aarjaneiro/planetmath huggingface dataset:
infinitely-differentiable function that is not analytic

infinitely-differentiable function that is not analytic

If

f

∈

𝒞

∞

, then we can certainly

write

a

Taylor series

for

f

. However, analyticity requires that this Taylor series actually

converge

(at least across some

radius of convergence

) to

f

. It is not necessary that the

power series

for

f

converge to

f

, as the following example shows.

Let

f

⁢

(

x

)

=

{

e

-

1

x

2

x

≠

0

0

x

=

0

.

Then

f

∈

𝒞

∞

, and for any

n

≥

0

,

f

(

n

)

⁢

(

0

)

=

0

(see below). So the Taylor series for

f

around 0 is 0; since

f

⁢

(

x

)

>

0

for all

x

≠

0

, clearly it does not converge to

f

.

Proof that

f

(

n

)

⁢

(

0

)

=

0

Let

p

⁢

(

x

)

,

q

⁢

(

x

)

∈

ℝ

⁢

[

x

]

be

polynomials

, and define

g

⁢

(

x

)

=

p

⁢

(

x

)

q

⁢

(

x

)

⋅

f

⁢

(

x

)

.

Then, for

x

≠

0

,

g

′

⁢

(

x

)

=

(

p

′

⁢

(

x

)

+

p

⁢

(

x

)

⁢

2

x

3

)

⁢

q

⁢

(

x

)

-

q

′

⁢

(

x

)

⁢

p

⁢

(

x

)

q

2

⁢

(

x

)

⋅

e

-

1

x

2

.

Computing (e.g. by applying L’Hôpital’s rule (

http://planetmath.org/LHpitalsRule

)), we see that

g

′

⁢

(

0

)

=

lim

x

→

0

⁡

g

′

⁢

(

x

)

=

0

.

Define

p

0

⁢

(

x

)

=

q

0

⁢

(

x

)

=

1

. Applying the above inductively, we see that we may write

f

(

n

)

⁢

(

x

)

=

p

n

⁢

(

x

)

q

n

⁢

(

x

)

⁢

f

⁢

(

x

)

. So

f

(

n

)

⁢

(

0

)

=

0

, as required.

Title

infinitely-differentiable function that is not analytic

Canonical name

InfinitelydifferentiableFunctionThatIsNotAnalytic

Date of creation

2013-03-22 12:46:15

Last modified on

2013-03-22 12:46:15

Owner

ariels (338)

Last modified by

ariels (338)

Numerical id

5

Author

ariels (338)

Entry type

Example

Classification

msc 30B10

Classification

msc 26A99

Generated on Fri Feb 9 20:15:26 2018 by

LaTeXML
Lambert series

Lambert series

The series

∑

n

=

1

∞

a

n

⁢

z

n

1

-

z

n

=

a

1

⁢

z

1

-

z

+

a

2

⁢

z

2

1

-

z

2

+

…

(1)

is called

Lambert series

.  We here consider more closely only the special case

∑

n

=

1

∞

x

n

1

-

x

n

=

x

1

-

x

+

x

2

1

-

x

2

+

…

(2)

for the real

x

.

I.

Convergence

1

∘

.

x

=

±

1

:  The series is not defined.

2

∘

.

|

x

|

>

1

:  We have

x

n

1

-

x

n

=

1

1

x

n

-

1

→

-

1

≠

0

as

⁢

n

→

∞

,

whence the series (2)

diverges

.

3

∘

.

0

≦

x

<

1

:  The series with nonnegative terms converges, since

x

n

1

-

x

n

n

=

x

1

-

x

n

n

→

x

<

1

as

⁢

n

→

∞

.

4

∘

.

-

1

<

x

<

0

:  We get an

alternating series

with

|

x

n

1

-

x

n

|

=

|

x

|

n

|

1

-

x

n

|

≦

|

x

|

n

1

-

|

x

|

n

≦

|

x

|

n

1

-

|

x

|

→

0

as

⁢

n

→

∞

,

and by

Leibniz theorem

, the series converges.

Thus we have the result that the Lambert series (2) converges, absolutely, when

|

x

|

<

1

.

Let

|

x

|

<

1

.  the terms to

geometric series

:

x

1

-

x

=

x

+

x

2

+

x

3

+

x

4

+

x

5

+

x

6

+

…

x

2

1

-

x

2

=

x

2

+

x

4

+

x

6

+

…

x

3

1

-

x

3

=

x

3

+

x

6

+

…

x

4

1

-

x

4

=

x

4

+

…

x

5

1

-

x

5

=

x

5

+

…

x

6

1

-

x

6

=

x

6

+

…

…

…

…

…

…

…

…

…

Those geometric series

converge absolutely

,

|

x

k

|

+

|

x

2

⁢

k

|

+

|

x

3

⁢

k

|

+

…

=

|

x

|

k

1

-

|

x

|

k

and the series

∑

k

=

1

∞

|

x

|

k

1

-

|

x

|

k

converges.  Thus we can sum the geometric series by the columns:

∑

n

=

1

∞

x

n

1

-

x

n

=

x

+

2

⁢

x

2

+

2

⁢

x

3

+

3

⁢

x

4

+

2

⁢

x

5

+

4

⁢

x

6

+

…

Apparently, the

coefficient

of any

x

k

in this

power series

expresses, by how many

positive integers

the number

k

is

divisible

, i.e. the coefficient is given by the

divisor function

τ

.  So we may write the power series form of the Lambert series as

∑

n

=

1

∞

x

n

1

-

x

n

=

τ

⁢

(

1

)

⁢

x

+

τ

⁢

(

2

)

⁢

x

2

+

τ

⁢

(

3

)

⁢

x

3

+

…

Title

Lambert series

Canonical name

LambertSeries

Date of creation

2013-03-22 18:46:42

Last modified on

2013-03-22 18:46:42

Owner

pahio (2872)

Last modified by

pahio (2872)

Numerical id

13

Author

pahio (2872)

Entry type

Example

Classification

msc 30B10

Classification

msc 40A05

Related topic

NecessaryConditionOfConvergence

Related topic

CauchysRootTest

Related topic

TauFunction

Generated on Fri Feb 9 20:15:29 2018 by

LaTeXML
Laurent expansion of rational function

Laurent expansion of rational function

The

Laurent series

expansion of a

rational function

may often be determined using the uniqueness of Laurent series

coefficients

in an

annulus

and applying geometric series.  We will determine the expansion of

f

⁢

(

z

)

:=

2

⁢

z

1

+

z

2

by the powers of

z

-

i

.

We first have the

partial fraction decomposition

f

⁢

(

z

)

=

1

z

-

i

+

1

z

+

i

(1)

whence the

principal part

of the Laurent expansion contains

1

z

-

i

.  Taking into account the poles

z

=

±

i

of

f

we see that there are two possible annuli for the Laurent expansion:

a)  The annulus

0

<

|

z

-

i

|

<

2

.  We can write

1

z

+

i

=

1

2

⁢

i

+

(

z

-

i

)

=

1

2

⁢

i

⋅

1

1

-

(

-

z

-

i

2

⁢

i

)

=

1

2

⁢

i

-

z

-

i

(

2

⁢

i

)

2

+

(

z

-

i

)

2

(

2

⁢

i

)

3

-

+

…

Thus

2

⁢

z

1

+

z

2

=

1

z

-

i

-

∑

n

=

0

∞

(

i

2

)

n

+

1

(

z

-

i

)

n

(

0

<

|

z

-

i

|

<

2

)

.

b)  The annulus

2

<

|

z

-

i

|

<

∞

.  Now we write

1

z

+

i

=

1

(

z

-

i

)

+

2

⁢

i

=

1

z

-

i

⋅

1

1

-

(

-

2

⁢

i

z

-

i

)

=

1

z

-

i

-

2

⁢

i

(

z

-

i

)

2

+

(

2

⁢

i

)

2

(

z

-

i

)

3

-

+

…

Accordingly

2

⁢

z

1

+

z

2

=

2

z

-

i

+

∑

n

=

2

∞

(

-

2

⁢

i

)

n

-

1

(

z

-

i

)

n

(

2

<

|

z

-

i

|

<

∞

)

.

This latter Laurent expansion consists of negative powers only, but

z

=

i

isn’t an

essential singularity

of

f

, though.

Title

Laurent expansion of rational function

Canonical name

LaurentExpansionOfRationalFunction

Date of creation

2013-03-11 19:16:06

Last modified on

2013-03-11 19:16:06

Owner

pahio (2872)

Last modified by

(0)

Numerical id

5

Author

pahio (0)

Entry type

Example

Classification

msc 30B10

Synonym

Related topic

Defines

Generated on Fri Feb 9 20:15:33 2018 by

LaTeXML
Laurent series

Laurent series

A

Laurent series

centered about

a

is a series of the form

∑

k

=

-

∞

∞

c

k

⁢

(

z

-

a

)

k

where

c

k

,

a

,

z

∈

ℂ

. The

principal part

of a Laurent series is the subseries

∑

k

=

-

∞

-

1

c

k

⁢

(

z

-

a

)

k

.

One can prove that the above series converges everywhere inside the (possibly empty) set

D

:=

{

z

∈

ℂ

∣

R

1

<

|

z

-

a

|

<

R

2

}

where

R

1

:=

lim sup

k

→

∞

⁡

|

c

-

k

|

1

/

k

and

R

2

:=

1

/

(

lim sup

k

→

∞

⁡

|

c

k

|

1

/

k

)

.

Every Laurent series has an associated

function

, given by

f

⁢

(

z

)

:=

∑

k

=

-

∞

∞

c

k

⁢

(

z

-

a

)

k

,

whose domain is the set of points in

ℂ

on which the series converges. This function is

analytic

inside the

annulus

D

, and conversely, every

analytic function

on an annulus is equal to some unique Laurent series. The

coefficients of Laurent series

for an analytic function can be determined using the

Cauchy integral formula

.

Title

Laurent series

Canonical name

LaurentSeries

Date of creation

2013-03-22 12:04:52

Last modified on

2013-03-22 12:04:52

Owner

djao (24)

Last modified by

djao (24)

Numerical id

12

Author

djao (24)

Entry type

Definition

Classification

msc 30B10

Synonym

Laurent expansion

Related topic

EssentialSingularity

Related topic

CoefficientsOfLaurentSeries

Defines

principal part

Generated on Fri Feb 9 20:15:35 2018 by

LaTeXML
limerick on Euler’s identity

limerick on Euler’s identity

e

raised to the

pi

times

i

,

And plus 1 leaves you nought but a sigh.

This fact amazed Euler

That genius toiler,

And still gives us pause, bye the bye.

References

1

Paul J. Nahin,

Dr. Euler’s Fabulous Formula: Cures Many Mathematical Ills

. Princeton: Princeton University Press (2006): 2

Title

limerick on Euler’s identity

Canonical name

LimerickOnEulersIdentity

Date of creation

2013-03-22 18:09:23

Last modified on

2013-03-22 18:09:23

Owner

PrimeFan (13766)

Last modified by

PrimeFan (13766)

Numerical id

4

Author

PrimeFan (13766)

Entry type

Feature

Classification

msc 30B10

Generated on Fri Feb 9 20:15:37 2018 by

LaTeXML
proof of identity theorem of power series

proof of identity theorem of power series

We start by proving a more modest result. Namely, we show that,
under the hypotheses of the

theorem

we are trying to prove, we
can conclude that

a

0

=

b

0

.

Let

R

be chosen such that both series

converge

when

|

z

-

z

0

|

<

R

.
From the set of points at which the two

power series

are equal, we may
choose a

sequence

{

w

k

}

k

=

0

∞

such that

•

|

w

k

-

z

0

|

<

R

/

2

for all

k

.

•

lim

k

→

∞

⁡

w

k

exists and equals

z

0

.

•

w

k

≠

z

0

for all

k

.

.

Since power series

converge uniformly

, we may interchange the
limit with the summation.

lim

k

→

∞

⁡

∑

n

=

0

∞

a

n

⁢

(

w

k

-

z

0

)

n

=

∑

n

=

0

∞

lim

k

→

∞

⁡

a

n

⁢

(

w

k

-

z

0

)

n

=

a

0

lim

k

→

∞

⁡

∑

n

=

0

∞

b

n

⁢

(

w

k

-

z

0

)

n

=

∑

n

=

0

∞

lim

k

→

∞

⁡

b

n

⁢

(

w

k

-

z

0

)

n

=

b

0

Because

∑

n

=

0

∞

a

n

⁢

(

w

k

-

z

0

)

n

=

s

⁢

u

⁢

m

n

=

0

∞

⁢

a

n

⁢

(

w

k

-

z

0

)

n

for all

k

,
this means that

a

0

=

b

0

.

We will now prove that

a

n

=

b

n

for all

n

by
an

induction

argument

. The intial step with

n

=

0

is, of course, the result demonstrated above.
Assume that

a

m

=

b

m

for all

m

less than
some integer

N

. Then we have

∑

n

=

N

∞

a

n

⁢

(

w

-

z

0

)

n

=

∑

n

=

N

∞

b

n

⁢

(

w

-

z

0

)

n

for all

w

∈

S

. Pulling out a common
factor and relabelling the index, we have

(

w

-

z

0

)

N

⁢

∑

n

=

0

∞

a

n

+

N

⁢

(

w

-

z

0

)

n

=

(

w

-

z

0

)

N

⁢

∑

n

=

0

∞

b

n

+

N

⁢

(

w

-

z

0

)

n

.

Because

z

0

∉

S

, the factor

w

-

z

0

will
not equal zero, so we may cancel it:

∑

n

=

0

∞

a

n

+

N

⁢

(

w

-

z

0

)

n

=

∑

n

=

0

∞

b

n

+

N

⁢

(

w

-

z

0

)

n

By our weaker result, we have

a

N

=

b

N

.
Hence, by induction, we have

a

n

=

b

n

for all

n

.

Title

proof of identity theorem of power series

Canonical name

ProofOfIdentityTheoremOfPowerSeries

Date of creation

2013-03-22 16:47:38

Last modified on

2013-03-22 16:47:38

Owner

rspuzio (6075)

Last modified by

rspuzio (6075)

Numerical id

7

Author

rspuzio (6075)

Entry type

Proof

Classification

msc 30B10

Classification

msc 40A30

Generated on Fri Feb 9 20:15:40 2018 by

LaTeXML
proof of radius of convergence of a complex function

proof of radius of convergence of a complex function

Without loss of generality, it may be assumed that

z

0

=

0

.

Let

c

n

denote the

coefficient

of the

n

-th term in the

Taylor series

of

f

about

0

. Let

r

be a

real number

such that

0

<

r

<

R

. Then

c

n

may be expressed as an

integral

using the

Cauchy integral formula

.

c

n

=

1

2

⁢

π

⁢

i

⁢

∮

|

z

|

=

r

f

⁢

(

z

)

z

n

+

1

⁢

𝑑

z

=

1

2

⁢

π

⁢

r

n

⁢

∫

-

π

+

π

e

-

n

⁢

θ

⁢

f

⁢

(

r

⁢

e

i

⁢

θ

)

⁢

𝑑

θ

Since

f

is

analytic

, it is also

continuous

. Since a

continuous function

on a compact set is

bounded

,

|

f

|

<

B

for some constant

B

>

0

on the circle

|

z

|

=

r

. Hence, we have

|

c

n

|

=

1

2

⁢

π

⁢

r

n

⁢

|

∫

-

π

+

π

e

-

n

⁢

θ

⁢

f

⁢

(

r

⁢

e

i

⁢

θ

)

⁢

𝑑

θ

|

≤

1

2

⁢

π

⁢

r

n

⁢

∫

-

π

+

π

|

e

-

n

⁢

θ

⁢

f

⁢

(

r

⁢

e

i

⁢

θ

)

|

⁢

𝑑

θ

≤

1

2

⁢

π

⁢

r

n

⁢

∫

-

π

+

π

B

⁢

𝑑

θ

=

B

r

n

Consequently,

c

n

n

≤

B

n

/

r

. Since

lim

n

→

∞

⁡

B

n

=

1

, the

radius of convergence

must be

greater

than or equal to

r

. Since this is true for all

r

<

R

, it follows that the radius of convergence is greater than or equal to

R

.

Title

proof of radius of convergence

of a

complex function

Canonical name

ProofOfRadiusOfConvergenceOfAComplexFunction

Date of creation

2013-03-22 14:40:35

Last modified on

2013-03-22 14:40:35

Owner

rspuzio (6075)

Last modified by

rspuzio (6075)

Numerical id

9

Author

rspuzio (6075)

Entry type

Proof

Classification

msc 30B10

Generated on Fri Feb 9 20:15:43 2018 by

LaTeXML
radius of convergence of a complex function

radius of convergence of a complex function

Let

f

be an

analytic function

defined in a disk of radius

R

about a point

z

0

∈

ℂ

. Then the

radius of convergence

of the

Taylor series

of

f

about

z

0

is at least

R

.

For example, the

function

a

⁢

(

z

)

=

1

/

(

1

-

z

)

2

is analytic inside the disk

|

z

|

<

1

. Hence its the radius of covergence of its Taylor series about

0

is at least

1

. By direct examination of the Taylor series we can see that its radius of convergence is, in fact, equal to

1

.

Colloquially, this theorem is stated in the sometimes imprecise but memorable form “The radius of convergence of the Taylor series is the distance to the nearest singularity.”

Title

radius of convergence of a

complex function

Canonical name

RadiusOfConvergenceOfAComplexFunction

Date of creation

2013-03-22 14:40:33

Last modified on

2013-03-22 14:40:33

Owner

rspuzio (6075)

Last modified by

rspuzio (6075)

Numerical id

6

Author

rspuzio (6075)

Entry type

Theorem

Classification

msc 30B10

Generated on Fri Feb 9 20:15:44 2018 by

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Taylor series of arcus tangent

Taylor series of arcus tangent

The

derivative

of the arcus

tangent

(

http://planetmath.org/CyclometricFunctions

)

function

,

1

1

+

x

2

, can be

expanded

as a

geometric series

1

1

+

x

2

=

1

-

x

2

+

x

4

-

x

6

+

-

…

,

the

radius of convergence

of which is

1

.  The series is valid only on the

open interval

-

1

<

x

<

1

,  because the series apparently

diverges

for

x

=

±

1

.  The

power series

may be integrated termwise on its interval of convergence, giving

∫

0

x

d

⁢

t

1

+

t

2

=

/

0

x

arctan

t

=

arctan

x

=

x

-

x

3

3

+

x

5

5

-

+

…

(

|

x

|

<

1

)

.

We can show that this

Taylor series of arcus tangent

is valid also for the

end points

x

=

±

1

of the

interval

.

We start from the identical equation

1

1

+

t

2

=

1

-

t

2

+

t

4

-

+

…

+

(

-

1

)

n

-

1

t

2

⁢

n

-

2

+

(

-

1

)

n

t

2

⁢

n

1

+

t

2

,

which can be verified by performing the

division

1

:

(

1

+

t

2

)

.  Integrating both sides from

0

to an arbitrary

x

, we obtain

arctan

⁡

x

=

x

-

x

3

3

+

x

5

5

-

+

…

+

(

-

1

)

n

-

1

x

2

⁢

n

-

1

2

⁢

n

-

1

⏟

S

2

⁢

n

-

1

+

(

-

1

)

n

⁢

∫

0

x

t

2

⁢

n

1

+

t

2

⁢

𝑑

t

⏟

R

2

⁢

n

-

1

.

We estimate

R

2

⁢

n

-

1

:

|

R

2

⁢

n

-

1

|

=

∫

0

|

x

|

t

2

⁢

n

1

+

t

2

⁢

𝑑

t

≦

∫

0

|

x

|

t

2

⁢

n

⁢

𝑑

t

=

|

x

|

2

⁢

n

-

1

2

⁢

n

-

1

→

0

as

⁢

n

→

∞

for

x

=

±

1

.  Accordingly, when

x

=

±

1

,  we see that

S

2

⁢

n

-

1

=

arctan

⁡

x

-

R

2

⁢

n

-

1

→

arctan

⁡

x

as

n

→

∞

.  This that

arctan

(

±

1

)

=

±

π

4

=

±

(

1

-

1

3

+

1

5

-

+

…

)

.

Title

Taylor series of arcus tangent

Canonical name

TaylorSeriesOfArcusTangent

Date of creation

2013-03-22 16:33:37

Last modified on

2013-03-22 16:33:37

Owner

pahio (2872)

Last modified by

pahio (2872)

Numerical id

7

Author

pahio (2872)

Entry type

Derivation

Classification

msc 30B10

Classification

msc 26A24

Classification

msc 41A58

Related topic

GregorySeries

Related topic

TaylorSeriesOfArcusSine

Related topic

SubstitutionNotation

Related topic

ExamplesOnHowToFindTaylorSeriesFromOtherKnownSeries

Related topic

CyclometricFunctions

Related topic

LogarithmSeries

Generated on Fri Feb 9 20:15:51 2018 by

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Taylor series of hyperbolic functions

Taylor series of hyperbolic functions

The differentiation rules

d

d

⁢

x

⁢

cosh

⁡

x

=

sinh

⁡

x

,

d

d

⁢

x

⁢

sinh

⁡

x

=

cosh

⁡

x

of the

hyperbolic functions

imply

d

2

⁢

n

d

⁢

x

2

⁢

n

cosh

x

=

cosh

x

,

d

2

⁢

n

+

1

d

⁢

x

2

⁢

n

+

1

cosh

x

=

sinh

x

(

n

=

0

,

1

,

2

,

…

)

.

In the origin

x

=

0

,  all even (

http://planetmath.org/Even

)-order

derivatives

of the

hyperbolic cosine

have the value 1, but the odd (

http://planetmath.org/Odd

)-order derivatives vanish.  Thus the

Taylor series

expansion

f

⁢

(

x

)

=

f

⁢

(

0

)

+

f

′

⁢

(

0

)

1

!

⁢

x

+

f

′′

⁢

(

0

)

2

!

⁢

x

2

+

f

′′′

⁢

(

0

)

3

!

⁢

x

3

+

…

of

f

⁢

(

x

)

:=

cosh

⁡

x

contains only the terms of even degree and writes simply

cosh

⁡

x

=

1

+

x

2

2

!

+

x

4

4

!

+

…

=

∑

n

=

0

∞

x

2

⁢

n

(

2

⁢

n

)

!

.

(1)

Similarly, one can derive for the

hyperbolic sine

the expansion

sinh

⁡

x

=

x

+

x

3

3

!

+

x

5

5

!

+

…

=

∑

n

=

0

∞

x

2

⁢

n

+

1

(

2

⁢

n

+

1

)

!

.

(2)

Both series

converge

(

http://planetmath.org/AbsoluteConvergence

) and the

functions

for all real (and

complex

) values of

x

.  Comparing the expansions (1) and (2) with the corresponding ones of the

circular functions

cosine and sine, one sees easily that

cosh

⁡

x

=

cos

⁡

i

⁢

x

,

sinh

⁡

x

=

-

i

⁢

sin

⁡

i

⁢

x

.

As for the

Taylor expansion

of the third important hyperbolic function tangens hyperbolica (

http://planetmath.org/HyperbolicFunctions

) tanh, it is obtained via division of the Taylor series (

http://planetmath.org/TaylorSeriesViaDivision

) (2) and (1); the begin of the quotient series is

tanh

x

=

x

-

1

3

x

3

+

2

15

x

5

-

17

315

x

7

+

-

…

(

|

x

|

<

π

2

)

.

(3)

The

coefficients

of this

power series

may be expressed with the

Bernoulli numbers

.

Title

Taylor series of hyperbolic functions

Canonical name

TaylorSeriesOfHyperbolicFunctions

Date of creation

2013-03-22 19:07:04

Last modified on

2013-03-22 19:07:04

Owner

pahio (2872)

Last modified by

pahio (2872)

Numerical id

7

Author

pahio (2872)

Entry type

Derivation

Classification

msc 30B10

Classification

msc 26A09

Related topic

HyperbolicIdentities

Related topic

HigherOrderDerivatives

Generated on Fri Feb 9 20:15:54 2018 by

LaTeXML
uniqueness of Laurent expansion

uniqueness of Laurent expansion

The

Laurent series

expansion of a

function

f

⁢

(

z

)

in an

annulus

r

<

|

z

-

z

0

|

<

R

is unique.

Proof.

Suppose that

f

⁢

(

z

)

has in the annulus two Laurent expansions:

f

⁢

(

z

)

=

∑

n

=

-

∞

∞

a

n

⁢

(

z

-

z

0

)

n

=

∑

n

=

-

∞

∞

b

n

⁢

(

z

-

z

0

)

n

It follows that

f

⁢

(

z

)

⁢

(

z

-

z

0

)

-

ν

-

1

=

∑

n

=

-

∞

∞

a

n

⁢

(

z

-

z

0

)

n

-

ν

-

1

=

∑

n

=

-

∞

∞

b

n

⁢

(

z

-

z

0

)

n

-

ν

-

1

where

ν

is an integer.  Let now

γ

be an arbitrary closed

contour

in the annulus, going once around

z

0

.  Since

γ

is a compact set of points, those two Laurent series

converge uniformly

(

http://planetmath.org/UniformConvergence

) on it and therefore they can be integrated termwise (

http://planetmath.org/SumFunctionOfSeries

) along

γ

, i.e.

∑

n

=

-

∞

∞

a

n

⁢

∮

γ

(

z

-

z

0

)

n

-

ν

-

1

⁢

𝑑

z

=

∑

n

=

-

∞

∞

b

n

⁢

∮

γ

(

z

-

z

0

)

n

-

ν

-

1

⁢

𝑑

z

.

(1)

But

∮

γ

(

z

-

z

0

)

n

-

ν

-

1

⁢

𝑑

z

=

{

2

⁢

i

⁢

π

if

⁢

n

=

ν

,

0

if

⁢

n

≠

ν

,

when integrated anticlockwise (see

calculation of contour integral

).  Thus (1) reads

2

⁢

i

⁢

π

⁢

a

ν

=

2

⁢

i

⁢

π

⁢

b

ν

,

i.e.

a

ν

=

b

ν

,  for any integer

ν

, whence both expansions are identical.

Title

uniqueness of Laurent expansion

Canonical name

UniquenessOfLaurentExpansion

Date of creation

2013-03-22 19:14:12

Last modified on

2013-03-22 19:14:12

Owner

pahio (2872)

Last modified by

pahio (2872)

Numerical id

11

Author

pahio (2872)

Entry type

Theorem

Classification

msc 30B10

Related topic

CoefficientsOfLaurentSeries

Related topic

UniquenessOfFourierExpansion

Related topic

UniquenessOfDigitalRepresentation

Generated on Fri Feb 9 20:16:01 2018 by

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Weierstrass double series theorem

Weierstrass double series theorem

If the

complex functions

f

0

,

f

1

,

f

2

,

…

are

holomorphic

in the disc

|

z

-

z

0

|

<

r

and thus

f

n

⁢

(

z

)

=

∑

ν

=

0

∞

a

n

⁢

ν

⁢

(

z

-

z

0

)

ν

,

a

n

⁢

ν

=

f

n

(

ν

)

⁢

(

z

0

)

ν

!

∀

n

,

ν

(1)

in this disc, and if the function series

∑

n

=

0

∞

f

n

=

f

0

+

f

1

+

f

2

+

…

(2)

converges uniformly

to the

function

F

in each disc

|

z

-

z

0

|

≦

ϱ

where

0

<

ϱ

<

r

,  then also all the series

∑

n

=

0

∞

a

n

⁢

ν

=

a

0

⁢

ν

+

a

1

⁢

ν

+

a

2

⁢

ν

+

…

(

ν

=

0

,

1

,

2

,

…

)

(3)

converge

, and in the disc

|

z

-

z

0

|

<

r

one has

F

⁢

(

z

)

=

∑

ν

=

0

∞

A

ν

⁢

(

z

-

z

0

)

ν

(4)

where the

A

ν

s are the sums of the series (3).

Proof.

Apparently, the series (2) converges uniformly also in every closed sub-disc of the open disc

|

z

-

z

0

|

<

r

.  Therefore the theorem 2 in the entry “

theorems on complex function series

(

http://planetmath.org/TheoremsOnComplexFunctionSeries

)” says that the sum

F

⁢

(

z

)

is holomorphic in

|

z

-

z

0

|

<

r

and

F

(

ν

)

(

z

)

=

f

0

(

ν

)

(

z

0

)

+

f

1

(

ν

)

(

z

0

)

+

f

2

(

ν

)

(

z

0

)

+

…

(

ν

=

0

,

1

,

2

,

…

)

.

Theorem 3 in the same entry thus guarantees that

F

⁢

(

z

)

has the

Taylor expansion

of the form (4) wherein

A

ν

=

1

ν

!

F

(

ν

)

(

z

0

)

(

ν

=

0

,

1

,

2

,

…

)

.

According to theorem 2 in the same entry the series (2) may be differentiated termwise,

A

ν

=

1

ν

!

⁢

∑

n

=

0

∞

f

n

(

ν

)

⁢

(

z

0

)

=

∑

n

=

0

∞

1

ν

!

⁢

f

n

(

ν

)

⁢

(

z

0

)

=

∑

n

=

0

∞

a

n

⁢

ν

Q.E.D.

Note.

In

Weierstrass double series theorem

it’s a question of changing the summing :

F

⁢

(

z

)

=

f

0

⁢

(

z

)

+

f

1

⁢

(

z

)

+

…

+

f

n

⁢

(

z

)

+

…

=

=

[

a

00

+

a

01

⁢

(

z

-

z

0

)

+

a

02

⁢

(

z

-

z

0

)

2

+

…

+

a

0

⁢

ν

⁢

(

z

-

z

0

)

ν

+

…

]

+

[

a

10

+

a

11

⁢

(

z

-

z

0

)

+

a

12

⁢

(

z

-

z

0

)

2

+

…

+

a

1

⁢

ν

⁢

(

z

-

z

0

)

ν

+

…

]

+

[

a

20

+

a

21

⁢

(

z

-

z

0

)

+

a

22

⁢

(

z

-

z

0

)

2

+

…

+

a

2

⁢

ν

⁢

(

z

-

z

0

)

ν

+

…

]

…

⁢

…

+

[

a

n

⁢

0

+

a

n

⁢

1

⁢

(

z

-

z

0

)

+

a

n

⁢

2

⁢

(

z

-

z

0

)

2

+

…

+

a

n

⁢

ν

⁢

(

z

-

z

0

)

ν

+

…

]

⋯

⋯

¯

=

A

0

+

A

1

⁢

(

z

-

z

0

)

+

A

2

⁢

(

z

-

z

0

)

2

+

…

+

A

ν

⁢

(

z

-

z

0

)

ν

+

…

Title

Weierstrass double series theorem

Canonical name

WeierstrassDoubleSeriesTheorem

Date of creation

2013-03-22 16:48:15

Last modified on

2013-03-22 16:48:15

Owner

pahio (2872)

Last modified by

pahio (2872)

Numerical id

7

Author

pahio (2872)

Entry type

Theorem

Classification

msc 30B10

Classification

msc 40A05

Classification

msc 30D30

Related topic

TheoremsOnComplexFunctionSeries

Generated on Fri Feb 9 20:16:04 2018 by

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