aarjaneiro/planetmath huggingface dataset:
proof of Young’s inequality by picture

proof of Young’s inequality by picture

The marked areas correspond to the

integrals

; it is clear that their sum exceeds the area of the rectangle with sides

a

and

b

.

Title

proof of Young’s

inequality

by picture

Canonical name

ProofOfYoungsInequalityByPicture

Date of creation

2013-03-22 14:09:20

Last modified on

2013-03-22 14:09:20

Owner

archibal (4430)

Last modified by

archibal (4430)

Numerical id

5

Author

archibal (4430)

Entry type

Proof

Classification

msc 26D15

Generated on Fri Feb 9 20:02:21 2018 by

LaTeXML
rearrangement inequality

rearrangement inequality

Let

x

1

,

x

2

,

…

,

x

n

and

y

1

,

y

2

,

…

,

y

n

two sequences of positive real numbers.
Then the sum

x

1

⁢

y

1

+

x

2

⁢

y

2

+

⋯

+

x

n

⁢

y

n

is maximized when the two sequences are ordered in the same way (i.e.

x

1

≤

x

2

≤

⋯

≤

x

n

and

y

1

≤

y

2

≤

⋯

≤

y

n

) and is minimized when the two sequences are ordered in the opposite way (i.e.

x

1

≤

x

2

≤

⋯

≤

x

n

and

y

1

≥

y

2

≥

⋯

≥

y

n

).

This can be seen intuitively as:
If

x

1

,

x

2

,

…

,

x

n

are the prices of

n

kinds of items, and

y

1

,

y

2

,

…

,

y

n

the number of units sold of each, then the highest profit is when you sell more items with high prices and fewer items with low prices (same

ordering

), and the lowest profit happens when you sell more items with lower prices and less items with high prices (opposite orders).

Title

rearrangement inequality

Canonical name

RearrangementInequality

Date of creation

2013-03-22 11:47:32

Last modified on

2013-03-22 11:47:32

Owner

drini (3)

Last modified by

drini (3)

Numerical id

9

Author

drini (3)

Entry type

Theorem

Classification

msc 26D15

Classification

msc 20L05

Classification

msc 22A22

Related topic

ChebyshevsInequality

Generated on Fri Feb 9 20:02:23 2018 by

LaTeXML
Schur’s inequality

Schur’s inequality

Theorem

(Schur’s inequality)
Let

A

be a square

n

×

n

matrix with
real (or possibly complex entries).
If

λ

1

,

…

,

λ

n

are the

eigenvalues

of

A

, and

D

is the

diagonal matrix

D

=

diag

⁡

(

λ

1

,

…

,

λ

n

)

,
then

∥

D

∥

F

≤

∥

A

∥

F

,

where

∥

⋅

∥

F

is the

Frobenius matrix norm

.
Equality holds if and only if

A

is a

normal matrix

.

References

1

V.V. Prasolov,

Problems and Theorems in

Linear Algebra

,

American Mathematical Society

, 1994.

Title

Schur’s inequality

Canonical name

SchursInequality

Date of creation

2013-03-22 13:43:30

Last modified on

2013-03-22 13:43:30

Owner

matte (1858)

Last modified by

matte (1858)

Numerical id

14

Author

matte (1858)

Entry type

Theorem

Classification

msc 26D15

Classification

msc 15A42

Related topic

TraceOfAMatrix

Related topic

WielandtHoffmanTheorem

Related topic

FrobeniusMatrixNorm

Generated on Fri Feb 9 20:02:24 2018 by

LaTeXML
Schur’s inequality

Schur’s inequality

If

a

,

b

, and

c

are non-negative

real numbers

and

k

≥

1

is real, then the following

inequality

holds:

a

k

⁢

(

a

-

b

)

⁢

(

a

-

c

)

+

b

k

⁢

(

b

-

c

)

⁢

(

b

-

a

)

+

c

k

⁢

(

c

-

a

)

⁢

(

c

-

b

)

≥

0

Proof.

We can assume without loss of generality that

c

≤

b

≤

a

via a

permutation

of the variables (as both sides are

symmetric

in those variables). Then collecting terms, we wish to show that

(

a

-

b

)

⁢

(

a

k

⁢

(

a

-

c

)

-

b

k

⁢

(

b

-

c

)

)

+

c

k

⁢

(

a

-

c

)

⁢

(

b

-

c

)

≥

0

which is clearly true as every term on the left is positive.∎

There are a couple of special cases worth noting:

•

Taking

k

=

1

, we get the well-known

a

3

+

b

3

+

c

3

+

3

⁢

a

⁢

b

⁢

c

≥

a

⁢

b

⁢

(

a

+

b

)

+

a

⁢

c

⁢

(

a

+

c

)

+

b

⁢

c

⁢

(

b

+

c

)

•

If

c

=

0

, we get

(

a

-

b

)

⁢

(

a

k

+

1

-

b

k

+

1

)

≥

0

.

•

If

b

=

c

=

0

, we get

a

k

+

2

≥

0

.

•

If

b

=

c

, we get

a

k

⁢

(

a

-

c

)

2

≥

0

.

Title

Schur’s inequality

Canonical name

SchursInequality

Date of creation

2013-03-22 13:19:30

Last modified on

2013-03-22 13:19:30

Owner

rspuzio (6075)

Last modified by

rspuzio (6075)

Numerical id

11

Author

rspuzio (6075)

Entry type

Theorem

Classification

msc 26D15

Generated on Fri Feb 9 20:02:26 2018 by

LaTeXML
Young’s inequality

Young’s inequality

Let

ϕ

:

ℝ

→

ℝ

be a

continuous

,

strictly
increasing function

such that

ϕ

⁢

(

0

)

=

0

. Then the following

inequality

holds:

a

⁢

b

≤

∫

0

a

ϕ

⁢

(

x

)

⁢

𝑑

x

+

∫

0

b

ϕ

-

1

⁢

(

y

)

⁢

𝑑

y

Equality only holds when

b

=

ϕ

⁢

(

a

)

.
This inequality can be demonstrated by drawing the graph of

ϕ

⁢

(

x

)

and by observing that the sum of the two areas represented by the

integrals

above is

greater

than the area of a

rectangle

of sides

a

and

b

, as
is illustrated in

http://planetmath.org/node/5575

an attachment.

Title

Young’s inequality

Canonical name

YoungsInequality

Date of creation

2013-03-22 13:19:25

Last modified on

2013-03-22 13:19:25

Owner

rspuzio (6075)

Last modified by

rspuzio (6075)

Numerical id

8

Author

rspuzio (6075)

Entry type

Theorem

Classification

msc 26D15

Related topic

YoungInequality

Generated on Fri Feb 9 20:02:28 2018 by

LaTeXML
another proof of Bernoulli’s inequality

another proof of Bernoulli’s inequality

For fixed

x

>

0

,

x

≠

1

the

function

w

x

⁢

(

r

)

=

∫

1

x

t

r

-

1

⁢

𝑑

t

=

{

x

r

-

1

r

r

≠

0

log

⁡

x

r

=

0

is

strictly increasing

.

Bernoulli inequality

is

equivalent

to

w

1

+

x

⁢

(

r

)

>

(

<

)

⁢

w

1

+

x

⁢

(

1

)

Title

another proof of Bernoulli’s

inequality

Canonical name

AnotherProofOfBernoullisInequality

Date of creation

2013-03-22 15:45:43

Last modified on

2013-03-22 15:45:43

Owner

a4karo (12322)

Last modified by

a4karo (12322)

Numerical id

7

Author

a4karo (12322)

Entry type

Proof

Classification

msc 26D99

Generated on Fri Feb 9 20:02:33 2018 by

LaTeXML
Bernoulli’s inequality

Bernoulli’s inequality

Let

x

and

r

be

real numbers

.
If

0

>

r

>

-

1

or

r

>

1

and

x

>

-

1

then

(

1

+

x

)

r

≥

1

+

x

⁢

r

.

The

inequality

also holds when

r

is an even integer.
For

0

<

r

<

1

the

inverse

inequality holds.

Title

Bernoulli’s inequality

Canonical name

BernoullisInequality

Date of creation

2013-03-22 11:47:28

Last modified on

2013-03-22 11:47:28

Owner

rspuzio (6075)

Last modified by

rspuzio (6075)

Numerical id

12

Author

rspuzio (6075)

Entry type

Theorem

Classification

msc 26D99

Classification

msc 55R40

Classification

msc 55U15

Classification

msc 55T25

Classification

msc 55M05

Classification

msc 55U30

Classification

msc 55U10

Related topic

InequalitiesForDifferencesAndQuotientsOfPowers

Generated on Fri Feb 9 20:02:35 2018 by

LaTeXML
characterizations of majorization

characterizations of majorization

Let

ℰ

n

be the set of all

n

×

n

permutation
matrices

that exchange two

components

. Such matrices have the form

[

⋱

0

1

⋱

1

0

⋱

]

A matrix

T

is called a

Pigou-Dalton transfer

(PDT) if

T

=

α

⁢

I

+

(

1

-

α

)

⁢

E

for some

α

between 0 and 1, and

E

∈

ℰ

n

.

The following are

equivalent

1.

x

is majorized (

http://planetmath.org/

Majorization

) by

y

.

2.

x

=

D

⁢

y

for a

doubly stochastic matrix

D

.

3.

x

=

T

1

⁢

T

2

⁢

⋯

⁢

T

k

⁢

y

for finitely many PDT

T

1

,

…

,

T

k

.

4.

∑

i

=

1

n

θ

⁢

(

x

i

)

≤

∑

i

=

1

n

θ

⁢

(

y

i

)

for
all

convex function

θ

.

5.

x

lies in the convex hull whose

vertex set

is

{

(

y

π

⁢

(

1

)

,

y

π

⁢

(

2

)

,

…

,

y

π

⁢

(

n

)

)

:

π

⁢

is a permutation of

⁢

{

1

,

…

,

n

}

}

.

6.

For any

n

non-negative

real numbers

a

1

,

…

,

a

n

,

∑

π

a

1

x

π

⁢

(

1

)

⁢

a

2

x

π

⁢

(

2

)

⁢

⋯

⁢

a

n

x

π

⁢

(

n

)

≤

∑

π

a

1

y

π

⁢

(

1

)

⁢

a

2

y

π

⁢

(

2

)

⁢

⋯

⁢

a

n

y

π

⁢

(

n

)

where

summation

is taken over all

permutations

of

{

1

,

…

,

n

}

.

The equivalence of the above conditions are due to Hardy,
Littlewood, Pólya, Birkhoff, von Neumann and Muirhead.

Reference

•

G. H. Hardy, J. E. Littlewood and G. Pólya,

Inequalities

, 2nd edition, 1952, Cambridge University Press,
London.

•

A. W. Marshall and I. Olkin,

Inequalities: Theory of
Majorization and Its Applications

, 1979, Acadamic Press, New
York.

Title

characterizations of majorization

Canonical name

CharacterizationsOfMajorization

Date of creation

2013-03-22 15:26:37

Last modified on

2013-03-22 15:26:37

Owner

Mathprof (13753)

Last modified by

Mathprof (13753)

Numerical id

7

Author

Mathprof (13753)

Entry type

Theorem

Classification

msc 26D99

Related topic

BirkoffVonNeumannTheorem

Related topic

MuirheadsTheorem

Defines

Pigou-Dalton transfer

Generated on Fri Feb 9 20:02:37 2018 by

LaTeXML
inequalities for differences of powers

inequalities for differences of powers

Oftentimes, one needs to estimate differences of powers of

real numbers

.
The following

inequalities

are useful for this purpose:

n

⁢

(

u

-

v

)

⁢

v

n

-

1

<

u

n

-

v

n

<

n

⁢

(

u

-

v

)

⁢

u

n

-

1

n

⁢

x

≤

(

1

+

x

)

n

-

1

(

1

+

x

)

n

-

1

≤

n

⁢

x

1

-

(

n

-

1

)

⁢

x

Here

n

is an integer

greater

than 1.
The first inequality holds when

0

<

v

<

u

, the second inequality holds
when

-

1

<

x

, and the third inequality holds when

-

1

<

x

<

1

/

(

n

-

1

)

.
Equality can only occur in the latter two inequalities when

x

=

0

.

Title

inequalities for differences of powers

Canonical name

InequalitiesForDifferencesOfPowers

Date of creation

2013-03-22 15:48:42

Last modified on

2013-03-22 15:48:42

Owner

rspuzio (6075)

Last modified by

rspuzio (6075)

Numerical id

11

Author

rspuzio (6075)

Entry type

Theorem

Classification

msc 26D99

Related topic

BernoullisInequality

Generated on Fri Feb 9 20:02:39 2018 by

LaTeXML
limit of (1+s_n)^n is one when limit of n⁢s_n is zero

limit of

(

1

+

s

n

)

n

is one when limit of

n

⁢

s

n

is zero

The

inequalities for differences of powers

may be used to show that

lim

n

→

∞

⁡

(

1

+

s

n

)

n

=

1

when

lim

n

→

∞

⁡

n

⁢

s

n

=

0

.
This fact plays an important role in the development of the theory of the

exponential function

as a limit of powers.

To derive this limit, we bound

1

+

s

n

using the inequalities for differences
of powers.

n

⁢

s

n

≤

(

1

+

s

n

)

n

-

1

≤

n

⁢

s

n

1

-

(

n

-

1

)

⁢

s

n

Since

lim

n

→

∞

⁡

n

⁢

s

n

=

0

, there must exist

N

such that

n

⁢

s

n

<

1

/

2

when

n

>

N

. Hence, when

n

>

N

,

|

(

1

+

s

n

)

n

-

1

|

<

2

⁢

|

n

⁢

s

n

|

so, as

n

→

∞

, we have

(

1

+

s

n

)

n

→

1

.

Title

limit of

(

1

+

s

n

)

n

is one when limit of

n

⁢

s

n

is zero

Canonical name

LimitOf1SnnIsOneWhenLimitOfNSnIsZero

Date of creation

2013-03-22 15:48:55

Last modified on

2013-03-22 15:48:55

Owner

rspuzio (6075)

Last modified by

rspuzio (6075)

Numerical id

7

Author

rspuzio (6075)

Entry type

Proof

Classification

msc 26D99

Generated on Fri Feb 9 20:02:40 2018 by

LaTeXML
majorization

majorization

For any real vector

x

=

(

x

1

,

x

2

,

…

,

x

n

)

∈

ℝ

n

, let

x

(

1

)

≥

x

(

2

)

≥

⋯

≥

x

(

n

)

denote the

components

of

x

in non-increasing order.

For

x

,

y

∈

ℝ

n

, we say that

x

is majorized by

y

, or

y

majorizes

x

, if

∑

i

=

1

m

x

(

i

)

≤

∑

i

=

1

m

y

(

i

)

,

for

m

=

1

,

…

,

n

-

1

, and

∑

i

=

1

n

x

(

i

)

=

∑

i

=

1

n

y

(

i

)

A common notation for “

x

is majorized by

y

” is

x

≺

y

.

Remark:

A canonical example is that, if

y

1

,

y

2

,

…

,

y

n

are non-negative

real numbers

such that their sum is equal to 1, then

(

1

n

,

…

,

1

n

)

≺

(

y

1

,

…

,

y

n

)

.

In general,

x

≺

y

vaguely means that the components of

x

is less spread out than are the components of

y

.

Reference

•

G. H. Hardy, J. E. Littlewood and G. Pólya,

Inequalities

, 2nd edition, 1952, Cambridge University Press, London.

•

A. W. Marshall and I. Olkin,

Inequalities: Theory of Majorization and Its Applications

, 1979, Acadamic Press, New York.

Title

majorization

Canonical name

Majorization

Date of creation

2013-03-22 14:30:22

Last modified on

2013-03-22 14:30:22

Owner

rspuzio (6075)

Last modified by

rspuzio (6075)

Numerical id

8

Author

rspuzio (6075)

Entry type

Definition

Classification

msc 26D99

Defines

majorize

Defines

majorization

Generated on Fri Feb 9 20:02:43 2018 by

LaTeXML
proof of Bernoulli’s inequality

proof of Bernoulli’s inequality

Let

I

be the

interval

(

-

1

,

∞

)

and

f

:

I

→

ℝ

the

function

defined as:

f

⁢

(

x

)

=

(

1

+

x

)

α

-

1

-

α

⁢

x

with

α

∈

ℝ

∖

{

0

,

1

}

fixed.
Then

f

is

differentiable

and its

derivative

is

f

′

⁢

(

x

)

=

α

⁢

(

1

+

x

)

α

-

1

-

α

,

for all

⁢

x

∈

I

,

from which it follows that

f

′

⁢

(

x

)

=

0

⇔

x

=

0

.

1.

If

0

<

α

<

1

then

f

′

⁢

(

x

)

<

0

for all

x

∈

(

0

,

∞

)

and

f

′

⁢

(

x

)

>

0

for all

x

∈

(

-

1

,

0

)

which means that

0

is a

global maximum

point for

f

.
Therefore

f

⁢

(

x

)

<

f

⁢

(

0

)

for all

x

∈

I

∖

{

0

}

which means that

(

1

+

x

)

α

<

1

+

α

⁢

x

for all

x

∈

(

-

1

,

0

)

.

2.

If

α

∉

[

0

,

1

]

then

f

′

⁢

(

x

)

>

0

for all

x

∈

(

0

,

∞

)

and

f

′

⁢

(

x

)

<

0

for all

x

∈

(

-

1

,

0

)

meaning that

0

is a

global
minimum

point for

f

.
This implies that

f

⁢

(

x

)

>

f

⁢

(

0

)

for all

x

∈

I

∖

{

0

}

which means that

(

1

+

x

)

α

>

1

+

α

⁢

x

for all

x

∈

(

-

1

,

0

)

.

Checking that the equality is satisfied for

x

=

0

or for

α

∈

{

0

,

1

}

ends the proof.

Title

proof of Bernoulli’s

inequality

Canonical name

ProofOfBernoullisInequality

Date of creation

2013-03-22 12:38:14

Last modified on

2013-03-22 12:38:14

Owner

danielm (240)

Last modified by

danielm (240)

Numerical id

6

Author

danielm (240)

Entry type

Proof

Classification

msc 26D99

Generated on Fri Feb 9 20:02:45 2018 by

LaTeXML
proof of Bernoulli’s inequality employing the mean value theorem

proof of Bernoulli’s inequality employing the mean value theorem

Let us take as our

assumption

that

x

∈

I

=

(

-

1

,

∞

)

and that

r

∈

J

=

(

0

,

∞

)

. Observe
that if

x

=

0

the

inequality

holds quite obviously. Let us now
consider the case where

x

≠

0

. Consider now the

function

f

:

I

⁢

x

⁢

J

→

ℝ

given by

f

⁢

(

x

,

r

)

=

(

1

+

x

)

r

-

1

-

r

⁢

x

Observe that for all

r

in

J

fixed,

f

is, indeed,

differentiable

on

I

. In particular,

∂

∂

⁡

x

⁢

f

⁢

(

x

,

r

)

=

r

⁢

(

1

+

x

)

r

-

1

-

r

Consider
two points

a

≠

0

in

I

and

0

in

I

. Then clearly by the
mean value theorem, for any arbitrary, fixed

α

in

J

, there
exists a

c

in

I

such that,

f

x

′

⁢

(

c

,

α

)

=

f

⁢

(

a

,

α

)

-

f

⁢

(

0

,

α

)

a

⇔

f

x

′

⁢

f

⁢

(

c

,

α

)

=

(

1

+

a

)

α

-

1

-

α

⁢

a

a

(1)

Since

α

is in

J

, it is clear that if

a

<

0

, then

f

x

′

⁢

(

a

,

α

)

<

0

and, accordingly, if

a

>

0

then

f

x

′

⁢

(

a

,

α

)

>

0

Thus, in either case, from

1

we deduce that

(

1

+

a

)

α

-

1

-

α

⁢

a

a

<

0

if

a

<

0

and

(

1

+

a

)

α

-

1

-

α

⁢

a

a

>

0

if

a

>

0

. From this we
conclude that, in either case,

(

1

+

a

)

α

-

1

-

α

⁢

a

>

0

. That
is,

(

1

+

a

)

α

>

1

+

α

⁢

a

for all choices of

a

in

I

-

{

0

}

and all choices of

α

in

J

. If

a

=

0

in

I

, we have

(

1

+

a

)

α

=

1

+

α

⁢

a

for all choices of

α

in

J

. Generally, for all

x

in

I

and all

r

in

J

we
have:

(

1

+

x

)

r

≥

1

+

r

⁢

x

This

completes

the proof.

Notice that if

r

is in

(

-

1

,

0

)

then the inequality would be reversed. That is:

(

1

+

x

)

r

≤

1

+

r

⁢

x

. This can be proved using exactly the same method, by fixing

α

in the proof above in

(

-

1

,

0

)

.

Title

proof of Bernoulli’s inequality employing the mean value theorem

Canonical name

ProofOfBernoullisInequalityEmployingTheMeanValueTheorem

Date of creation

2013-03-22 15:49:53

Last modified on

2013-03-22 15:49:53

Owner

rspuzio (6075)

Last modified by

rspuzio (6075)

Numerical id

10

Author

rspuzio (6075)

Entry type

Proof

Classification

msc 26D99

Generated on Fri Feb 9 20:02:47 2018 by

LaTeXML
proof of inequalities for difference of powers

proof of inequalities for difference of powers

1

First Inequality

We have the factorization

u

n

-

v

n

=

(

u

-

v

)

⁢

∑

k

=

0

n

-

1

u

k

⁢

v

n

-

k

-

1

.

Since the largest term in the sum is is

u

n

-

1

and the smallest is

v

n

-

1

, and there are

n

terms in the sum, we deduce the following

inequalities

:

n

⁢

(

u

-

v

)

⁢

v

n

-

1

<

u

n

-

v

n

<

n

⁢

(

u

-

v

)

⁢

u

n

-

1

2

Second Inequality

This inequality is trivial when

x

=

0

. We split the rest of
the proof into two cases.

2.1

-

1

<

x

<

0

In this case, we set

u

=

1

and

v

=

1

+

x

in the second
inequality above:

1

-

(

1

+

x

)

n

<

n

⁢

(

-

x

)

Reversing the signs of both sides yields

n

⁢

x

<

(

1

+

x

)

n

-

1

2.2

0

<

x

In this case, we set

u

=

1

+

x

and

v

=

1

in the first
inequality above:

n

⁢

x

<

(

1

+

x

)

n

-

1

3

Third Inequality

This inequality is trivial when

x

=

0

. We split the rest of
the proof into two cases.

3.1

-

1

<

x

<

0

Start with the first

inequality for differences of powers

, expand
the left-hand side,

n

⁢

u

⁢

v

n

-

1

-

n

⁢

v

n

<

u

n

-

v

n

,

move the

v

n

to the other side of the inequality,

n

⁢

u

⁢

v

n

-

1

-

(

n

-

1

)

⁢

v

n

<

u

n

,

and divide by

v

n

to obtain

n

⁢

u

v

-

n

+

1

<

(

u

v

)

n

.

Taking the

reciprocal

, we obtain

(

v

u

)

n

<

v

v

+

n

⁢

(

u

-

v

)

=

1

-

n

⁢

(

u

-

v

)

v

+

n

⁢

(

u

-

v

)

Setting

u

=

1

and

v

=

1

+

x

, and moving a term from one side
to the other, this becomes

(

1

+

x

)

n

-

1

<

n

⁢

x

1

-

(

n

-

1

)

⁢

x

.

3.2

0

<

x

<

1

/

(

n

-

1

)

Start with the second inequality for differences of powers, expand
the right-hand side,

u

n

-

v

n

<

n

⁢

u

n

-

n

⁢

u

n

-

1

⁢

v

move terms from one side of the inequality to the other,

n

⁢

u

n

-

1

⁢

v

-

(

n

-

1

)

⁢

u

n

<

v

n

and divide by

u

n

to obtain

n

⁢

v

u

-

n

+

1

<

(

v

u

)

n

When the left-hand side is positive, (i.e.

n

⁢

v

>

(

n

-

1

)

⁢

u

)
we can take the reciprocal:

(

u

v

)

n

<

u

u

-

n

⁢

(

u

-

v

)

=

1

+

n

⁢

(

u

-

v

)

u

-

n

⁢

(

u

-

v

)

Setting

u

=

1

+

x

and

v

=

1

, and moving a term from one side
to the other, this becomes

(

1

+

x

)

n

-

1

<

n

⁢

x

1

-

(

n

-

1

)

⁢

x

and the positivity condition mentioned above becomes

(

n

-

1

)

⁢

x

<

1

.

Title

proof of inequalities for difference of powers

Canonical name

ProofOfInequalitiesForDifferenceOfPowers

Date of creation

2013-03-22 15:48:45

Last modified on

2013-03-22 15:48:45

Owner

Mathprof (13753)

Last modified by

Mathprof (13753)

Numerical id

14

Author

Mathprof (13753)

Entry type

Proof

Classification

msc 26D99

Generated on Fri Feb 9 20:02:50 2018 by

LaTeXML
test newcommand and nested lists

test newcommand and nested lists

Here’s my list:

I.

Item 1.

I. a

List 2, Item 1

I. b

List 2, Item 2

II.

Item 2.

III.

Item 3.

Title

test newcommand and nested lists

Canonical

name

TestNewcommandAndNestedLists

Date of creation

2013-03-22 18:51:38

Last modified on

2013-03-22 18:51:38

Owner

Mathprof (13753)

Last modified by

Mathprof (13753)

Numerical id

4

Author

Mathprof (13753)

Entry type

Definition

Classification

msc 26D99

Generated on Fri Feb 9 20:02:52 2018 by

LaTeXML
example of infinite hyperreal number

example of infinite hyperreal number

The

hyperreal number

{

n

}

n

∈

ℕ

∈

ℝ

*

is (or ).

Proof :

Let

ℱ

be the

nonprincipal ultrafilter

in the entry (

http://planetmath.org/Hyperreal

).

Given any positive

a

∈

ℝ

we have that

{

n

∈

ℕ

:

n

≤

a

}

is finite,
so

{

n

∈

ℕ

:

a

<

n

}

∈

ℱ

and therefore

{

a

}

n

∈

ℕ

<

{

n

}

n

∈

ℕ

.

Thus

{

n

}

n

∈

ℕ

is

infinite

.

□

Title

example of infinite hyperreal number

Canonical name

ExampleOfInfiniteHyperrealNumber

Date of creation

2013-03-22 17:25:59

Last modified on

2013-03-22 17:25:59

Owner

asteroid (17536)

Last modified by

asteroid (17536)

Numerical id

5

Author

asteroid (17536)

Entry type

Example

Classification

msc 26E35

Generated on Fri Feb 9 20:03:14 2018 by

LaTeXML
example of infinitesimal hyperreal number

example of infinitesimal hyperreal number

The

hyperreal number

{

1

n

}

n

∈

ℕ

∈

ℝ

*

is

infinitesimal

.

Proof -

Let

ℱ

be the

nonprincipal ultrafilter

in the entry (

http://planetmath.org/Hyperreal

).

{

n

∈

ℕ

:

0

<

1

n

}

=

ℕ

∈

ℱ

so

0

<

{

1

n

}

n

∈

ℕ

.

Given any

positive

a

∈

ℝ

we have that

{

n

∈

ℕ

:

a

≤

1

n

}

is finite,
so

{

n

∈

ℕ

:

1

n

<

a

}

∈

ℱ

and therefore

{

1

n

}

n

∈

ℕ

<

{

a

}

n

∈

ℕ

.

Thus

0

<

{

1

n

}

n

∈

ℕ

<

{

a

}

n

∈

ℕ

for every positive real number

{

a

}

n

∈

ℕ

∈

ℝ

, and so

{

1

n

}

n

∈

ℕ

is infinitesimal.

□

Title

example of infinitesimal hyperreal number

Canonical name

ExampleOfInfinitesimalHyperrealNumber

Date of creation

2013-03-22 17:25:57

Last modified on

2013-03-22 17:25:57

Owner

asteroid (17536)

Last modified by

asteroid (17536)

Numerical id

4

Author

asteroid (17536)

Entry type

Example

Classification

msc 26E35

Generated on Fri Feb 9 20:03:16 2018 by

LaTeXML