞

xN

f(x) −

N1

X

n=0

an xn

{1, x1, x12 , x13 , . . .}

is an asymptotic sequence

(7)

Finding the coefficients

Further, from

f(x) = a0 + ax1 + ax22 + O

x13

a2 = lim

x→∞

x2

f(x) − a0ax1

In general

aN = lim

x→∞

xN

f(x) −

N1

X

n=0

an xn

{1, x1, x12 , x13 , . . .}

is an asymptotic sequence

(8)

Asymptotic sequences

If we write

ε = x1

for

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

0(ε), δ1(ε), δ2(ε), . . .}

δn+1(ε) = o(δn(ε)) ε → 0

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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, . . .}

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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

ε0

f(ε) − PN1

n=0 anδn(ε) δN(ε)

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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

ε0

f(ε) − PN1

n=0 anδn(ε) δN(ε)

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Positive example

Let us decompose the function

eε

using

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))

,

εlim0

eε − 1 ln(1 + ε)

= a1 + lim

ε0

o(ln(1 + ε)) ln(1 + ε)

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Positive example

eε − 1 − ln(1 + ε) = a2 ln(1 + ε2) + o(ln(1 + ε2))

,

a2 = lim

ε0

eε − 1 − ln(1 + ε) ln(1 + ε2)

= lim

ε0

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))

(14)

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

ε0

cos(ε1/2 + ε) − 1 ln(1 + ε)

= lim 1 − (ε1/2 + ε)2/2 + O(ε2) − 1

= −1

(15)

Negative example

a2 = lim

ε0

cos(ε1/2 + ε) − 1 ln(1 + ε)

= lim

ε0

1 − (ε + 2ε3/2 + ε2)/2 + O(ε2) − 1 + ε/2 − ε2/4 ε2 + O(ε4)

= lim

ε0

− 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.

(16)

Uniqueness of expansions

Any function

f(x)

produce unique expansion.

Let us consider

f1 = sin ε

and

f2 = sin ε + exp(−1/ε2)

. If

sinε + exp(−1/ε2) = a0 + a1ε + a2ε2 + a3ε3 + . . .

then

a0 = 0

,

a1 = lim

ε0

sin ε + exp(−1/ε2) ε

= lim

ε0

sinε ε

+ lim

ε0

+ exp(−1/ε2) ε

= 1 + 0 = 1

(17)

Uniqueness of expansions

Any function

f(x)

produce unique expansion.

Let us consider

f1 = sin ε

and

f2 = sin ε + exp(−1/ε2)

. If

sinε + exp(−1/ε2) = a0 + a1ε + a2ε2 + a3ε3 + . . .

then

a0 = 0

,

a1 = lim

ε0

sin ε + exp(−1/ε2) ε

= lim

ε0

sinε ε

+ exp(−1/ε2)

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Uniqueness of expansions

a2 = lim

ε0

sin ε + exp(−1/ε2) − ε ε2

= 0

a3 = lim

ε0

sin ε + exp(−1/ε2) − ε ε3

= −1 6 exp(−1/ε2) = 0 · 1 + 0 · ε + 0 · ε2 + 0 · ε3 + . . .

(19)

Taylor series is asymptotic expansion

Why the Taylor series at

0

are 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.

xlim0

RN

xN+1 = 1

(N + 1)!f(N+1)(0) = 0

(20)

Taylor series is asymptotic expansion

Why the Taylor series at

0

are 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

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The end

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