aarjaneiro/planetmath huggingface dataset:
speediest inclined plane

speediest inclined plane

We set the problem, how great must be the

difference

in altitude of the top and the bottom of an inclined plane in that a little ball would frictionlessly roll the whole length of the plane as soon as possible
(cf. the

brachistochrone problem

(

http://planetmath.org/CalculusOfVariations

)).  It is assumed that the

http://planetmath.org/node/9475

projection of the length on a horizontal plane has a given value

b

.

Using notations of mechanics, we can write

F

=

m

⁢

a

=

m

⁢

g

⁢

sin

⁡

α

=

m

⁢

g

⁢

x

x

2

+

b

2

,

x

2

+

b

2

=

s

=

1

2

⁢

t

2

⁢

a

=

t

2

2

⋅

g

⁢

x

x

2

+

b

2

.

Thus we get the

function

t

2

=

2

g

⋅

x

2

+

b

2

x

=

:

f

(

x

)

(

x

>

0

)

,

the

absolute minimum

point of which is to be found.  This function is

differentiable

, and its

derivative

is

f

′

⁢

(

x

)

=

2

g

⋅

x

2

-

b

2

x

2

.

The only zero of

f

′

⁢

(

x

)

is

x

=

b

,  where the sign changes from minus to plus as

x

increases.  It means that

x

=

b

is the searched

minimum

point.  The difference in altitude is thus equal to the

http://planetmath.org/node/11642

base, and the inclination

α

must be

45

∘

.

Title

speediest inclined plane

Canonical name

SpeediestInclinedPlane

Date of creation

2013-03-22 19:19:11

Last modified on

2013-03-22 19:19:11

Owner

pahio (2872)

Last modified by

pahio (2872)

Numerical id

9

Author

pahio (2872)

Entry type

Example

Classification

msc 26A09

Classification

msc 26A06

Related topic

Extremum

Related topic

CalculusOfVariations

Related topic

BrachistochroneCurve

Generated on Fri Feb 9 19:47:42 2018 by

LaTeXML
tangent of halved angle

tangent of halved angle

The formulae

cos

⁡

2

⁢

α

=

1

-

2

⁢

sin

2

⁡

α

,

cos

⁡

2

⁢

α

=

2

⁢

cos

2

⁡

α

-

1

may be solved for

sin

⁡

α

and

cos

⁡

α

, respectively.  One gets the equations

sin

⁡

α

=

±

1

-

cos

⁡

2

⁢

α

2

,

cos

⁡

α

=

±

1

+

cos

⁡

2

⁢

α

2

,

where the signs have to be chosen according to the quadrant where the angle

α

is.  Changing

α

to

x

2

and dividing these equations gives us the formula

tan

⁡

x

2

=

±

1

-

cos

⁡

x

1

+

cos

⁡

x

.

(1)

Also here one must chose the sign according to the quadrant of

x

2

.

We obtain two alternative forms of (1) when we multiply both the numerator and
the denominator of the radicand the first time by

1

-

cos

⁡

x

and the
second time by

1

+

cos

⁡

x

; note that

1

-

cos

2

⁡

x

=

sin

2

⁡

x

:

tan

⁡

x

2

=

1

-

cos

⁡

x

sin

⁡

x

,

(2)

tan

⁡

x

2

=

sin

⁡

x

1

+

cos

⁡

x

(3)

Here,

sin

⁡

x

determines the sign of the hand sides;
it can be justified that it has always the same sign as

tan

⁡

x

2

.

Title

tangent of halved angle

Canonical name

TangentOfHalvedAngle

Date of creation

2013-03-22 17:00:32

Last modified on

2013-03-22 17:00:32

Owner

pahio (2872)

Last modified by

pahio (2872)

Numerical id

9

Author

pahio (2872)

Entry type

Derivation

Classification

msc 26A09

Related topic

DerivationOfHalfAngleFormulaeForTangent

Related topic

GoniometricFormulae

Generated on Fri Feb 9 19:47:44 2018 by

LaTeXML
Taylor expansion of sqrt(1+x)

Taylor expansion of

1

+

x

The

Taylor series

for

f

⁢

(

x

)

=

1

+

x

using the

T

⁢

(

x

)

=

∑

k

=

0

∞

f

(

k

)

⁢

(

a

)

k

!

⁢

(

x

-

a

)

k

is given in the table below for the first few .

k

expansion

simplified

at

a

=

0

0

f

⁢

(

a

)

(

1

+

a

)

1

/

2

1

1

f

′

⁢

(

a

)

⁢

(

x

-

a

)

1

2

⁢

(

1

+

a

)

-

1

/

2

⁢

(

x

-

a

)

1

2

⁢

x

2

f

(

2

)

⁢

(

a

)

2

!

⁢

(

x

-

a

)

2

-

1

8

⁢

(

1

+

a

)

-

3

/

2

⁢

(

x

-

a

)

2

-

1

8

⁢

x

2

3

f

(

3

)

⁢

(

a

)

3

!

⁢

(

x

-

a

)

3

3

48

⁢

(

1

+

a

)

-

5

/

2

⁢

(

x

-

a

)

3

1

16

⁢

x

3

4

f

(

4

)

⁢

(

a

)

4

!

⁢

(

x

-

a

)

4

-

15

384

⁢

(

1

+

a

)

-

7

/

2

⁢

(

x

-

a

)

4

-

5

128

⁢

x

4

5

f

(

5

)

⁢

(

a

)

5

!

⁢

(

x

-

a

)

5

105

3840

⁢

(

1

+

a

)

-

9

/

2

⁢

(

x

-

a

)

5

7

256

⁢

x

5

6

f

(

6

)

⁢

(

a

)

6

!

⁢

(

x

-

a

)

6

-

945

46080

⁢

(

1

+

a

)

-

11

/

2

⁢

(

x

-

a

)

6

-

21

1024

⁢

x

6

Table 1:

Taylor Series for

f

⁢

(

x

)

=

1

+

x

The general

coefficient

of the expansion, for

n

≥

2

is:

f

(

n

)

⁢

(

a

)

n

!

=

(

1

2

n

)

⁢

(

1

+

a

)

1

2

-

n

=

1

2

⁢

(

-

1

2

)

⁢

(

-

3

2

)

⁢

⋯

⁢

(

1

2

-

(

n

-

1

)

)

n

⁢

(

n

-

1

)

⁢

(

n

-

2

)

⁢

⋯

⁢

1

(

1

+

a

)

-

2

⁢

n

-

1

2

=

1

2

⁢

(

-

1

2

)

n

-

1

1

⋅

3

⁢

⋯

⁢

(

2

⁢

n

-

3

)

n

⁢

(

n

-

1

)

⁢

(

n

-

2

)

⁢

⋯

⁢

1

(

1

+

a

)

-

2

⁢

n

-

1

2

=

(

-

1

)

n

-

1

2

n

⁢

n

!

⁢

(

2

⁢

n

-

3

)

!

(

2

⁢

n

-

4

)

⁢

(

2

⁢

n

-

6

)

⁢

⋯

⁢

2

(

1

+

a

)

-

2

⁢

n

-

1

2

=

(

-

1

)

n

-

1

⁢

(

2

⁢

n

-

3

)

!

2

2

⁢

n

-

2

⁢

n

!

⁢

(

n

-

2

)

!

(

1

+

a

)

-

2

⁢

n

-

1

2

.

Title

Taylor expansion

of

1

+

x

Canonical name

TaylorExpansionOfsqrt1x

Date of creation

2013-03-22 15:45:52

Last modified on

2013-03-22 15:45:52

Owner

stevecheng (10074)

Last modified by

stevecheng (10074)

Numerical id

11

Author

stevecheng (10074)

Entry type

Example

Classification

msc 26A09

Related topic

ExamplesOnHowToFindTaylorSeriesFromOtherKnownSeries

Generated on Fri Feb 9 19:47:47 2018 by

LaTeXML
trigonometric identity involving product of sines of roots of unity

trigonometric identity involving product of sines of roots of unity

Let

n

>

1

be a

positive integer

, and

ζ

n

=

e

2

⁢

i

⁢

π

/

n

, a

primitive

n

th

root of unity

.

The purpose of this article is to prove

Theorem 1

.

Let

m

=

⌊

n

2

⌋

. Then

∏

k

=

1

m

sin

2

⁡

(

π

⁢

k

n

)

=

∏

k

=

1

n

-

1

sin

⁡

(

π

⁢

k

n

)

=

n

2

n

-

1

(1)

The theorem follows easily from the following simple lemma:

Lemma 2

.

Let

n

>

1

be a positive integer. Then

∏

k

=

1

n

-

1

(

1

-

ζ

n

k

)

=

n

Proof.

We have

x

n

-

1

=

∏

k

=

1

n

(

x

-

ζ

n

k

)

. Dividing both sides by

x

-

1

gives

x

n

-

1

x

-

1

=

1

+

x

+

x

2

+

…

⁢

x

n

-

1

=

∏

k

=

1

n

-

1

(

x

-

ζ

n

k

)

Substitute

x

=

1

to get the result.
∎

Proof of Theorem

1

.

Using the definition of

ζ

n

and the half-angle formulas, we have

1

-

ζ

n

k

=

1

-

cos

⁡

(

2

⁢

π

⁢

k

n

)

-

i

⁢

sin

⁡

(

2

⁢

π

⁢

k

n

)

=

2

⁢

sin

2

⁡

(

π

⁢

k

n

)

-

2

⁢

i

⁢

sin

⁡

(

π

⁢

k

n

)

⁢

cos

⁡

(

π

⁢

k

n

)

=

2

⁢

sin

⁡

(

π

⁢

k

n

)

⁢

(

sin

⁡

(

π

⁢

k

n

)

-

i

⁢

cos

⁡

(

π

⁢

k

n

)

)

Note that

|

sin

⁡

θ

-

i

⁢

cos

⁡

θ

|

=

sin

2

⁡

θ

+

cos

2

⁡

θ

=

1

, so taking

absolute values

, we get

|

1

-

ζ

n

k

|

=

2

⁢

|

sin

⁡

(

π

⁢

k

n

)

|

Now, for

1

≤

k

≤

n

-

1

,

sin

⁡

(

π

⁢

k

n

)

>

0

so is equal to its absolute value. Thus (using, for

n

even, the fact that

sin

⁡

π

2

=

1

),

∏

k

=

1

m

sin

2

⁡

(

π

⁢

k

n

)

=

∏

k

=

1

m

sin

⁡

(

π

⁢

k

n

)

⁢

sin

⁡

(

π

-

π

⁢

k

n

)

=

∏

k

=

1

m

sin

⁡

(

π

⁢

k

n

)

⁢

sin

⁡

(

π

⁢

(

n

-

k

)

n

)

=

∏

k

=

1

n

-

1

sin

⁡

(

π

⁢

k

n

)

=

∏

k

=

1

n

-

1

|

sin

⁡

(

π

⁢

k

n

)

|

=

1

2

n

-

1

⁢

|

∏

k

=

1

n

-

1

(

1

-

ζ

n

k

)

|

=

1

2

n

-

1

⁢

|

n

|

=

n

2

n

-

1

∎

(Thanks to dh2718 for greatly simplifying the original proof.)

Title

trigonometric identity involving product of sines of roots of unity

Canonical name

TrigonometricIdentityInvolvingProductOfSinesOfRootsOfUnity

Date of creation

2013-03-22 19:00:03

Last modified on

2013-03-22 19:00:03

Owner

rm50 (10146)

Last modified by

rm50 (10146)

Numerical id

11

Author

rm50 (10146)

Entry type

Theorem

Classification

msc 26A09

Classification

msc 33B10

Generated on Fri Feb 9 19:47:53 2018 by

LaTeXML
formal definition of Landau notation

formal definition of Landau notation

Let us consider a domain

D

and an

accumulation point

x

0

∈

D

¯

. Important examples are

D

=

ℝ

and

x

0

∈

D

or

D

=

ℕ

and

x

0

=

+

∞

. Let

f

:

D

→

ℝ

be any

function

. We are going to define the spaces

o

⁢

(

f

)

and

O

⁢

(

f

)

which are families of

real functions

defined on

D

and which depend on the point

x

0

∈

D

¯

.

Suppose first that there exists a

neighbourhood

U

of

x

0

such that

f

restricted to

U

∩

D

is always different from zero.
We say that

g

∈

o

⁢

(

f

)

as

x

→

x

0

if

lim

x

→

x

0

⁡

g

⁢

(

x

)

f

⁢

(

x

)

=

0

.

We say that

g

∈

O

⁢

(

f

)

as

x

→

x

0

if there exists a neighbourhood

U

of

x

0

such that

g

⁢

(

x

)

f

⁢

(

x

)

⁢

is bounded if restricted to

D

∩

U

.

In the case when

f

≡

0

in a neighbourhood of

x

0

, we define

o

⁢

(

f

)

=

O

⁢

(

f

)

as the set of all functions

g

which are null in a neighbourhood of

0

.

The families

o

and

O

are usually called ”

small-o

” and ”big-o” or, sometimes,
”small ordo”, ”big ordo”.

Title

formal definition of Landau notation

Canonical name

FormalDefinitionOfLandauNotation

Date of creation

2013-03-22 15:15:48

Last modified on

2013-03-22 15:15:48

Owner

paolini (1187)

Last modified by

paolini (1187)

Numerical id

6

Author

paolini (1187)

Entry type

Definition

Classification

msc 26A12

Synonym

Landau notation

Synonym

small o

Synonym

big o

Synonym

order of infinity

Synonym

order of zero

Related topic

PropertiesOfOAndO

Generated on Fri Feb 9 19:47:59 2018 by

LaTeXML
growth of exponential function

growth of exponential function

Lemma.

lim

x

→

∞

⁡

x

a

e

x

=

0

for all values of

a

.

Proof.

Let

ε

be any positive number.  Then we get:

0

<

x

a

e

x

≦

x

⌈

a

⌉

e

x

<

x

⌈

a

⌉

x

⌈

a

⌉

+

1

(

⌈

a

⌉

+

1

)

!

=

(

⌈

a

⌉

+

1

)

!

x

<

ε

as soon as

x

>

max

⁡

{

1

,

(

⌈

a

⌉

+

1

)

!

ε

}

.  Here,

⌈

⋅

⌉

the

ceiling function

;

e

x

has been estimated downwards by taking only one of the all positive

e

x

=

1

+

x

1

!

+

x

2

2

!

+

⋯

+

x

n

n

!

+

⋯

Theorem

.

The of the real

exponential function

x

↦

b

x

exceeds all

power functions

, i.e.

lim

x

→

∞

⁡

x

a

b

x

=

0

with

a

and

b

any ,

b

>

1

.

Proof.

Since

ln

⁡

b

>

0

,  we obtain by using the lemma the result

lim

x

→

∞

⁡

x

a

b

x

=

lim

x

→

∞

⁡

(

x

a

ln

⁡

b

e

x

)

ln

⁡

b

=

0

ln

⁡

b

=

0

.

Corollary 1.

lim

x

→

0

+

⁡

x

⁢

ln

⁡

x

=

0

.

Proof.

According to the lemma we get

0

=

lim

u

→

∞

⁡

-

u

e

u

=

lim

x

→

0

+

⁡

-

ln

⁡

1

x

1

x

=

lim

x

→

0

+

⁡

x

⁢

ln

⁡

x

.

Corollary 2.

lim

x

→

∞

⁡

ln

⁡

x

x

=

0

.

Proof.

Change in the lemma

x

to

ln

⁡

x

.

Corollary 3.

lim

x

→

∞

⁡

x

1

x

=

1

.

(Cf.

limit of nth root of n

.)

Proof.

By corollary 2, we can write:

x

1

x

=

e

ln

⁡

x

x

⟶

e

0

=

1

as

x

→

∞

(see also theorem 2 in

limit rules of functions

).

Title

growth of exponential function

Canonical name

GrowthOfExponentialFunction

Date of creation

2013-03-22 14:51:32

Last modified on

2013-03-22 14:51:32

Owner

pahio (2872)

Last modified by

pahio (2872)

Numerical id

18

Author

pahio (2872)

Entry type

Theorem

Classification

msc 26A12

Classification

msc 26A06

Related topic

MaximalNumber

Related topic

LimitRulesOfFunctions

Related topic

NaturalLogarithm

Related topic

AsymptoticBoundsForFactorial

Related topic

MinimalAndMaximalNumber

Related topic

FunctionXx

Related topic

Growth

Related topic

LimitsOfNaturalLogarithm

Related topic

DerivativeOfLimitFunctionDivergesFromLimitOfDerivatives

Generated on Fri Feb 9 19:48:01 2018 by

LaTeXML
Landau notation

Landau notation

Given two

functions

f

and

g

from

ℝ

+

to

ℝ

+

,
the notation

f

=

O

⁢

(

g

)

means that the ratio

f

⁢

(

x

)

g

⁢

(

x

)

stays

bounded

as

x

→

∞

. If moreover that ratio approaches zero,
we write

f

=

o

⁢

(

g

)

.

It is legitimate to write, say,

2

⁢

x

=

O

⁢

(

x

)

=

O

⁢

(

x

2

)

, with the understanding
that we are using the equality sign in an unsymmetric (and informal) way,
in that we do not have, for example,

O

⁢

(

x

2

)

=

O

⁢

(

x

)

.

The notation

f

=

Ω

⁢

(

g

)

means that the ratio

f

⁢

(

x

)

g

⁢

(

x

)

is bounded away from zero as

x

→

∞

, or equivalently

g

=

O

⁢

(

f

)

.

If both

f

=

O

⁢

(

g

)

and

f

=

Ω

⁢

(

g

)

, we write

f

=

Θ

⁢

(

g

)

.

One more notational convention in this group is

f

⁢

(

x

)

∼

g

⁢

(

x

)

,

meaning

lim

x

→

∞

⁡

f

⁢

(

x

)

g

⁢

(

x

)

=

1

.

In

analysis

, such notation is useful in describing error estimates (

http://planetmath.org/AsymptoticEstimate

).
For example, the

Riemann hypothesis

is equivalent to the

conjecture

π

⁢

(

x

)

=

li

⁡

x

+

O

⁢

(

x

⁢

log

⁡

x

)

,

where

li

⁡

x

denotes the

logarithmic integral

.

Landau notation

is also handy in applied mathematics, e.g. in describing
the time complexity of an algorithm. It is common to say that an algorithm
requires

O

⁢

(

x

3

)

steps, for example, without needing to specify exactly what
is a step; for if

f

=

O

⁢

(

x

3

)

, then

f

=

O

⁢

(

A

⁢

x

3

)

for any

positive

constant

A

.

Title

Landau notation

Canonical

name

LandauNotation

Date of creation

2013-03-22 11:42:56

Last modified on

2013-03-22 11:42:56

Owner

Mathprof (13753)

Last modified by

Mathprof (13753)

Numerical id

28

Author

Mathprof (13753)

Entry type

Definition

Classification

msc 26A12

Classification

msc 20H15

Classification

msc 20B30

Synonym

O notation

Synonym

omega notation

Synonym

theta notation

Synonym

big-O notation

Related topic

LowerBoundForSorting

Related topic

ConvergenceOfIntegrals

Defines

big-o

Defines

small-o

Defines

small-omega

Generated on Fri Feb 9 19:48:03 2018 by

LaTeXML
∼ is an equivalence relation

∼

is an equivalence relation

Note that

∼

as defined in the entry

Landau notation

is an

equivalence relation

on the set of all

functions

from

ℝ

+

to

ℝ

+

. This set of functions will be denoted in this entry as

F

.

Reflexive

(

http://planetmath.org/Reflexive

):

For any

f

∈

F

,

lim

x

→

∞

⁡

f

⁢

(

x

)

f

⁢

(

x

)

=

1

, and

f

∼

f

.

Symmetric

:

If

f

,

g

∈

F

with

f

∼

g

, then

lim

x

→

∞

⁡

f

⁢

(

x

)

g

⁢

(

x

)

=

1

. Thus:

lim

x

→

∞

⁡

g

⁢

(

x

)

f

⁢

(

x

)

=

lim

x

→

∞

⁡

1

(

f

⁢

(

x

)

g

⁢

(

x

)

)

=

1

1

=

1

Therefore,

g

∼

f

.

Transitive

(

http://planetmath.org/Transitive3

)

:

If

f

,

g

,

h

∈

F

with

f

∼

g

and

g

∼

h

, then

lim

x

→

∞

⁡

f

⁢

(

x

)

g

⁢

(

x

)

=

1

and

lim

x

→

∞

⁡

g

⁢

(

x

)

h

⁢

(

x

)

=

1

. Thus:

lim

x

→

∞

⁡

f

⁢

(

x

)

h

⁢

(

x

)

=

lim

x

→

∞

⁡

(

f

⁢

(

x

)

g

⁢

(

x

)

⋅

g

⁢

(

x

)

h

⁢

(

x

)

)

=

1

⋅

1

=

1

Therefore,

f

∼

h

.

Title

∼

is an equivalence relation

Canonical name

simIsAnEquivalenceRelation

Date of creation

2013-03-22 16:13:16

Last modified on

2013-03-22 16:13:16

Owner

Wkbj79 (1863)

Last modified by

Wkbj79 (1863)

Numerical id

9

Author

Wkbj79 (1863)

Entry type

Result

Classification

msc 26A12

Generated on Fri Feb 9 19:48:10 2018 by

LaTeXML
compact spaces with group structure

compact spaces with group structure

Proposition

.

Assume that

(

G

,

M

)

is a group (with

multiplication

M

:

G

×

G

→

G

) and

G

is also a

topological space

. If

G

is

compact

Hausdorff

and

M

:

G

×

G

→

G

is

continuous

, then

(

G

,

M

)

is a

topological group

.

Proof.

Indeed, all we need to show is that

function

f

:

G

→

G

given by

f

⁢

(

g

)

=

g

-

1

is continuous. Note, that the following holds for the graph of

f

:

Γ

(

f

)

=

{

(

g

,

f

(

g

)

)

∈

G

×

G

}

=

{

(

g

,

g

-

1

)

∈

G

×

G

}

=

M

-

1

(

e

)

,

where

e

denotes the neutral element in

G

. It follows (from

continuity

of

M

) that

Γ

⁢

(

f

)

is closed in

G

×

G

. It is well known (see the parent object for details) that this implies that

f

is continuous, which

completes

the proof.

□

Title

compact spaces with group structure

Canonical name

CompactSpacesWithGroupStructure

Date of creation

2013-03-22 19:15:13

Last modified on

2013-03-22 19:15:13

Owner

joking (16130)

Last modified by

joking (16130)

Numerical id

6

Author

joking (16130)

Entry type

Corollary

Classification

msc 26A15

Classification

msc 54C05

Generated on Fri Feb 9 19:48:12 2018 by

LaTeXML
composition of continuous mappings is continuous

composition of continuous mappings is continuous

Theorem 1

.

The

composition

of two

continuous mappings

(when
defined) is

continuous

.

Proof.

Let

X

,

Y

,

Z

be

topological space

, and let

f

,

g

be

mappings

f

:

X

→

Y

,

g

:

Y

→

Z

.

We wish to prove that

g

∘

f

is continuous. Suppose

B

is
an

open set

in

Z

.
Since

g

is continuous,

g

-

1

⁢

(

B

)

is an open set in

Y

,
and
since

f

is continuous,

f

-

1

⁢

(

g

-

1

⁢

(

B

)

)

is an open set in

X

.
Since

f

-

1

⁢

(

g

-

1

⁢

(

B

)

)

=

(

g

∘

f

)

-

1

⁢

(

B

)

, it follows
that

(

g

∘

f

)

-

1

⁢

(

B

)

is open and the composition if continuous.
∎

Title

composition of continuous mappings is continuous

Canonical name

CompositionOfContinuousMappingsIsContinuous

Date of creation

2013-03-22 15:16:52

Last modified on

2013-03-22 15:16:52

Owner

mathcam (2727)

Last modified by

mathcam (2727)

Numerical id

8

Author

mathcam (2727)

Entry type

Theorem

Classification

msc 26A15

Classification

msc 54C05

Generated on Fri Feb 9 19:48:14 2018 by

LaTeXML
continuity of sine and cosine

continuity of sine and cosine

Theorem

.

The real functions

x

↦

sin

⁡

x

and

x

↦

cos

⁡

x

are

continuous at

every

real number

x

.

Proof.

Let

ε

be an arbitrary positive number.
Denote

Δ

sin

x

=

:

sin

z

-

sin

x

,

Δ

cos

x

=

:

cos

z

-

cos

x

where we suppose that

|

z

-

x

|

<

π

2

.  We may interpret

|

z

-

x

|

as an arc
of the

unit circle

of the

x

⁢

y

-plane.  Let’s think in the
circle the

right triangle

with

hypotenuse

the chord of the arc and
the

catheti

(i.e. the shorter sides) vertical and horizontal.  Then

|

Δ

⁢

sin

⁡

x

|

and

|

Δ

⁢

cos

⁡

x

|

are just these cathets; so we have

|

Δ

⁢

sin

⁡

x

|

≦

|

z

-

x

|

,

|

Δ

⁢

cos

⁡

x

|

≦

|

z

-

x

|

.

If we make

|

z

-

x

|

<

ε

,  then also

|

Δ

⁢

sin

⁡

x

|

and

|

Δ

⁢

cos

⁡

x

|

are less than

ε

.  It means that both
functions are continuous at

x

.

References

1

E. Lindelöf:

Johdatus korkeampaan analyysiin

. Neljäs painos.  Werner Söderström Osakeyhtiö, Porvoo ja Helsinki (1956).

Title

continuity of sine and cosine

Canonical

name

ContinuityOfSineAndCosine

Date of creation

2013-03-22 19:15:37

Last modified on

2013-03-22 19:15:37

Owner

pahio (2872)

Last modified by

pahio (2872)

Numerical id

4

Author

pahio (2872)

Entry type

Theorem

Classification

msc 26A15

Generated on Fri Feb 9 19:48:16 2018 by

LaTeXML
continuous

continuous

Let

X

and

Y

be

topological spaces

. A function

f

:

X

→

Y

is

continuous

if, for every

open set

U

⊂

Y

, the

inverse image

f

-

1

⁢

(

U

)

is an open subset of

X

.

In the case where

X

and

Y

are

metric spaces

(e.g.

Euclidean space

, or the space of

real numbers

), a function

f

:

X

→

Y

is continuous if and only if for every

x

∈

X

and every real number

ϵ

>

0

, there exists a real number

δ

>

0

such that whenever a point

z

∈

X

has

distance

less than

δ

to

x

, the point

f

⁢

(

z

)

∈

Y

has distance less than

ϵ

to

f

⁢

(

x

)

.

Continuity

at a point

A related notion is that of local continuity, or continuity at a point (as opposed to the whole space

X

at once). When

X

and

Y

are topological spaces, we say

f

is

continuous at a point

x

∈

X

if, for every open subset

V

⊂

Y

containing

f

⁢

(

x

)

, there is an open subset

U

⊂

X

containing

x

whose image

f

⁢

(

U

)

is contained in

V

. Of course, the function

f

:

X

→

Y

is continuous in the first sense if and only if

f

is continuous at every point

x

∈

X

in the second sense (for students who haven’t seen this before, proving it is a worthwhile exercise).

In the common case where

X

and

Y

are metric spaces (e.g., Euclidean spaces), a function

f

is continuous at

x

∈

X

if and only if for every real number

ϵ

>

0

, there exists a real number

δ

>

0

satisfying the property that

d

Y

⁢

(

f

⁢

(

x

)

,

f

⁢

(

z

)

)

<

ϵ

for all

z

∈

X

with

d

X

⁢

(

x

,

z

)

<

δ

.
Alternatively, the function

f

is continuous at

a

∈

X

if and only if the limit of

f

⁢

(

x

)

as

x

→

a

satisfies the equation

lim

x

→

a

⁡

f

⁢

(

x

)

=

f

⁢

(

a

)

.

Title

continuous

Canonical name

Continuous

Date of creation

2013-03-22 11:51:55

Last modified on

2013-03-22 11:51:55

Owner

djao (24)

Last modified by

djao (24)

Numerical id

12

Author

djao (24)

Entry type

Definition

Classification

msc 26A15

Classification

msc 54C05

Classification

msc 81-00

Classification

msc 82-00

Classification

msc 83-00

Classification

msc 46L05

Synonym

continuous function

Synonym

continuous map

Synonym

continuous mapping

Related topic

Limit

Defines

continuous at

Generated on Fri Feb 9 19:48:18 2018 by

LaTeXML
Dirichlet’s function

Dirichlet’s function

Dirichlet’s function

f

:

ℝ

→

ℝ

is defined as

f

⁢

(

x

)

=

{

1

q

if

⁢

x

=

p

q

⁢

is a rational number in lowest terms,

0

if

⁢

x

⁢

is an irrational number.

This function has the

property

that it is

continuous at

every

irrational number

and

discontinuous

at every

rational

one.

Another function that often goes by the same name is the function

f

⁢

(

x

)

=

{

1

if

⁢

x

⁢

is an rational number.

0

if

⁢

x

⁢

is an irrational number.

This nowhere-continuous function has the surprising

expression

f

⁢

(

x

)

=

lim

m

→

∞

⁡

lim

n

→

∞

⁡

cos

2

⁢

n

⁡

(

m

!

⁢

π

⁢

x

)

.

This is often given as the (amazing!) example of a

sequence

of everywhere-continuous functions whose limit function is nowhere

continuous

.

Title

Dirichlet’s function

Canonical

name

DirichletsFunction

Date of creation

2013-03-22 13:11:14

Last modified on

2013-03-22 13:11:14

Owner

mathcam (2727)

Last modified by

mathcam (2727)

Numerical id

9

Author

mathcam (2727)

Entry type

Definition

Classification

msc 26A15

Related topic

FunctionContinuousAtOnlyOnePoint

Related topic

APathologicalFunctionOfRiemann

Generated on Fri Feb 9 19:48:19 2018 by

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