how small is small? - russ hermanrussherman.com/talks/latex_ex/seriestalk.pdf · 2013-09-11 ·...

How Small is Small?Asymptotic Series in Physics

Dr. Russell Herman

Department of Mathematics & Statistics, UNCW

April 11, 2007

−0.5 −0.4 −0.3 −0.2 −0.1 0 0.1 0.2 0.3 0.4 0.50.6

2Comparison of 1/(1−x) 1+x+x2 and 1+x+x2+x3

Dr. Herman (UNCW) How Small is Small? April 11, 2007 1 / 37

Abstract

How often have you used the small angle approximation sin θ ≈ θ or the binomial

expansion (1 + x)p ≈ 1 + px? How accurate is the formula T = 2π√

Lg giving the

period of a pendulum for angles that are not small? How fast must you have tomove for relativistic effects to be important? Do series expansions have toconverge to be useful? In this lecture we will explore some of these questions aswe investigate the role of series expansions in our approximations. In particular,we describe the typical use of series expansions in undergraduate physics, providesome examples and if there is time we may see how divergent series are oftenuseful.

Outline of Talk

1 Geometric Series

2 Binomial SeriesSpecial Relativity Example

3 Taylor Series

4 ApplicationsSmall AnglesThe Nonlinear Pendulum

5 Asymptotic Series

6 Homework - The World on a String

Geometric Series

DefinitionA geometric series is of the form

∞∑n=0

arn = a + ar + ar2 + ar3 + · · ·+ arn + . . . . (1)

Here a is the first term and r is called the ratio.

Examples∞∑

3n= 1 +

33+ · · ·

∞∑n=2

3(2n) = 3(22) + 3(23) + 3(24) + · · ·

∞∑n=0

(−1)n

2n= 1− 1

4− 1

8+ · · ·

Geometric Series

DefinitionA geometric series is of the form

∞∑n=0

arn = a + ar + ar2 + ar3 + · · ·+ arn + . . . . (1)

Here a is the first term and r is called the ratio.

Examples∞∑

3n= 1 +

33+ · · ·

∞∑n=2

3(2n) = 3(22) + 3(23) + 3(24) + · · ·

∞∑n=0

(−1)n

2n= 1− 1

4− 1

8+ · · ·

nth Partial Sum

Sum of Geometric Progression

Consider the nth partial sum:

sn = a + ar + · · ·+ arn−2 + arn−1. (2)

Now, multiply this equation by r .

rsn = ar + ar2 + · · ·+ arn−1 + arn. (3)

Subtracting(1− r)sn = a− arn. (4)

Thus, the nth partial sum is

sn =a(1− rn)

1− r. (5)

nth Partial Sum

sn = a + ar + · · ·+ arn−2 + arn−1. (2)

rsn = ar + ar2 + · · ·+ arn−1 + arn. (3)

sn =a(1− rn)

1− r. (5)

nth Partial Sum

sn = a + ar + · · ·+ arn−2 + arn−1. (2)

rsn = ar + ar2 + · · ·+ arn−1 + arn. (3)

sn =a(1− rn)

1− r. (5)

nth Partial Sum

sn = a + ar + · · ·+ arn−2 + arn−1. (2)

rsn = ar + ar2 + · · ·+ arn−1 + arn. (3)

sn =a(1− rn)

1− r. (5)

nth Partial Sum

sn = a + ar + · · ·+ arn−2 + arn−1. (2)

rsn = ar + ar2 + · · ·+ arn−1 + arn. (3)

sn =a(1− rn)

1− r. (5)

Sum of a Geometric Series

Limit of Partial SumsRecall that the sum, if it exists, is given by

S = limn→∞

sn = limn→∞

a(1− rn)

1− r.

Geometric Series ResultThe sum of the geometric series is

∞∑n=0

arn =a

1− r, |r | < 1. (6)

Sum of a Geometric Series

Limit of Partial SumsRecall that the sum, if it exists, is given by

S = limn→∞

sn = limn→∞

a(1− rn)

1− r.

Geometric Series ResultThe sum of the geometric series is

∞∑n=0

arn =a

1− r, |r | < 1. (6)

Examples

Example 1.∑∞

n=012n

In this case we have that a = 1 and r = 12 . Therefore, this infinite series converges

and the sum is

1− 12

Example 2.∑∞

k=243k

In this example we note that the first term occurs for k = 2. So, a = 49 . Also,

r = 13 . So,

1− 13

Examples

Example 1.∑∞

n=012n

and the sum is

1− 12

Example 2.∑∞

k=243k

r = 13 . So,

1− 13

Examples

Example 1.∑∞

n=012n

and the sum is

1− 12

Example 2.∑∞

k=243k

r = 13 . So,

1− 13

Examples

Example 1.∑∞

n=012n

and the sum is

1− 12

Example 2.∑∞

k=243k

r = 13 . So,

1− 13

Binomial Series

Definition

A binomial series is the power series expansion of (1 + x)p when p is not anonnegative integer and |x | < 1.

The First Terms

(1 + x)p = 1 + px +p(p − 1)

2!x2 +

p(p − 1)(p − 2)

3!x3 + . . . . (7)

More General Expressions

Consider the approximation (r << R) :

1√r2 + R2

= (r2 + R2)−1/2 =1

)2)−1/2

(1− r2

Binomial Series

Definition

The First Terms

(1 + x)p = 1 + px +p(p − 1)

2!x2 +

p(p − 1)(p − 2)

3!x3 + . . . . (7)

1√r2 + R2

= (r2 + R2)−1/2 =1

)2)−1/2

(1− r2

Binomial Series

Definition

The First Terms

(1 + x)p = 1 + px +p(p − 1)

2!x2 +

p(p − 1)(p − 2)

3!x3 + . . . . (7)

1√r2 + R2

= (r2 + R2)−1/2 =1

)2)−1/2

(1− r2

Binomial Series

Definition

The First Terms

(1 + x)p = 1 + px +p(p − 1)

2!x2 +

p(p − 1)(p − 2)

3!x3 + . . . . (7)

1√r2 + R2

= (r2 + R2)−1/2

)2)−1/2

(1− r2

Binomial Series

Definition

The First Terms

(1 + x)p = 1 + px +p(p − 1)

2!x2 +

p(p − 1)(p − 2)

3!x3 + . . . . (7)

1√r2 + R2

= (r2 + R2)−1/2 =1

)2)−1/2

(1− r2

Binomial Series

Definition

The First Terms

(1 + x)p = 1 + px +p(p − 1)

2!x2 +

p(p − 1)(p − 2)

3!x3 + . . . . (7)

1√r2 + R2

= (r2 + R2)−1/2 =1

)2)−1/2

(1− r2

Nonnegative Integer Powers

Looking for Patterns ...

Consider (a + b)p for nonnegative integer p’s:

(a + b)0 = 1

(a + b)1 = a + b

(a + b)2 = a2 + 2ab + b2

(a + b)3 = a3 + 3a2b + 3ab2 + b3

(a + b)4 = a4 + 4a3b + 6a2b2 + 4ab3 + b4

· · ·

Do you see any patterns?

(a + b)0 = 1

(a + b)1 = a + b

(a + b)2 = a2 + 2ab + b2

(a + b)3 = a3 + 3a2b + 3ab2 + b3

(a + b)4 = a4 + 4a3b + 6a2b2 + 4ab3 + b4

· · ·

(a + b)0 = 1

(a + b)1 = a + b

(a + b)2 = a2 + 2ab + b2

(a + b)3 = a3 + 3a2b + 3ab2 + b3

(a + b)4 = a4 + 4a3b + 6a2b2 + 4ab3 + b4

· · ·

(a + b)0 = 1

(a + b)1 = a + b

(a + b)2 = a2 + 2ab + b2

(a + b)3 = a3 + 3a2b + 3ab2 + b3

(a + b)4 = a4 + 4a3b + 6a2b2 + 4ab3 + b4

· · ·

(a + b)0 = 1

(a + b)1 = a + b

(a + b)2 = a2 + 2ab + b2

(a + b)3 = a3 + 3a2b + 3ab2 + b3

(a + b)4 = a4 + 4a3b + 6a2b2 + 4ab3 + b4

· · ·

(a + b)0 = 1

(a + b)1 = a + b

(a + b)2 = a2 + 2ab + b2

(a + b)3 = a3 + 3a2b + 3ab2 + b3

(a + b)4 = a4 + 4a3b + 6a2b2 + 4ab3 + b4

· · ·

(a + b)0 = 1

(a + b)1 = a + b

(a + b)2 = a2 + 2ab + b2

(a + b)3 = a3 + 3a2b + 3ab2 + b3

(a + b)4 = a4 + 4a3b + 6a2b2 + 4ab3 + b4

· · ·

General Patterns

Observations

Each term consists of a product of a power of a and a power of b.

The powers of a are decreasing from n to 0

The powers of b increase from 0 to n.

The sums of the exponents in each term is n.

ResultSo, we can write the k + 1st term in the expansion as

C rk−1a

n−kbk .

Example - 6th term in (a + b)51

a51−5b5 = a46b5. What is the numerical coefficient?

General Patterns

ObservationsEach term consists of a product of a power of a and a power of b.

C rk−1a

n−kbk .

General Patterns

C rk−1a

n−kbk .

General Patterns

C rk−1a

n−kbk .

General Patterns

C rk−1a

n−kbk .

General Patterns

C rk−1a

n−kbk .

General Patterns

C rk−1a

n−kbk .

General Patterns

C rk−1a

n−kbk .

Patterns in the Coefficients - Pascal’s Triangle

n = 0 : 1n = 1 : 1 1n = 2 : 1 2 1n = 3 : 1 3 3 1n = 4 : 1 4 6 4 1

Observations

Each row begins and ends with a one.

The second and next to last terms have coefficient = n.

Add consecutive pairs in each row to obtain next row.

n = 2 : 1 2 1↘ ↙ ↘ ↙

n = 3 : 1 3 3 1(9)

n = 0 : 1n = 1 : 1 1n = 2 : 1 2 1n = 3 : 1 3 3 1n = 4 : 1 4 6 4 1

ObservationsEach row begins and ends with a one.

n = 2 : 1 2 1↘ ↙ ↘ ↙

n = 3 : 1 3 3 1(9)

n = 0 : 1n = 1 : 1 1n = 2 : 1 2 1n = 3 : 1 3 3 1n = 4 : 1 4 6 4 1

n = 2 : 1 2 1↘ ↙ ↘ ↙

n = 3 : 1 3 3 1(9)

n = 0 : 1n = 1 : 1 1n = 2 : 1 2 1n = 3 : 1 3 3 1n = 4 : 1 4 6 4 1

n = 2 : 1 2 1↘ ↙ ↘ ↙

n = 3 : 1 3 3 1(9)

n = 0 : 1n = 1 : 1 1n = 2 : 1 2 1n = 3 : 1 3 3 1n = 4 : 1 4 6 4 1

n = 2 : 1 2 1↘ ↙ ↘ ↙

n = 3 : 1 3 3 1(9)

Coefficient Patterns

n = 2 : 1 2 1↘ ↙ ↘ ↙

n = 3 : 1 3 3 1(10)

We can generate the next several rows of our triangle.

n = 3 : 1 3 3 1n = 4 : 1 4 6 4 1n = 5 : 1 5 10 10 5 1n = 6 : 1 6 15 20 15 6 1

The kth term in the expansion of (a + b)n.

Let r = k − 1. Then this term is of the form C nr an−rbr , where C n

r = C n−1r + C n−1

r−1 .

Coefficient Patterns

n = 2 : 1 2 1↘ ↙ ↘ ↙

n = 3 : 1 3 3 1(10)

We can generate the next several rows of our triangle.

n = 3 : 1 3 3 1n = 4 : 1 4 6 4 1n = 5 : 1 5 10 10 5 1n = 6 : 1 6 15 20 15 6 1

The kth term in the expansion of (a + b)n.

Let r = k − 1. Then this term is of the form C nr an−rbr , where C n

r = C n−1r + C n−1

r−1 .

Combinatorial Coefficients

Actually, the coefficients have been found to take a simple form.

C nr =

(n − r)!r !=

For example, the r = 2 case for n = 4 involves the six products: aabb, abab,abba, baab, baba, and bbaa.

The Binomial Series for Nonnegative Integer Powers

So, we have found that

(a + b)n =n∑

)an−rbr . (12)

Combinatorial Coefficients

Actually, the coefficients have been found to take a simple form.

C nr =

(n − r)!r !=

For example, the r = 2 case for n = 4 involves the six products: aabb, abab,abba, baab, baba, and bbaa.

The Binomial Series for Nonnegative Integer Powers

So, we have found that

(a + b)n =n∑

)an−rbr . (12)

Binomial Expansion - Other Powers

Note: Sums may be infinite.

(1− x)−1 =1

1− x= 1 + x + x2 + . . . .

(1 + x)p =∞∑r=0

)x r for p real. (13)

Is There a Problem?

Consider the coefficient for r = 1 in an expansion of (1 + x)−1.(−11

(−1)!

(−1− 1)!1!=

(−1)!

(−2)!1!.

where (−1)! = (−1)(−2)(−3) · · · =????

(1− x)−1 =1

1− x= 1 + x + x2 + . . . .

(1 + x)p =∞∑r=0

Is There a Problem?

(−1)!

(−1− 1)!1!=

(−1)!

(−2)!1!.

where (−1)! = (−1)(−2)(−3) · · · =????

(1− x)−1 =1

1− x= 1 + x + x2 + . . . .

(1 + x)p =∞∑r=0

Is There a Problem?

(−1)!

(−1− 1)!1!=

(−1)!

(−2)!1!.

where (−1)! = (−1)(−2)(−3) · · · =????

(1− x)−1 =1

1− x= 1 + x + x2 + . . . .

(1 + x)p =∞∑r=0

Is There a Problem?

(−1)!

(−1− 1)!1!=

(−1)!

(−2)!1!.

where (−1)! = (−1)(−2)(−3) · · · =????

Eliminate the factorials

Exercising a little care:

(−1)!

(−2)!=

(−1)(−2)!

(−2)!= −1.

In General

(p − r)!r !

=p(p − 1) · · · (p − r + 1)(p − r)!

(p − r)!r !

=p(p − 1) · · · (p − r + 1)

r !. (14)

Eliminate the factorials

Exercising a little care:

(−1)!

(−2)!=

(−1)(−2)!

(−2)!= −1.

In General

(p − r)!r !

=p(p − 1) · · · (p − r + 1)(p − r)!

(p − r)!r !

=p(p − 1) · · · (p − r + 1)

r !. (14)

General Binomial Expansion

Theorem

The general binomial expansion of (1 + x)p for p real is

(1 + x)p =∞∑r=0

p(p − 1) · · · (p − r + 1)

r !x r . (15)

Approximations

Often we need the first few terms for the case that x � 1 :

(1 + x)p = 1 + px +p(p − 1)

2x2 + O(x3). (16)

General Binomial Expansion

Theorem

The general binomial expansion of (1 + x)p for p real is

(1 + x)p =∞∑r=0

p(p − 1) · · · (p − r + 1)

r !x r . (15)

Approximations

Often we need the first few terms for the case that x � 1 :

(1 + x)p = 1 + px +p(p − 1)

2x2 + O(x3). (16)

Example

Relativistic Kinetic Energy

K = E − E0

= γmc2 −mc2

=mc2√

)2−mc2

) (−v2

)+ · · ·

)− 1

Example

K = E − E0

= γmc2 −mc2

=mc2√

)2−mc2

) (−v2

)+ · · ·

)− 1

Example

K = E − E0

= γmc2 −mc2

=mc2√

)2−mc2

) (−v2

)+ · · ·

)− 1

Example

K = E − E0

= γmc2 −mc2

=mc2√

)2−mc2

) (−v2

)+ · · ·

)− 1

Example

K = E − E0

= γmc2 −mc2

=mc2√

)2−mc2

) (−v2

)+ · · ·

)− 1

Example

K = E − E0

= γmc2 −mc2

=mc2√

)2−mc2

) (−v2

)+ · · ·

)− 1

Taylor and Maclaurin Series

Power SeriesA power series expansion about x = a with coefficients cn is given by∑∞

n=0 cn(x − a)n.

Taylor Series

A Taylor series expansion of f (x) about x = a is the series

f (x) ∼∞∑

cn(x − a)n, (17)

cn =f (n)(a)

n!. (18)

Maclaurin Series - a = 0

n=0 cn(x − a)n.

Taylor Series

f (x) ∼∞∑

cn(x − a)n, (17)

cn =f (n)(a)

n!. (18)

n=0 cn(x − a)n.

Taylor Series

f (x) ∼∞∑

cn(x − a)n, (17)

cn =f (n)(a)

n!. (18)

Sample Coefficient Computation

Expand f (x) = ex about x = 0.

n f (n)(x) cn = f (n)(0)n!

0 ex e0

0! = 1

1 ex e0

1! = 1

2 ex e0

2! = 12!

3 ex e0

3! = 13!

In this case, we have that thepattern is obvious:

ex ∼∞∑

More Expansions

Common Series Expansions

cos x = 1− x2

4!− · · · =

∞∑n=0

(−1)nx2n

sin x = x − x3

5!− · · · =

∞∑n=0

(−1)nx2n+1

(2n + 1)!

1 + x= 1− x + x2 − x3 + · · · =

∞∑n=0

(−1)nxn

tan−1 x = x − x3

5− x7

7+ · · · =

∞∑n=0

(−1)nx2n+1

2n + 1

ln(1 + x) = −x +x2

2− x3

3+ · · · =

∞∑n=1

(−1)nxn

Small Angle Approximation

sin θ = θ − θ3

5!+ . . . ⇒ Relative Error =

∥∥∥∥ sin θ − θ

sin θ

∥∥∥∥ .

The relative error in percent when approximating sin θ by θ.

A one percent relativeerror occurs for θ ≈ 0.24radians= 0.24rad 180o

πrad < 14o .

sin θ = θ − θ3

5!+ . . .

⇒ Relative Error =

∥∥∥∥ sin θ − θ

sin θ

∥∥∥∥ .

πrad < 14o .

sin θ = θ − θ3

∥∥∥∥ sin θ − θ

sin θ

∥∥∥∥ .

πrad < 14o .

sin θ = θ − θ3

∥∥∥∥ sin θ − θ

sin θ

∥∥∥∥ .

πrad < 14o .

The Simple Pendulum

Consider Newton’s Second Law for RotationalMotion:

τ = Iα

−(mg sin θ)L = mL2θ̈ ⇒ Lθ̈ + g sin θ = 0.

Pendulum Equations

Nonlinear Pendulum: Lθ̈ + g sin θ = 0

Linear Pendulum: Lθ̈ + gθ = 0.

The Simple Pendulum

τ = Iα

Pendulum Equations

The Simple Pendulum

τ = Iα

Pendulum Equations

The Simple Pendulum

τ = Iα

Pendulum Equations

The Simple Pendulum

τ = Iα

Pendulum Equations

Linear Pendulum, Lθ̈ + gθ = 0

The general solution

θ(t) = c1 cos(ωt) + c2 sin(ωt) (19)

ω ≡√

The period is found to be

T =2π

ω= 2π

g. (20)

ω ≡√

T =2π

ω= 2π

g. (20)

ω ≡√

T =2π

ω= 2π

g. (20)

The Nonlinear Pendulum, θ̈ + ω2 sin θ = 0

Multiply Equation by θ̇ :

θ̈θ̇ + ω2 sin θθ̇

= 0 ⇒ d

2θ̇2 − ω2 cos θ

Therefore,1

2θ̇2 − ω2 cos θ = c . (21)

Solving for θ̇, we obtaindθ

√2(c + ω2 cos θ).

Rearrange and integrate:

∫dt =

∫dθ√

2(c + ω2 cos θ).

θ̈θ̇ + ω2 sin θθ̇

= 0 ⇒ d

Therefore,1

2θ̇2 − ω2 cos θ = c . (21)

√2(c + ω2 cos θ).

∫dt =

∫dθ√

2(c + ω2 cos θ).

θ̈θ̇ + ω2 sin θθ̇ = 0 ⇒ d

Therefore,1

2θ̇2 − ω2 cos θ = c . (21)

√2(c + ω2 cos θ).

∫dt =

∫dθ√

2(c + ω2 cos θ).

θ̈θ̇ + ω2 sin θθ̇ = 0 ⇒ d

Therefore,1

2θ̇2 − ω2 cos θ = c . (21)

√2(c + ω2 cos θ).

∫dt =

∫dθ√

2(c + ω2 cos θ).

θ̈θ̇ + ω2 sin θθ̇ = 0 ⇒ d

Therefore,1

2θ̇2 − ω2 cos θ = c . (21)

√2(c + ω2 cos θ).

∫dt =

∫dθ√

2(c + ω2 cos θ).

θ̈θ̇ + ω2 sin θθ̇ = 0 ⇒ d

Therefore,1

2θ̇2 − ω2 cos θ = c . (21)

√2(c + ω2 cos θ).

∫dt =

∫dθ√

2(c + ω2 cos θ).

Energy Analysis

The kinetic energy,

potential energy, and total mechanical energy.

2mv2 =

2mL2θ̇2.

U = mgh = mgL(1− cos θ).

2mL2θ̇2 + mgL(1− cos θ). (22)

A little rearranging:

2θ̇2 − ω2 cos θ =

mL2E − ω2 = c .

and for θmax = θ0,

E = mgL(1− cos θ0) = mL2ω2(1− cos θ0).

Energy Analysis

The kinetic energy, potential energy,

and total mechanical energy.

2mv2 =

2mL2θ̇2.

2mL2θ̇2 + mgL(1− cos θ). (22)

2θ̇2 − ω2 cos θ =

mL2E − ω2 = c .

Energy Analysis

The kinetic energy, potential energy, and total mechanical energy.

2mv2 =

2mL2θ̇2.

2mL2θ̇2 + mgL(1− cos θ). (22)

2θ̇2 − ω2 cos θ =

mL2E − ω2 = c .

Energy Analysis

2mv2 =

2mL2θ̇2.

2mL2θ̇2 + mgL(1− cos θ). (22)

2θ̇2 − ω2 cos θ =

mL2E − ω2 = c .

Energy Analysis

2mv2 =

2mL2θ̇2.

2mL2θ̇2 + mgL(1− cos θ). (22)

2θ̇2 − ω2 cos θ =

mL2E − ω2 = c .

Returning to Nonlinear Pendulum Computation

Therefore, we have found that

2θ̇2 − ω2 cos θ = ω2(1− cos θ0). (23)

Using the half angle formula,sin2 θ2 = 1

2 (1− cos θ), we have

2θ̇2 = 2ω2

[sin2 θ0

2− sin2 θ

]. (24)

Solving for θ̇,

dt= 2ω

[sin2 θ0

2− sin2 θ

. (25)

And separating:

2ωdt =dθ[

sin2 θ0

2 − sin2 θ2

2θ̇2 − ω2 cos θ = ω2(1− cos θ0). (23)

2θ̇2 = 2ω2

[sin2 θ0

2− sin2 θ

]. (24)

Solving for θ̇,

dt= 2ω

[sin2 θ0

2− sin2 θ

. (25)

And separating:

2ωdt =dθ[

sin2 θ0

2 − sin2 θ2

2θ̇2 − ω2 cos θ = ω2(1− cos θ0). (23)

2θ̇2 = 2ω2

[sin2 θ0

2− sin2 θ

]. (24)

Solving for θ̇,

dt= 2ω

[sin2 θ0

2− sin2 θ

. (25)

And separating:

2ωdt =dθ[

sin2 θ0

2 − sin2 θ2

2θ̇2 − ω2 cos θ = ω2(1− cos θ0). (23)

2θ̇2 = 2ω2

[sin2 θ0

2− sin2 θ

]. (24)

Solving for θ̇,

dt= 2ω

[sin2 θ0

2− sin2 θ

. (25)

And separating:

2ωdt =dθ[

sin2 θ0

2 − sin2 θ2

The Period of Oscillation

Consider a quarter of a cycle (θ = 0 to θ = θ0):

∫ θ0

dφ√sin2 θ0

2 − sin2 θ2

. (26)

Defining z =sin θ

sinθ02

and k = sin θ0

2 , we obtain

T = 4ω

0dz√

(1−z2)(1−k2z2)

This is done using

dz = 12k cos θ

2 dθ = 12k (1− k2z2)1/2 dθ and sin2 θ0

2 − sin2 θ2 = k2(1− z2).

Period - Small Angle Approximation

For small angles, k = sin θ0

2 is small.

(1− k2z2)−1/2 = 1 +1

2k2z2 +

8k2z4 + O((kz)6)

dz√(1− z2)(1− k2z2)

64k4 + . . .

]. (27)

... and finally!

T = 2π√

[1 + 1

4k2 + 964k4 + . . .

]The Relative Error - Using 1,2,3 Terms

Usefulness of Divergent Series

Convergent Power Series

To date you have learned that convergent power series are good and divergentseries are bad. Recall the ratio test: For

∑cn(x − a)n, the ratio test

ρ = limn→∞

∣∣∣∣cn+1

∣∣∣∣ |x − a| < 1

Simple Example

Recall

ln(1 + x) = −x +x2

2− x3

3+ · · · =

∞∑n=1

(−1)nxn

This converges absolutely when

ρ = limn→∞

∣∣∣∣ (−1)n+1n

(−1)n(n + 1)

∣∣∣∣ |x | < 1,

or |x | < 1.

Usefulness of Divergent Series

Convergent Power Series

To date you have learned that convergent power series are good and divergentseries are bad. Recall the ratio test: For

∑cn(x − a)n, the ratio test

ρ = limn→∞

∣∣∣∣cn+1

∣∣∣∣ |x − a| < 1

Simple Example

Recall

ln(1 + x) = −x +x2

2− x3

3+ · · · =

∞∑n=1

(−1)nxn

This converges absolutely when

ρ = limn→∞

∣∣∣∣ (−1)n+1n

(−1)n(n + 1)

∣∣∣∣ |x | < 1,

or |x | < 1.

Asymptotic Series

Definition

f (x) ' φ(x)∑∞

xn provided

lim|x|→∞

[f (x)

φ(x)−

N∑n=0

]→ 0

for a given N, the sum of N + 1 terms of the series can be as close to f (x)φ(x) as

one desires for sufficiently large x .

For each x and N the error is of the order 1/xN+1

However, the series is divergent and thus there are an optimal number ofterms needed.

A Simple Asymptotic Series

Example∫ ∞

e−zt

1 + t2dt =

z− 2!

z5− · · ·+ (−1)n−1(2n − 2)!

z2n−1+ Rn(z)

where |Rn(z)| ≤ (2n)!z2n+1 and for fixed n, lim|z|→∞ Rn(z) = 0.

The Relative Error for x = 5, 10

Example∫ ∞

e−zt

1 + t2dt =

z− 2!

z5− · · ·+ (−1)n−1(2n − 2)!

z2n−1+ Rn(z)

Example∫ ∞

e−zt

1 + t2dt =

z− 2!

z5− · · ·+ (−1)n−1(2n − 2)!

z2n−1+ Rn(z)

Asymptotic Series from Integrals

Exponential Integral

Ei(x) =

∫ ∞

Integration by parts (u = x−1 , dv = e−x dx) gives

Ei(x) =e−x

∫ ∞

Further integration by parts gives

Ei(x) =e−x

[1− 1

x2− 3!

x3+ · · ·+ (−1)nn!

]+(−1)n+1(n+1)!

∫ ∞

tn+2dt.

In general,

Ei(x) =e−x

∞∑n=0

(−1)nn!

Ei(x) =

∫ ∞

Ei(x) =e−x

∫ ∞

Ei(x) =e−x

[1− 1

x2− 3!

x3+ · · ·+ (−1)nn!

]+(−1)n+1(n+1)!

∫ ∞

tn+2dt.

In general,

Ei(x) =e−x

∞∑n=0

(−1)nn!

Ei(x) =

∫ ∞

Ei(x) =e−x

∫ ∞

Ei(x) =e−x

[1− 1

x2− 3!

x3+ · · ·+ (−1)nn!

]+(−1)n+1(n+1)!

∫ ∞

tn+2dt.

In general,

Ei(x) =e−x

∞∑n=0

(−1)nn!

Ei(x) =

∫ ∞

Ei(x) =e−x

∫ ∞

Ei(x) =e−x

[1− 1

x2− 3!

x3+ · · ·+ (−1)nn!

]+(−1)n+1(n+1)!

∫ ∞

tn+2dt.

In general,

Ei(x) =e−x

∞∑n=0

(−1)nn!

Convergence

Convergence of the Series Ei(x) = e−x

∑∞n=0

(−1)nn!xn

ρ = limn→∞

∣∣∣∣ (n + 1)!

∣∣∣∣ |x | =∞,

This series is divergent! Is it useless?

The Relative Error - Using 1,5,10 Terms

Convergence

∑∞n=0

(−1)nn!xn

ρ = limn→∞

∣∣∣∣ (n + 1)!

∣∣∣∣ |x | =∞,

Convergence

∑∞n=0

(−1)nn!xn

ρ = limn→∞

∣∣∣∣ (n + 1)!

∣∣∣∣ |x | =∞,

Convergence

∑∞n=0

(−1)nn!xn

ρ = limn→∞

∣∣∣∣ (n + 1)!

∣∣∣∣ |x | =∞,

The World on a String Problem

The ProblemTie a string around the Earths equator so that it is tight. Now, add ten feet to thestring. Pull it at one point until it is tight but comes up to a point. How far fromthe Earths surface is this? (How long a pole would you need to support it?) Tohow many digits can you give your answer?

Key Equations

tan θ =x

2+ Rθ

h =√

R2 + x2 − R

Note: θ is small Solve

tan θ − θ =d

2R≈ 5

21008452.4881

√R2 + (

2+ Rθ)2 − R

Key Equations

tan θ =x

2+ Rθ

h =√

R2 + x2 − R

tan θ − θ =d

2R≈ 5

21008452.4881

√R2 + (

2+ Rθ)2 − R

Key Equations

tan θ =x

2+ Rθ

h =√

R2 + x2 − R

tan θ − θ =d

2R≈ 5

21008452.4881

√R2 + (

2+ Rθ)2 − R

Key Equations

tan θ =x

2+ Rθ

h =√

R2 + x2 − R

tan θ − θ =d

2R≈ 5

21008452.4881

√R2 + (

2+ Rθ)2 − R

Key Equations

tan θ =x

2+ Rθ

h =√

R2 + x2 − R

Note: θ is small

tan θ − θ =d

2R≈ 5

21008452.4881

√R2 + (

2+ Rθ)2 − R

Key Equations

tan θ =x

2+ Rθ

h =√

R2 + x2 − R

tan θ − θ =d

2R≈ 5

21008452.4881

√R2 + (

2+ Rθ)2 − R

Key Equations

tan θ =x

2+ Rθ

h =√

R2 + x2 − R

tan θ − θ =d

2R≈ 5

21008452.4881

√R2 + (

2+ Rθ)2 − R

Summary of Talk

1 Geometric Series

2 Binomial SeriesSpecial Relativity Example

3 Taylor Series

4 ApplicationsSmall AnglesThe Nonlinear Pendulum

5 Asymptotic Series

6 Homework - The World on a String

how small is small? - russ hermanrussherman.com/talks/latex_ex/seriestalk.pdf · 2013-09-11 ·...

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