aarjaneiro/planetmath huggingface dataset:
residues of tangent and cotangent

residues of tangent and cotangent

We will determine the

residues

of the

tangent

and the

cotangent

at their poles, which by the

http://planetmath.org/node/9074

parent entry are simple (

http://planetmath.org/SimplePole

).

By the rule in the entry

coefficients of Laurent series

, in a

simple pole

z

=

a

of

f

one has

Res

⁢

(

f

;

a

)

=

lim

z

→

a

⁡

(

z

-

a

)

⁢

f

⁢

(

z

)

.

•

We get first

Res

⁢

(

cot

;

0

)

=

lim

z

→

0

⁡

z

⁢

cot

⁡

z

=

lim

z

→

0

⁡

cos

⁡

z

sin

⁡

z

z

=

1

1

=

1

.

(1)

•

All the poles of cotangent are

n

⁢

π

with

n

∈

ℤ

.  Since

π

is the period of cotangent, we could infer that the residues in all poles are the same as (1).  We may also calculate (with the change of variable

z

-

n

⁢

π

=

w

) directly

Res

⁢

(

cot

;

n

⁢

π

)

=

lim

z

→

n

⁢

π

⁡

(

z

-

n

⁢

π

)

⁢

cot

⁡

z

=

lim

w

→

0

⁡

w

⁢

cot

⁡

(

w

+

n

⁢

π

)

=

lim

w

→

0

⁡

w

⁢

cot

⁡

w

=

1

.

•

In the parent entry (

http://planetmath.org/ComplexTangentAndCotangent

), the

complement formula

for the

tangent function

is derived.  Using it, we can find the residues of tangent at its poles

π

2

+

n

⁢

π

, which are .  For example,

Res

⁢

(

tan

;

π

2

)

=

lim

z

→

π

2

⁡

(

z

-

π

2

)

⁢

cot

⁡

(

π

2

-

z

)

=

lim

w

→

0

⁡

w

⁢

cot

⁡

(

-

w

)

=

-

Res

⁢

(

cot

;

0

)

=

-

1

.

Similarly as above, the residues in other poles are

-

1

.

Consequently, the residues of cotangent are equal to 1 and the residues of tangent equal to

-

1

.

Title

residues of tangent and cotangent

Canonical name

ResiduesOfTangentAndCotangent

Date of creation

2013-03-22 18:57:35

Last modified on

2013-03-22 18:57:35

Owner

pahio (2872)

Last modified by

pahio (2872)

Numerical id

6

Author

pahio (2872)

Entry type

Example

Classification

msc 33B10

Classification

msc 30D10

Classification

msc 30A99

Related topic

Residue

Related topic

TechniqueForComputingResidues

Related topic

ResiduesOfGammaFunction

Generated on Fri Feb 9 20:35:10 2018 by

LaTeXML
alternate integral representation of beta function

alternate integral representation of beta function

By making the change of variable

x

p

=

y

, we see that

∫

0

1

x

p

-

1

⁢

(

1

-

x

)

q

-

1

⁢

𝑑

x

=

1

p

⁢

∫

0

1

(

1

-

y

1

p

)

q

-

1

⁢

𝑑

y

.

Hence, we have

∫

0

1

(

1

-

y

1

p

)

q

-

1

⁢

𝑑

y

=

p

⁢

Γ

⁢

(

p

)

⁢

Γ

⁢

(

q

)

Γ

⁢

(

p

+

q

)

.

Title

alternate integral representation of beta function

Canonical name

AlternateIntegralRepresentationOfBetaFunction

Date of creation

2013-03-22 17:10:08

Last modified on

2013-03-22 17:10:08

Owner

rspuzio (6075)

Last modified by

rspuzio (6075)

Numerical id

4

Author

rspuzio (6075)

Entry type

Result

Classification

msc 33B15

Generated on Fri Feb 9 20:35:27 2018 by

LaTeXML
alternate integral representation of beta function (2)

alternate integral representation of beta function (2)

Substitute

x

:=

1

1

+

s

,

d

⁢

x

=

-

1

(

1

+

s

)

2

⁢

d

⁢

s

:

∫

0

1

x

p

-

1

⁢

(

1

-

x

)

q

-

1

⁢

𝑑

x

=

∫

0

∞

1

(

1

+

s

)

p

+

1

⁢

(

s

1

+

s

)

q

-

1

⁢

𝑑

s

=

∫

0

∞

s

q

-

1

(

1

+

s

)

p

+

q

⁢

𝑑

s

Since

B

⁢

(

p

,

q

)

=

B

⁢

(

q

,

p

)

this gives:

∫

0

∞

s

p

-

1

(

1

+

s

)

p

+

q

⁢

𝑑

s

=

Γ

⁢

(

p

)

⁢

Γ

⁢

(

q

)

Γ

⁢

(

p

+

q

)

Title

alternate integral representation of beta function

(2)

Canonical name

AlternateIntegralRepresentationOfBetaFunction2

Date of creation

2013-03-22 18:34:31

Last modified on

2013-03-22 18:34:31

Owner

karstenb (16623)

Last modified by

karstenb (16623)

Numerical id

8

Author

karstenb (16623)

Entry type

Result

Classification

msc 33B15

Generated on Fri Feb 9 20:35:29 2018 by

LaTeXML
beta function

beta function

The

beta function

is defined as

B

⁢

(

p

,

q

)

=

∫

0

1

x

p

-

1

⁢

(

1

-

x

)

q

-

1

⁢

𝑑

x

for any

real numbers

p

,

q

>

0

.
For other

complex

values of

p

and

q

,
we can define

B

⁢

(

p

,

q

)

by

analytic continuation

.

The beta function has the

property

B

⁢

(

p

,

q

)

=

Γ

⁢

(

p

)

⁢

Γ

⁢

(

q

)

Γ

⁢

(

p

+

q

)

for all

complex numbers

p

and

q

for which the right-hand side is defined.
Here

Γ

is the

gamma function

.

Also,

B

⁢

(

p

,

q

)

=

B

⁢

(

q

,

p

)

and

B

⁢

(

1

2

,

1

2

)

=

π

.

The beta function was first defined
by L. Euler (

http://planetmath.org/EulerLeonhard

) in 1730,
and the name was given by J. Binet.

Title

beta function

Canonical

name

BetaFunction

Date of creation

2013-03-22 13:26:23

Last modified on

2013-03-22 13:26:23

Owner

yark (2760)

Last modified by

yark (2760)

Numerical id

21

Author

yark (2760)

Entry type

Definition

Classification

msc 33B15

Generated on Fri Feb 9 20:35:31 2018 by

LaTeXML
Bohr-Mollerup theorem

Bohr-Mollerup theorem

Let

f

:

ℝ

+

→

ℝ

+

be a

function

with the following

properties

:

1.

log

⁡

f

⁢

(

x

)

is a

convex function

(i.e.

f

is

logarithmically convex

);

2.

f

⁢

(

x

+

1

)

=

x

⁢

f

⁢

(

x

)

for all

x

>

0

;

3.

f

⁢

(

1

)

=

1

.

Then

f

⁢

(

x

)

=

Γ

⁢

(

x

)

for all

x

>

0

.

That is, the only function satisfying those properties is the

gamma function

(restricted to the

positive

reals.)

Title

Bohr-Mollerup theorem

Canonical

name

BohrMollerupTheorem

Date of creation

2013-03-22 13:15:10

Last modified on

2013-03-22 13:15:10

Owner

Koro (127)

Last modified by

Koro (127)

Numerical id

5

Author

Koro (127)

Entry type

Theorem

Classification

msc 33B15

Synonym

characterization

of the gamma function

Related topic

GammaFunction

Related topic

LogarithmicallyConvexFunction

Generated on Fri Feb 9 20:35:32 2018 by

LaTeXML
evaluation of beta function using Laplace transform

evaluation of beta function using Laplace transform

The

beta integral

can be evaluated elegantly using the

convolution theorem

(

http://planetmath.org/LaplaceTransform

) for

Laplace transforms

.

Start with the following Laplace transform:

s

-

α

=

ℒ

⁢

[

t

α

-

1

Γ

⁢

(

α

)

]

=

∫

0

∞

e

-

s

⁢

t

⁢

t

α

-

1

Γ

⁢

(

α

)

⁢

𝑑

t

Since

s

-

q

⁢

s

-

p

=

s

-

q

-

p

, the convolution theorem imples that

t

q

-

1

Γ

⁢

(

q

)

*

t

p

-

1

Γ

⁢

(

p

)

=

t

q

+

p

-

1

Γ

⁢

(

q

+

p

)

Writing out the

definition

of

convolution

, this becomes

∫

0

t

(

t

-

s

)

q

-

1

Γ

⁢

(

q

)

⁢

s

p

-

1

Γ

⁢

(

p

)

⁢

𝑑

s

=

t

q

+

p

-

1

Γ

⁢

(

p

+

q

)

Setting

t

=

1

and simplifying, we conclude that

∫

0

1

x

p

-

1

⁢

(

1

-

x

)

q

-

1

⁢

𝑑

x

=

Γ

⁢

(

p

)

⁢

Γ

⁢

(

q

)

Γ

⁢

(

p

+

q

)

QED

Title

evaluation of

beta function

using Laplace transform

Canonical

name

EvaluationOfBetaFunctionUsingLaplaceTransform

Date of creation

2013-03-22 14:37:36

Last modified on

2013-03-22 14:37:36

Owner

rspuzio (6075)

Last modified by

rspuzio (6075)

Numerical id

10

Author

rspuzio (6075)

Entry type

Derivation

Classification

msc 33B15

Generated on Fri Feb 9 20:35:34 2018 by

LaTeXML
Gauss’ digamma theorem

Gauss’ digamma theorem

Theorem (Gauss)

.

Ψ

⁢

(

x

+

n

)

=

1

x

+

1

x

+

1

+

⋯

+

1

x

+

n

-

1

+

Ψ

⁢

(

x

)

,

n

=

1

,

2

,

3

,

…

Ψ

⁢

(

p

q

)

=

-

γ

-

π

2

⁢

cot

⁡

π

⁢

p

q

-

ln

⁡

q

+

∑

n

=

1

q

-

1

cos

⁡

2

⁢

π

⁢

n

⁢

p

q

⁢

ln

⁡

(

2

⁢

sin

⁡

π

⁢

n

q

)

where

0

<

p

<

q

.

Title

Gauss’ digamma theorem

Canonical name

GaussDigammaTheorem

Date of creation

2013-03-22 16:23:45

Last modified on

2013-03-22 16:23:45

Owner

rm50 (10146)

Last modified by

rm50 (10146)

Numerical id

6

Author

rm50 (10146)

Entry type

Theorem

Classification

msc 33B15

Classification

msc 30D30

Generated on Fri Feb 9 20:35:36 2018 by

LaTeXML
multiplication formula for gamma function

multiplication formula for gamma function

For any integer

n

>

0

, the

gamma function

satisfes the following relation:

Γ

⁢

(

n

⁢

z

)

=

(

2

⁢

π

)

n

-

1

2

⁢

n

n

⁢

z

-

1

2

⁢

∏

k

=

0

n

-

1

Γ

⁢

(

z

+

k

n

)

This equation is true for all

complex

values of

z

for which both sides are defined.

Title

multiplication formula for gamma function

Canonical name

MultiplicationFormulaForGammaFunction

Date of creation

2013-03-22 14:44:08

Last modified on

2013-03-22 14:44:08

Owner

rspuzio (6075)

Last modified by

rspuzio (6075)

Numerical id

8

Author

rspuzio (6075)

Entry type

Theorem

Classification

msc 33B15

Classification

msc 30D30

Generated on Fri Feb 9 20:35:38 2018 by

LaTeXML
partial fraction series for digamma function

partial fraction series for digamma function

Theorem 1

ψ

⁢

(

z

)

=

-

γ

-

1

z

+

∑

k

=

1

∞

(

1

k

-

1

z

+

k

)

=

-

γ

+

∑

k

=

0

∞

(

1

k

+

1

-

1

z

+

k

)

Proof:

Start with

Γ

⁢

(

z

)

=

e

-

γ

⁢

z

z

⁢

∏

k

=

1

∞

(

1

+

z

k

)

-

1

⁢

e

z

/

k

,

so

ln

⁡

Γ

⁢

(

z

)

=

-

γ

⁢

z

-

ln

⁡

z

+

∑

k

=

1

∞

(

-

ln

⁡

(

1

+

z

k

)

+

z

k

)

and thus, taking derivatives,

ψ

⁢

(

z

)

=

-

γ

-

1

z

+

∑

k

=

1

∞

(

-

1

/

k

1

+

z

k

+

1

k

)

=

-

γ

-

1

z

+

∑

k

=

1

∞

(

1

k

-

1

z

+

k

)

The second formula follows after rearranging terms (the rearrangement is legal since we are simply exchanging

adjacent

terms, so partial sums remain the same).

Title

partial fraction series for digamma function

Canonical name

PartialFractionSeriesForDigammaFunction

Date of creation

2013-03-22 16:23:40

Last modified on

2013-03-22 16:23:40

Owner

rm50 (10146)

Last modified by

rm50 (10146)

Numerical id

6

Author

rm50 (10146)

Entry type

Theorem

Classification

msc 33B15

Classification

msc 30D30

Generated on Fri Feb 9 20:35:39 2018 by

LaTeXML
proof of Bohr-Mollerup theorem

proof of Bohr-Mollerup theorem

We prove this

theorem

in two stages: first, we establish that the

gamma function

satisfies the given conditions and then we prove that these conditions uniquely determine a

function

on

(

0

,

∞

)

.

By its

definition

,

Γ

⁢

(

x

)

is

positive

for positive

x

. Let

x

,

y

>

0

and

0

≤

λ

≤

1

.

log

⁡

Γ

⁢

(

λ

⁢

x

+

(

1

-

λ

)

⁢

y

)

=

log

⁢

∫

0

∞

e

-

t

⁢

t

λ

⁢

x

+

(

1

-

λ

)

⁢

y

-

1

⁢

𝑑

t

=

log

⁢

∫

0

∞

(

e

-

t

⁢

t

x

-

1

)

λ

⁢

(

e

-

t

⁢

t

y

-

1

)

1

-

λ

⁢

𝑑

t

≤

log

⁡

(

(

∫

0

∞

e

-

t

⁢

t

x

-

1

⁢

𝑑

t

)

λ

⁢

(

∫

0

∞

e

-

t

⁢

t

y

-

1

⁢

𝑑

t

)

1

-

λ

)

=

λ

⁢

log

⁡

Γ

⁢

(

x

)

+

(

1

-

λ

)

⁢

log

⁡

Γ

⁢

(

y

)

The

inequality

follows from Hölder’s inequality, where

p

=

1

λ

and

q

=

1

1

-

λ

.

This proves that

Γ

is

log-convex

. Condition 2 follows from the definition by applying

integration by parts

. Condition 3 is a trivial verification from the definition.

Now we show that the 3 conditions uniquely determine a function. By condition 2, it suffices to show that the conditions uniquely determine a function on

(

0

,

1

)

.

Let

G

be a function satisfying the 3 conditions,

0

≤

x

≤

1

and

n

∈

ℕ

.

n

+

x

=

(

1

-

x

)

⁢

n

+

x

⁢

(

n

+

1

)

and by log-convexity of

G

,

G

⁢

(

n

+

x

)

≤

G

⁢

(

n

)

1

-

x

⁢

G

⁢

(

n

+

1

)

x

=

G

⁢

(

n

)

1

-

x

⁢

G

⁢

(

n

)

x

⁢

n

x

=

(

n

-

1

)

!

⁢

n

x

.

Similarly

n

+

1

=

x

⁢

(

n

+

x

)

+

(

1

-

x

)

⁢

(

n

+

1

+

x

)

gives

n

!

≤

G

⁢

(

n

+

x

)

⁢

(

n

+

x

)

1

-

x

.

Combining these two we get

n

!

⁢

(

n

+

x

)

x

-

1

≤

G

⁢

(

n

+

x

)

≤

(

n

-

1

)

!

⁢

n

x

and by using condition 2 to express

G

⁢

(

n

+

x

)

in terms of

G

⁢

(

x

)

we find

a

n

:=

n

!

⁢

(

n

+

x

)

x

-

1

x

⁢

(

x

+

1

)

⁢

…

⁢

(

x

+

n

-

1

)

≤

G

(

x

)

≤

(

n

-

1

)

!

⁢

n

x

x

⁢

(

x

+

1

)

⁢

…

⁢

(

x

+

n

-

1

)

=

:

b

n

.

Now these inequalities hold for every positive integer

n

and the terms on the left and right side have a common limit (

lim

n

→

∞

⁡

a

n

b

n

=

1

) so we find this determines

G

.

As a corollary we find another

expression

for

Γ

.

For

0

≤

x

≤

1

,

Γ

⁢

(

x

)

=

lim

n

→

∞

⁡

n

!

⁢

n

x

x

⁢

(

x

+

1

)

⁢

…

⁢

(

x

+

n

)

.

In fact, this equation, called Gauß’s product, goes for the whole

complex plane

minus the negative integers.

Title

proof of Bohr-Mollerup theorem

Canonical

name

ProofOfBohrMollerupTheorem

Date of creation

2013-03-22 13:18:14

Last modified on

2013-03-22 13:18:14

Owner

lieven (1075)

Last modified by

lieven (1075)

Numerical id

6

Author

lieven (1075)

Entry type

Proof

Classification

msc 33B15

Defines

Gauß’s product

Generated on Fri Feb 9 20:35:45 2018 by

LaTeXML
proof of multiplication formula for gamma function

proof of multiplication formula for gamma function

Define the

function

f

as

f

⁢

(

z

)

=

n

n

⁢

z

⁢

∏

k

=

0

n

-

1

Γ

⁢

(

z

+

k

n

)

Γ

⁢

(

n

⁢

z

)

By the functional equation of the

gamma function

,

f

⁢

(

z

+

1

)

=

n

n

⁢

n

n

⁢

z

⁢

(

∏

m

=

0

n

-

1

Γ

⁢

(

z

+

m

n

)

)

⁢

∏

k

=

0

n

-

1

(

z

+

k

n

)

Γ

⁢

(

n

⁢

z

)

⁢

∏

k

=

0

n

-

1

(

n

⁢

z

+

k

)

=

f

⁢

(

z

)

Hence

f

is a

periodic function

of

z

. However, for large values
of

z

, we can apply the

Stirling approximation formula

to conclude

f

⁢

(

z

)

=

(

2

⁢

π

)

n

/

2

⁢

n

n

⁢

z

⁢

∏

k

=

0

n

-

1

[

e

-

z

-

k

/

n

⁢

(

z

+

k

/

n

)

z

+

k

/

n

-

1

/

2

+

O

⁢

(

e

-

z

⁢

(

z

+

k

/

n

)

z

+

k

/

n

-

3

/

2

)

]

(

2

⁢

π

)

1

/

2

⁢

e

-

n

⁢

z

⁢

(

n

⁢

z

)

n

⁢

z

-

1

/

2

+

O

⁢

(

e

-

n

⁢

z

⁢

(

n

⁢

z

)

n

⁢

z

-

3

/

2

)

=

(

2

⁢

π

)

(

n

-

1

)

/

2

⁢

n

1

/

2

⁢

∏

k

=

0

n

-

1

[

e

-

k

/

n

⁢

(

z

+

k

/

n

)

z

+

k

/

n

-

1

/

2

+

O

⁢

(

(

z

+

k

/

n

)

z

+

k

/

n

-

3

/

2

)

]

z

n

⁢

z

-

1

/

2

+

O

⁢

(

z

n

⁢

z

-

3

/

2

)

=

(

2

⁢

π

)

(

n

-

1

)

/

2

⁢

n

1

/

2

⁢

z

1

/

2

⁢

∏

k

=

0

n

-

1

[

e

-

k

/

n

⁢

(

1

+

k

n

⁢

z

)

z

+

k

/

n

-

1

/

2

⁢

z

k

/

n

-

1

/

2

+

O

⁢

(

(

z

+

k

/

n

)

k

/

n

-

3

/

2

)

]

1

+

O

⁢

(

z

-

1

)

Note that

∏

k

=

0

n

-

1

e

-

k

/

n

=

e

-

∑

k

=

0

n

-

1

k

/

n

=

e

(

1

-

n

)

/

2

z

1

/

2

⁢

∏

k

=

0

n

-

1

z

k

/

n

-

1

/

2

=

z

1

/

2

⁢

z

∑

k

=

0

n

-

1

(

k

/

n

-

1

/

2

)

=

z

1

/

2

+

(

n

-

1

)

/

2

-

n

/

2

=

1

Also,

(

1

+

k

n

⁢

z

)

z

=

e

k

/

n

+

O

⁢

(

z

-

1

)

Hence,

f

⁢

(

z

)

=

(

2

⁢

π

)

(

n

-

1

)

/

2

⁢

n

1

/

2

+

O

⁢

(

z

-

1

)

. Now, the only way for a function to be

periodic

and have a definite limit is for that function to be constant. Therefore,

f

⁢

(

z

)

=

(

2

⁢

π

)

(

n

-

1

)

/

2

⁢

n

1

/

2

. Writing out the definition of

f

and rearranging gives the multiplication formula.

Title

proof of multiplication formula for gamma function

Canonical name

ProofOfMultiplicationFormulaForGammaFunction

Date of creation

2013-03-22 14:44:10

Last modified on

2013-03-22 14:44:10

Owner

rspuzio (6075)

Last modified by

rspuzio (6075)

Numerical id

9

Author

rspuzio (6075)

Entry type

Proof

Classification

msc 33B15

Classification

msc 30D30

Generated on Fri Feb 9 20:35:49 2018 by

LaTeXML
error function

error function

The

error function

erf

:

ℂ

→

ℂ

is defined as follows:

erf

⁢

(

z

)

=

2

π

⁢

∫

0

z

e

-

t

2

⁢

𝑑

t

The

complementary error function

erfc

:

ℂ

→

ℂ

is defined as

erfc

⁢

(

z

)

=

2

π

⁢

∫

z

∞

e

-

t

2

⁢

𝑑

t

The name “error function” comes from the role that these

functions

play in the theory of the

normal random variable

. It is also worth noting that the error function is a special case of the confluent hypergeometric functions and of the

Mittag-Leffler function

.

Note.

By

Cauchy integral theorem

(

http://planetmath.org/SecondFormOfCauchyIntegralTheorem

), the choice path of integration in the definition of

erf

is irrelevant since the integrand is an

entire function

. In the definition of

erfc

, the path may be taken to be a half-line parallel to the positive real axis with

endpoint

z

.

Title

error function

Canonical name

ErrorFunction

Date of creation

2013-03-22 14:46:51

Last modified on

2013-03-22 14:46:51

Owner

rspuzio (6075)

Last modified by

rspuzio (6075)

Numerical id

10

Author

rspuzio (6075)

Entry type

Definition

Classification

msc 33B20

Related topic

AreaUnderGaussianCurve

Related topic

ListOfImproperIntegrals

Related topic

UsingConvolutionToFindLaplaceTransform

Defines

complementary error function

Generated on Fri Feb 9 20:35:51 2018 by

LaTeXML
Lambert W function

Lambert W function

Lambert’s

W

function

is the

inverse

of the function

f

:

ℂ

→

ℂ

given by

f

⁢

(

x

)

:=

x

⁢

e

x

. That is,

W

⁢

(

x

)

is the complex valued function that satisfies

W

⁢

(

x

)

⁢

e

W

⁢

(

x

)

=

x

,

for all

x

∈

ℂ

. In practice the

definition

of

W

⁢

(

x

)

requires a branch cut, which is usually taken along the negative real axis. Lambert’s W function is sometimes also called product log function.

This function allow us to solve the

functional equation

g

⁢

(

x

)

g

⁢

(

x

)

=

x

since

g

⁢

(

x

)

=

e

W

⁢

(

ln

⁡

(

x

)

)

.

1

References

A site with good information on Lambert’s W function is Corless’ page

http://kong.apmaths.uwo.ca/ rcorless/frames/PAPERS/LambertW/

“On the

Lambert W Function

”

Title

Lambert W function

Canonical

name

LambertWFunction

Date of creation

2013-03-22 12:40:48

Last modified on

2013-03-22 12:40:48

Owner

drini (3)

Last modified by

drini (3)

Numerical id

8

Author

drini (3)

Entry type

Definition

Classification

msc 33B30

Synonym

product log

Generated on Fri Feb 9 20:35:53 2018 by

LaTeXML
(1+1/n)^n is an increasing sequence

(

1

+

1

/

n

)

n

is an increasing sequence

Theorem 1

.

The

sequence

(

1

+

1

/

n

)

n

is

increasing

.

Proof.

To see this, rewrite

1

+

(

1

/

n

)

=

(

1

+

n

)

/

n

and divide two
consecutive terms of the sequence:

(

1

+

1

n

)

n

(

1

+

1

n

-

1

)

n

-

1

=

(

n

+

1

n

)

n

(

n

n

-

1

)

n

-

1

=

(

(

n

-

1

)

⁢

(

n

+

1

)

n

2

)

n

-

1

⁢

n

+

1

n

=

(

1

-

1

n

2

)

n

-

1

⁢

(

1

+

1

n

)

Since

(

1

-

x

)

n

≥

1

-

n

⁢

x

, we have

(

1

+

1

n

)

n

(

1

+

1

n

-

1

)

n

-

1

≥

(

1

-

n

-

1

n

2

)

⁢

(

1

+

1

n

)

=

1

+

1

n

3

>

1

,

hence the sequence is increasing.
∎

Theorem 2

.

The sequence

(

1

+

1

/

n

)

n

+

1

is decreasing.

Proof.

As before, rewrite

1

+

(

1

/

n

)

=

(

1

+

n

)

/

n

and divide two
consecutive terms of the sequence:

(

1

+

1

n

)

n

+

1

(

1

+

1

n

-

1

)

n

=

(

n

+

1

n

)

n

+

1

(

n

n

-

1

)

n

=

(

(

n

-

1

)

⁢

(

n

+

1

)

n

2

)

n

⁢

n

+

1

n

=

(

1

-

1

n

2

)

n

⁢

(

1

+

1

n

)

Writing

1

+

1

/

n

as

1

+

n

/

n

2

and applying the

inequality

1

+

n

/

n

2

≤

(

1

+

1

/

n

2

)

n

, we obtain

(

1

+

1

n

)

n

+

1

(

1

+

1

n

-

1

)

n

≤

(

1

-

1

n

2

)

n

⁢

(

1

+

1

n

2

)

n

=

(

1

-

1

n

4

)

n

<

1

,

hence the sequence is decreasing.

∎

Theorem 3

.

For all

positive integers

m

and

n

, we have

(

1

+

1

/

m

)

m

<

(

1

+

1

/

n

)

n

+

1

.

Proof.

We consider three cases.

Suppose that

m

=

n

. Since

n

>

0

, we have

1

/

n

>

0

and

1

<

1

+

1

/

n

. Hence,

(

1

+

1

/

n

)

n

<

(

1

+

1

/

n

)

n

+

1

.

Suppose that

m

<

n

. By the previous case,

(

1

+

1

/

n

)

n

<

(

1

+

1

/

n

)

n

+

1

.
By theorem 1,

(

1

+

1

/

m

)

m

<

(

1

+

1

/

n

)

n

. Combining,

(

1

+

1

/

m

)

m

<

(

1

+

1

/

n

)

n

+

1

.

Suppose that

m

>

n

.
By the first case,

(

1

+

1

/

m

)

m

<

(

1

+

1

/

m

)

m

+

1

By theorem 2,

(

1

+

1

/

m

)

m

+

1

<

(

1

+

1

/

n

)

n

+

1

.
Combining,

(

1

+

1

/

m

)

m

<

(

1

+

1

/

n

)

n

+

1

.
∎

Title

(

1

+

1

/

n

)

n

is an

increasing sequence

Canonical name

11nnIsAnIncreasingSequence

Date of creation

2013-03-22 15:48:39

Last modified on

2013-03-22 15:48:39

Owner

rspuzio (6075)

Last modified by

rspuzio (6075)

Numerical id

14

Author

rspuzio (6075)

Entry type

Theorem

Classification

msc 33B99

Generated on Fri Feb 9 20:35:58 2018 by

LaTeXML
bounds for e

bounds for e

If

n

and

m

are

positive integers

and

n

>

m

, we have
the following

inequality

:

(

1

+

1

n

)

n

<

n

n

+

1

⁢

(

1

+

1

m

)

m

+

1

Taking the limit as

n

→

∞

, we obtain an

upper bound

for

e

.
Combining this with the fact that the

(

1

+

1

/

n

)

n

is an

increasing
sequence

, we have the following bounds for

e

:

(

1

+

1

m

)

m

<

e

<

(

1

+

1

m

)

m

+

1

This can be used to show that

e

is not an

integer

– if we take

m

=

5

, we obtain

2.48832

<

e

<

2.985984

, for instance.

Title

bounds for e

Canonical name

BoundsForE

Date of creation

2013-03-22 15:48:48

Last modified on

2013-03-22 15:48:48

Owner

rspuzio (6075)

Last modified by

rspuzio (6075)

Numerical id

7

Author

rspuzio (6075)

Entry type

Theorem

Classification

msc 33B99

Generated on Fri Feb 9 20:36:04 2018 by

LaTeXML
convergence of the sequence (1+1/n)^n

convergence of the sequence (1+1/n)^n

Theorem 1

.

The following

sequence

:

a

n

=

(

1

+

1

n

)

n

(1)

is

convergent

.

Proof.

The proof will be given by demonstrating that the sequence (

1

) is:

1.

monotonic

(increasing), that is

a

n

<

a

n

+

1

2.

bounded above, that is

∀

n

∈

ℕ

,

a

n

<

M

for some

M

>

0

In order to prove part 1, consider the

binomial expansion

for

a

n

:

a

n

=

∑

k

=

0

n

(

n

k

)

⁢

1

n

k

=

∑

k

=

0

n

1

k

!

⁢

n

n

⁢

n

-

1

n

⁢

…

⁢

n

-

(

k

-

1

)

n

=

∑

k

=

0

n

1

k

!

⁢

(

1

-

1

n

)

⁢

…

⁢

(

1

-

k

-

1

n

)

.

Since

∀

i

∈

{

1

,

2

⁢

…

⁢

(

k

-

1

)

}

:

(

1

-

i

n

)

<

(

1

-

i

n

+

1

)

, and since the sum

a

n

+

1

has one term more than

a

n

, it is demonstrated that the sequence (

1

) is monotonic.

In order to prove part 2, consider again the binomial expansion:

a

n

=

1

+

n

n

+

1

2

!

⁢

n

⁢

(

n

-

1

)

n

2

+

1

3

!

⁢

n

⁢

(

n

-

1

)

⁢

(

n

-

2

)

n

3

+

…

+

1

n

!

⁢

n

⁢

(

n

-

1

)

⁢

…

⁢

(

n

-

n

+

1

)

n

n

.

Since

∀

k

∈

{

2

,

3

⁢

…

⁢

n

}

:

1

k

!

<

1

2

k

-

1

and

n

⁢

(

n

-

1

)

⁢

…

⁢

(

n

-

(

k

-

1

)

)

n

k

<

1

:

a

n

<

1

+

(

1

+

1

2

+

1

2

×

2

+

…

+

1

2

n

-

1

)

<

1

+

(

1

-

1

2

n

1

-

1

2

)

<

3

-

1

2

n

-

1

<

3

where the

formula

giving the sum of the

geometric progression

with ratio

1

/

2

has been used.
∎

In

conclusion

, we can say that the sequence (

1

) is convergent and its limit corresponds to the

supremum

of the set

{

a

n

}

⊂

[

2

,

3

)

, denoted by

e

, that is:

lim

n

→

∞

⁡

(

1

+

1

n

)

n

=

sup

n

∈

ℕ

⁡

{

(

1

+

1

n

)

n

}

≜

e

,

which is the definition of the Napier’s constant.

Title

convergence

of the sequence (1+1/n)^n

Canonical name

ConvergenceOfTheSequence11nn

Date of creation

2013-03-22 17:43:26

Last modified on

2013-03-22 17:43:26

Owner

kfgauss70 (18761)

Last modified by

kfgauss70 (18761)

Numerical id

7

Author

kfgauss70 (18761)

Entry type

Theorem

Classification

msc 33B99

Related topic

NondecreasingSequenceWithUpperBound

Generated on Fri Feb 9 20:36:07 2018 by

LaTeXML