aarjaneiro/planetmath huggingface dataset:
proof of Green’s theorem

proof of Green’s theorem

Consider the region

R

bounded

by the closed curve

P

in a simply connected space.

P

can be given by a vector valued

function

F

→

⁢

(

x

,

y

)

=

(

f

⁢

(

x

,

y

)

,

g

⁢

(

x

,

y

)

)

.
The region

R

can then be described by

∫

∫

R

(

∂

⁡

g

∂

⁡

x

-

∂

⁡

f

∂

⁡

y

)

⁢

𝑑

A

=

∫

∫

R

∂

⁡

g

∂

⁡

x

⁢

𝑑

A

-

∫

∫

R

∂

⁡

f

∂

⁡

y

⁢

𝑑

A

The

double integrals

above can be evaluated separately. Let’s look at

∫

∫

R

∂

⁡

g

∂

⁡

x

⁢

𝑑

A

=

∫

a

b

∫

A

⁢

(

y

)

B

⁢

(

y

)

∂

⁡

g

∂

⁡

x

⁢

𝑑

x

⁢

𝑑

y

Evaluating the above double integral, we get

∫

a

b

(

g

⁢

(

A

⁢

(

y

)

,

y

)

-

g

⁢

(

B

⁢

(

y

)

,

y

)

)

⁢

𝑑

y

=

∫

a

b

g

⁢

(

A

⁢

(

y

)

,

y

)

⁢

𝑑

y

-

∫

a

b

g

⁢

(

B

⁢

(

y

)

,

y

)

⁢

𝑑

y

According to the

fundamental theorem of line integrals

, the above equation is actually

equivalent

to the evaluation of the

line integral

of the function

F

→

1

⁢

(

x

,

y

)

=

(

0

,

g

⁢

(

x

,

y

)

)

over a path

P

=

P

1

+

P

2

, where

P

1

=

(

A

⁢

(

y

)

,

y

)

and

P

2

=

(

B

⁢

(

y

)

,

y

)

.

∫

a

b

g

⁢

(

A

⁢

(

y

)

,

y

)

⁢

𝑑

y

-

∫

a

b

g

⁢

(

B

⁢

(

y

)

,

y

)

⁢

𝑑

y

=

∫

P

1

F

1

→

⋅

𝑑

t

→

+

∫

P

2

F

1

→

⋅

𝑑

t

→

=

∮

P

F

1

→

⋅

𝑑

t

→

Thus we have

∫

∫

R

∂

⁡

g

∂

⁡

x

⁢

𝑑

A

=

∮

P

F

1

→

⋅

𝑑

t

→

By a similar

argument

, we can show that

∫

∫

R

∂

⁡

f

∂

⁡

y

⁢

𝑑

A

=

-

∮

P

F

2

→

⋅

𝑑

t

→

where

F

→

2

=

(

f

⁢

(

x

,

y

)

,

0

)

. Putting all of the above together, we can see that

∫

∫

R

(

∂

⁡

g

∂

⁡

x

-

∂

⁡

f

∂

⁡

y

)

⁢

𝑑

A

=

∮

P

F

1

→

⋅

𝑑

t

→

+

∮

P

F

2

→

⋅

𝑑

t

→

=

∮

P

(

F

→

1

+

F

→

2

)

⋅

𝑑

t

→

=

∮

P

(

f

⁢

(

x

,

y

)

,

g

⁢

(

x

,

y

)

)

⋅

𝑑

t

→

which is Green’s

theorem

.

Title

proof of Green’s theorem

Canonical name

ProofOfGreensTheorem

Date of creation

2013-03-22 12:28:47

Last modified on

2013-03-22 12:28:47

Owner

mathcam (2727)

Last modified by

mathcam (2727)

Numerical id

11

Author

mathcam (2727)

Entry type

Proof

Classification

msc 26B12

Generated on Fri Feb 9 19:58:27 2018 by

LaTeXML
proof of least and greatest value of function

proof of least and greatest value of function

f

is

continuous

, so it will transform

compact sets

into compact sets.
Thus since

[

a

,

b

]

is

compact

,

f

⁢

(

[

a

,

b

]

)

is also compact.

f

will thus attain on the

interval

[

a

,

b

]

a

maximum

and a

minimum

value because real compact sets are closed and

bounded

.

Consider the maximum and later use the same

argument

for

-

f

to consider the minimum.

By a known theorem (

http://planetmath.org/FermatsTheoremStationaryPoints

) if the maximum is attained in the

interior

of the domain,

c

∈

]

a

,

b

[

then

f

⁢

(

c

)

⁢

is a maximum

⟹

f

′

⁢

(

c

)

=

0

, since

f

is

differentiable

.

If the maximum isn’t attained in

]

a

,

b

[

and since it must be attained in

[

a

,

b

]

either

f

⁢

(

a

)

or

f

⁢

(

b

)

is a maximum.

For the minimum consider

-

f

and note that

-

f

will verify all conditions of the theorem and that a maximum of

-

f

corresponds to a minimum of

f

and that

-

f

′

⁢

(

c

)

=

0

⇔

f

′

⁢

(

c

)

=

0

.

Title

proof of least and greatest value of function

Canonical name

ProofOfLeastAndGreatestValueOfFunction

Date of creation

2013-03-22 15:52:09

Last modified on

2013-03-22 15:52:09

Owner

cvalente (11260)

Last modified by

cvalente (11260)

Numerical id

5

Author

cvalente (11260)

Entry type

Proof

Classification

msc 26B12

Related topic

FermatsTheoremStationaryPoints

Related topic

HeineBorelTheorem

Related topic

CompactnessIsPreservedUnderAContinuousMap

Generated on Fri Feb 9 19:58:29 2018 by

LaTeXML
relations between Hessian matrix and local extrema

relations between Hessian matrix and local extrema

Let

x

be a vector, and let

H

⁢

(

x

)

be the

Hessian

for

f

at a point

x

. Let

f

have

continuous

partial derivatives

of first and

second order

in a

neighborhood

of

x

. Let

∇

⁡

f

⁢

(

x

)

=

0

.

If

H

⁢

(

x

)

is

positive definite

(

http://planetmath.org/PositiveDefinite

), then

x

is a

strict local minimum

for

f

.

If

x

is a

local minimum

for

x

, then

H

⁢

(

x

)

is

positive semidefinite

.

If

H

⁢

(

x

)

is

negative definite

(

http://planetmath.org/NegativeDefinite

), then

x

is a strict local maximum for

f

.

If

x

is a

local maximum

for

x

, then

H

⁢

(

x

)

is

negative semidefinite

.

If

H

⁢

(

x

)

is indefinite,

x

is a

nondegenerate

saddle point

.

If the case when the

dimension

of

x

is 1 (i.e.

f

:

ℝ

→

ℝ

), this reduces to the

Second Derivative Test

, which is as follows:

Let the neighborhood of

x

be in the domain for

f

, and let

f

have continuous partial derivatives of first and second order.
Let

f

′

⁢

(

x

)

=

0

. If

f

′′

⁢

(

x

)

>

0

, then

x

is a strict local minimum. If

f

′′

⁢

(

x

)

<

0

, then

x

is a strict local maximum. In the case that

f

′′

⁢

(

x

)

=

0

, being

f

′′′

⁢

(

x

)

≠

0

,

x

is said to be an

inflexion point

(also called turning point). A typical example is

f

⁢

(

x

)

=

sin

⁡

x

,

f

′′

⁢

(

x

)

=

-

sin

⁡

x

=

0

,

x

=

n

⁢

π

,

n

=

0

,

±

1

,

±

2

,

…

,

f

′′′

⁢

(

x

)

=

-

cos

⁡

x

,

f

′′′

⁢

(

n

⁢

π

)

=

-

cos

⁡

n

⁢

π

=

(

-

1

)

n

+

1

≠

0

.

Title

relations between Hessian matrix and local extrema

Canonical name

RelationsBetweenHessianMatrixAndLocalExtrema

Date of creation

2013-03-22 12:59:52

Last modified on

2013-03-22 12:59:52

Owner

bshanks (153)

Last modified by

bshanks (153)

Numerical id

14

Author

bshanks (153)

Entry type

Result

Classification

msc 26B12

Related topic

Extrema

Related topic

Extremum

Related topic

HessianForm

Related topic

TestsForLocalExtremaForLagrangeMultiplierMethod

Defines

second derivative test

Generated on Fri Feb 9 19:58:35 2018 by

LaTeXML
solenoidal field

solenoidal field

A solenoidal vector field is one that satisfies

∇

⋅

𝐁

=

0

at every point where the

vector field

𝐁

is defined. Here

∇

⋅

𝐁

is the

divergence

.

This condition actually implies that there exists a vector

𝐀

, such that

𝐁

=

∇

×

𝐀

.

For a function

f

satisfying Laplace’s equation

∇

2

⁡

f

=

0

,

it follows that

∇

⁡

f

is solenoidal.

Title

solenoidal field

Canonical

name

SolenoidalField

Date of creation

2013-03-22 13:09:02

Last modified on

2013-03-22 13:09:02

Owner

giri (919)

Last modified by

giri (919)

Numerical id

9

Author

giri (919)

Entry type

Definition

Classification

msc 26B12

Synonym

solenoidal

Related topic

SourcesAndSinksOfVectorField

Generated on Fri Feb 9 19:58:37 2018 by

LaTeXML
vanishing of gradient in domain

vanishing of gradient in domain

Theorem

.

If the

function

f

is defined in a domain (

http://planetmath.org/Domain2

)

D

of

ℝ

n

and all the

partial derivatives

of a

f

vanish identically in

D

, i.e.

∇

⁡

f

≡

0

→

in

⁢

D

,

then the function has a

constant

value in the whole domain.

Proof.

For the sake of simpler notations, think that

n

=

3

; thus we have

f

x

′

⁢

(

x

,

y

,

z

)

=

f

y

′

⁢

(

x

,

y

,

z

)

=

f

z

′

⁢

(

x

,

y

,

z

)

=

0

for all

⁢

(

x

,

y

,

z

)

∈

D

.

(1)

Make the antithesis that there are the points

P

0

=

(

x

0

,

y

0

,

z

0

)

and

P

1

=

(

x

1

,

y

1

,

z

1

)

of

D

such that

f

⁢

(

x

0

,

y

0

,

z

0

)

≠

f

⁢

(

x

1

,

y

1

,

z

1

)

.
Since

D

is

connected

, one can form the

broken line

P

0

⁢

Q

1

⁢

Q

2

⁢

…

⁢

Q

k

⁢

P

1

contained

in

D

.  When one now goes along this broken line from

P

0

to

P

1

, one mets the first corner where the value of

f

does not equal

f

⁢

(

x

0

,

y

0

,

z

0

)

.  Thus

D

contains a

line segment

, the

end points

of which give unequal values to

f

.  When necessary, we change the notations such that this line segment is

P

0

⁢

P

1

.  Now,

f

x

′

,

f

y

′

,

f

z

′

are

continuous

in

D

because they vanish.  The

mean-value theorem for several variables

guarantees an

interior point

(

a

,

b

,

c

)

of the

segment

such that

0

≠

f

⁢

(

x

1

,

y

1

,

z

1

)

-

f

⁢

(

x

0

,

y

0

,

z

0

)

=

f

x

′

⁢

(

a

,

b

,

c

)

⁢

(

x

1

-

x

0

)

+

f

y

′

⁢

(

a

,

b

,

c

)

⁢

(

y

1

-

y

0

)

+

f

z

′

⁢

(

a

,

b

,

c

)

⁢

(

z

1

-

z

0

)

.

But by (1), the last sum must vanish.  This contradictory result shows that the antithesis is wrong, which settles the proof.

Title

vanishing of gradient in domain

Canonical name

VanishingOfGradientInDomain

Date of creation

2013-03-22 19:11:58

Last modified on

2013-03-22 19:11:58

Owner

pahio (2872)

Last modified by

pahio (2872)

Numerical id

8

Author

pahio (2872)

Entry type

Theorem

Classification

msc 26B12

Synonym

partial derivatives vanish

Related topic

FundamentalTheoremOfIntegralCalculus

Related topic

ExtremumPointsOfFunctionOfSeveralVariables

Generated on Fri Feb 9 19:58:43 2018 by

LaTeXML
vector potential

vector potential

Let

U

→

=

U

→

⁢

(

x

,

y

,

z

)

be a

vector field

in

ℝ

3

with

continuous

partial derivatives

.  Then the following three conditions are

equivalent

(

http://planetmath.org/Equivalent3

):

•

The

surface integrals

of

U

→

over all contractible closed surfaces (

http://planetmath.org/Sphere

)

S

vanish:

∮

S

U

→

⋅

𝑑

S

→

=

0

•

The

divergence

of

U

→

vanishes

everywhere in the field (

http://planetmath.org/VectorField

):

∇

⋅

U

→

=

0

•

There exists the

vector potential

A

→

=

A

→

⁢

(

x

,

y

,

z

)

of

U

→

:

∇

×

A

→

=

U

→

References

1

K. Väisälä:

Vektorianalyysi

.  Werner Söderström Osakeyhtiö, Helsinki (1961).

Title

vector potential

Canonical name

VectorPotential

Date of creation

2013-03-22 15:42:54

Last modified on

2013-03-22 15:42:54

Owner

pahio (2872)

Last modified by

pahio (2872)

Numerical id

8

Author

pahio (2872)

Entry type

Definition

Classification

msc 26B12

Related topic

IntegrationWithRespectToSurfaceArea

Related topic

LaminarField

Related topic

KalleVaisala

Generated on Fri Feb 9 19:58:45 2018 by

LaTeXML
arc length

arc length

Arclength

is the of a

section

of a

differentiable

curve. Finding the length of an arc is useful in many applications, for the length of a curve can represent

distance

traveled, work, etc. It is commonly represented as

S

or the

differential

d

⁢

s

if one is differentiating or integrating with respect to change in arclength.

If one knows the vector function or

parametric equations

of a curve, finding the arclength is , as it can be given by the sum of the lengths of the

tangent vectors

to the curve or

∫

a

b

|

F

→

′

⁢

(

t

)

|

⁢

𝑑

t

=

S

Note that

t

is an

independent

parameter. In

Cartesian coordinates

, arclength can be calculated by the formula

S

=

∫

a

b

1

+

(

f

′

⁢

(

x

)

)

2

⁢

𝑑

x

This formula is derived by viewing arclength as the

Riemann sum

lim

Δ

⁢

x

→

∞

⁡

∑

i

=

1

n

1

+

f

′

⁢

(

x

i

)

2

⁢

Δ

⁢

x

The term being summed is the length of an approximating secant to the curve over the distance

Δ

⁢

x

. As

Δ

⁢

x

vanishes

, the sum approaches the arclength, as desired. Arclength can also be derived for

polar coordinates

from the general formula for vector functions given above. The result is

L

=

∫

a

b

r

⁢

(

θ

)

2

+

(

r

′

⁢

(

θ

)

)

2

⁢

𝑑

θ

Title

arc length

Canonical name

ArcLength

Date of creation

2013-03-22 12:02:43

Last modified on

2013-03-22 12:02:43

Owner

mathcam (2727)

Last modified by

mathcam (2727)

Numerical id

14

Author

mathcam (2727)

Entry type

Algorithm

Classification

msc 26B15

Synonym

length of a curve

Related topic

Rectifiable

Related topic

IntegralRepresentationOfLengthOfSmoothCurve

Related topic

StraightLineIsShortestCurveBetweenTwoPoints

Related topic

PerimeterOfEllipse

Related topic

Evolute2

Related topic

Cycloid

Generated on Fri Feb 9 19:58:47 2018 by

LaTeXML
area of spherical zone

area of spherical zone

Let us consider the circle

(

x

-

r

)

2

+

y

2

=

r

2

with radius

r

and centre

(

r

,

0

)

.  A spherical zone may be thought to be formed when an arc of the circle rotates around the

x

-axis.  For finding the are of the zone, we can use the formula

A

=

2

⁢

π

⁢

∫

a

b

y

⁢

1

+

(

d

⁢

y

d

⁢

x

)

2

⁢

𝑑

x

(1)

of the entry

area of surface of revolution

.  Let the ends of the arc correspond the values

a

and

b

of the

abscissa

such that

b

-

a

=

h

is the of the spherical zone.  In the formula, we must use the solved form

y

=

(

±

)

⁢

r

⁢

x

-

x

2

of the equation of the circle.  The formula then yields

A

=

2

⁢

π

⁢

∫

a

b

r

⁢

x

-

x

2

⁢

1

+

(

r

-

x

r

⁢

x

-

x

2

)

2

⁢

𝑑

x

=

2

⁢

π

⁢

∫

a

b

r

⁢

𝑑

x

=

2

⁢

π

⁢

r

⁢

(

b

-

a

)

.

Hence the area of a spherical zone (and also of a

spherical calotte

) is

A

=

2

⁢

π

⁢

r

⁢

h

.

(2)

From here one obtains as a special case

h

=

2

⁢

r

the area of the whole sphere:

A

=

4

⁢

π

⁢

r

2

.

(3)

Remark.

The formula (2) implies that the

centre of mass

of a half-sphere is at the halfway point of the axis of

symmetry

(

h

=

r

2

).

Title

area of spherical zone

Canonical name

AreaOfSphericalZone

Date of creation

2013-03-22 18:19:05

Last modified on

2013-03-22 18:19:05

Owner

pahio (2872)

Last modified by

pahio (2872)

Numerical id

9

Author

pahio (2872)

Entry type

Derivation

Classification

msc 26B15

Classification

msc 53A05

Classification

msc 51M04

Synonym

area of spherical calotte

Related topic

AreaOfTheNSphere

Related topic

CentreOfMassOfHalfDisc

Generated on Fri Feb 9 19:58:58 2018 by

LaTeXML
area under Gaussian curve

area under Gaussian curve

Theorem

.

The area between the curve

y

=

e

-

x

2

and the

x

-axis equals

π

,  i.e.

∫

-

∞

∞

e

-

x

2

⁢

𝑑

x

=

π

.

Proof.

The square of the area is

(

∫

-

∞

∞

e

-

x

2

⁢

𝑑

x

)

2

=

lim

a

→

∞

⁡

(

∫

-

a

a

e

-

x

2

⁢

𝑑

x

)

2

=

lim

a

→

∞

⁡

∫

-

a

a

e

-

x

2

⁢

𝑑

x

⋅

∫

-

a

a

e

-

y

2

⁢

𝑑

y

=

lim

a

→

∞

⁡

∫

-

a

a

∫

-

a

a

e

-

(

x

2

+

y

2

)

⁢

𝑑

x

⁢

𝑑

y

=

lim

R

→

∞

⁡

∫

0

R

∫

0

2

⁢

π

e

-

r

2

⁢

r

⁢

𝑑

r

⁢

𝑑

φ

=

lim

R

→

∞

⁡

2

⁢

π

⁢

∫

0

R

e

-

r

2

⁢

r

⁢

𝑑

r

=

-

π

⁢

lim

R

→

∞

⁡

/

0

R

⁡

e

-

r

2

=

π

⁢

lim

R

→

∞

⁡

(

1

-

e

-

R

2

)

=

π

.

Here, the limit of the

double integral

over a square has been replaced by the limit of the double integral over a disc, because both limits are equal.  That both limits are equal can be demonstrated by the elementary

0

≤

∫

-

a

a

∫

-

a

a

e

-

(

x

2

+

y

2

)

⁢

𝑑

x

⁢

𝑑

y

-

∫

0

a

∫

0

2

⁢

π

e

-

r

2

⁢

r

⁢

𝑑

r

⁢

𝑑

φ

≤

e

-

a

2

⏟

g

⁢

r

⁢

e

⁢

a

⁢

t

⁢

e

⁢

s

⁢

t

⁢

v

⁢

a

⁢

l

⁢

u

⁢

e

⋅

(

4

⁢

a

2

-

π

⁢

a

2

)

⏟

a

⁢

r

⁢

e

⁢

a

=

(

4

-

π

)

⋅

a

2

e

a

2

,

and

a

2

e

a

2

→

0

when

a

→

∞

(see

growth of exponential function

).

Remark.

Since

e

-

x

2

is an

even function

,

∫

0

∞

e

-

x

2

d

x

=

π

2

⋅

Title

area under Gaussian curve

Canonical name

AreaUnderGaussianCurve

Date of creation

2013-03-22 15:16:36

Last modified on

2013-03-22 15:16:36

Owner

pahio (2872)

Last modified by

pahio (2872)

Numerical id

22

Author

pahio (2872)

Entry type

Theorem

Classification

msc 26B15

Classification

msc 26A36

Synonym

Gaussian integral

Synonym

area under the

bell curve

Related topic

SubstitutionNotation

Related topic

ProofThatNormalDistributionIsADistribution

Related topic

Distribution

Related topic

ErrorFunction

Related topic

EvaluatingTheGammaFunctionAt12

Related topic

NormalRandomVariable

Related topic

TableOfProbabilitiesOfStandardNormalDistribution

Related topic

ApplyingGeneratingFunction

Related topic

FresnelFormulas

Generated on Fri Feb 9 19:59:01 2018 by

LaTeXML
flux of vector field

flux of vector field

Let

U

→

=

U

x

⁢

i

→

+

U

y

⁢

j

→

+

U

z

⁢

k

→

be a

vector field

in

ℝ

3

and let

a

be a portion of some surface in the vector field.  Define one ; if

a

is a closed surface, then the of it.  For any surface element

d

⁢

a

of

a

, the corresponding

vectoral surface element

is

d

⁢

a

→

=

n

→

⁢

d

⁢

a

,

where

n

→

is the

unit normal

vector on the of

d

⁢

a

.

The

flux

of the vector

U

→

through the surface

a

is the

∫

a

U

→

⋅

𝑑

a

→

.

Remark.

One can imagine that

U

→

represents the

velocity vector

of a flowing liquid; suppose that the flow is , i.e. the velocity

U

→

depends only on the location, not on the time.  Then the

scalar product

U

→

⋅

d

⁢

a

→

is the volume of the liquid flown per time-unit through the surface element

d

⁢

a

; it is positive or negative depending on whether the flow is from the negative to the positive or contrarily.

Example.

Let

U

→

=

x

⁢

i

→

+

2

⁢

y

⁢

j

→

+

3

⁢

z

⁢

k

→

and

a

be the portion of the plane

x

+

y

+

x

=

1

in the

first octant

(

x

≧

0

,

y

≧

0

,

z

≧

0

) with the away from the origin.

One has the constant unit normal vector:

n

→

=

1

3

⁢

i

→

+

1

3

⁢

j

→

+

1

3

⁢

k

→

.

The flux of

U

→

through

a

is

φ

=

∫

a

U

→

⋅

𝑑

a

→

=

1

3

⁢

∫

a

(

x

+

2

⁢

y

+

3

⁢

z

)

⁢

𝑑

a

.

However, this

surface integral

may be converted to one in which

a

is replaced by its

projection

(

http://planetmath.org/ProjectionOfPoint

)

A

on the

x

⁢

y

-plane, and

d

⁢

a

is then similarly replaced by its projection

d

⁢

A

;

d

⁢

A

=

cos

⁡

α

⁢

d

⁢

a

where

α

is the angle between the normals of both surface elements, i.e. the angle between

n

→

and

k

→

:

cos

⁡

α

=

n

→

⋅

k

→

=

1

3

.

Then we also express

z

on

a

with the

coordinates

x

and

y

:

φ

=

1

3

⁢

∫

A

(

x

+

2

⁢

y

+

3

⁢

(

1

-

x

-

y

)

)

⁢

3

⁢

𝑑

A

=

∫

0

1

(

∫

0

1

-

x

(

3

-

2

⁢

x

-

y

)

⁢

𝑑

y

)

⁢

𝑑

x

=

1

Title

flux of vector field

Canonical name

FluxOfVectorField

Date of creation

2013-03-22 18:45:25

Last modified on

2013-03-22 18:45:25

Owner

pahio (2872)

Last modified by

pahio (2872)

Numerical id

14

Author

pahio (2872)

Entry type

Definition

Classification

msc 26B15

Classification

msc 26B12

Synonym

flux of vector

Related topic

GaussGreenTheorem

Related topic

MutualPositionsOfVectors

Related topic

AngleBetweenTwoVectors

Defines

flux

Generated on Fri Feb 9 19:59:03 2018 by

LaTeXML
generalisation of Gaussian integral

generalisation of Gaussian integral

The

integral

∫

0

∞

e

-

x

2

⁢

cos

⁡

t

⁢

x

⁢

d

⁢

x

:=

w

⁢

(

t

)

is a generalisation of the

Gaussian integral

w

⁢

(

0

)

=

π

2

.  For evaluating it we first form its

derivative

which may be done by differentiating under the

integral sign

(

http://planetmath.org/DifferentiationUnderIntegralSign

):

w

′

⁢

(

t

)

=

∫

0

∞

e

-

x

2

⁢

(

-

x

)

⁢

sin

⁡

t

⁢

x

⁢

d

⁢

x

=

1

2

⁢

∫

0

∞

e

-

x

2

⁢

(

-

2

⁢

x

)

⁢

sin

⁡

t

⁢

x

⁢

d

⁢

x

Using

integration by parts

this yields

w

′

⁢

(

t

)

=

1

2

⁢

/

x

=

0

∞

⁡

e

-

x

2

⁢

sin

⁡

t

⁢

x

-

t

2

⁢

∫

0

∞

e

-

x

2

⁢

cos

⁡

t

⁢

x

⁢

d

⁢

x

=

1

2

⁢

(

0

-

0

)

-

t

2

⁢

∫

0

∞

e

-

x

2

⁢

cos

⁡

t

⁢

x

⁢

d

⁢

x

=

-

t

2

⁢

w

⁢

(

t

)

.

Thus

w

⁢

(

t

)

satisfies

the

linear differential equation

d

⁢

w

d

⁢

t

=

-

1

2

⁢

t

⁢

w

,

where one can separate the

variables

(

http://planetmath.org/SeparationOfVariables

) and integrate:

∫

d

⁢

w

w

=

-

1

2

⁢

∫

t

⁢

𝑑

t

.

So,

ln

⁡

w

=

-

1

4

⁢

t

2

+

ln

⁡

C

,  i.e.

w

=

w

⁢

(

t

)

=

C

⁢

e

-

1

4

⁢

t

2

,
and since there is the

initial condition

w

⁢

(

0

)

=

π

2

, we obtain the result

w

⁢

(

t

)

=

π

2

⁢

e

-

1

4

⁢

t

2

.

Title

generalisation of Gaussian integral

Canonical

name

GeneralisationOfGaussianIntegral

Date of creation

2013-03-22 18:43:36

Last modified on

2013-03-22 18:43:36

Owner

pahio (2872)

Last modified by

pahio (2872)

Numerical id

6

Author

pahio (2872)

Entry type

Derivation

Classification

msc 26B15

Classification

msc 26A36

Related topic

SubstitutionNotation

Generated on Fri Feb 9 19:59:13 2018 by

LaTeXML
length of curve in a metric space

length of curve in a metric space

Suppose that

(

X

,

d

)

is a

metric space

. Let

f

be a curve, so that

f

:

[

0

,

1

]

→

X

is a

continuous function

, and let

0

=

t

0

<

t

1

<

⋯

<

t

n

=

1

and

x

i

=

f

⁢

(

t

i

)

for

0

≤

i

≤

n

.
The set

{

x

0

,

x

1

,

…

,

x

n

}

is called a

partition

of the curve.
The

of the curve

is defined to be
the

supremum

over all partitions of the quantity

∑

i

=

1

n

d

⁢

(

x

i

,

x

i

-

1

)

.

Title

length of curve

in a metric space

Canonical name

LengthOfCurveInAMetricSpace

Date of creation

2013-03-22 16:50:27

Last modified on

2013-03-22 16:50:27

Owner

Mathprof (13753)

Last modified by

Mathprof (13753)

Numerical id

8

Author

Mathprof (13753)

Entry type

Definition

Classification

msc 26B15

Defines

length of a curve

Generated on Fri Feb 9 19:59:15 2018 by

LaTeXML
perimeter of astroid

perimeter of astroid

The

astroid

x

2

3

+

y

2

3

=

a

2

3

can be presented in the

parametric form

x

=

a

⁢

cos

3

⁡

φ

,

y

=

a

⁢

sin

3

⁡

φ

,

where the polar angle

φ

gets the values from

0

to

2

⁢

π

. The curve consists of four

congruent

arcs, one of which is obtained letting

0

≦

φ

≦

π

2

. Thus the whole perimeter of the astroid is

s

=

4

⁢

∫

0

π

2

(

d

⁢

x

d

⁢

φ

)

2

+

(

d

⁢

y

d

⁢

φ

)

2

⁢

𝑑

φ

.

This

expression

is easily simplified to

s

=

12

⁢

a

⁢

∫

0

π

2

sin

⁡

φ

⁢

cos

⁡

φ

⁢

d

⁢

φ

giving the result

s

=

6

⁢

a

⁢

∫

0

π

2

sin

⁡

2

⁢

φ

⁢

d

⁢

φ

=

-

3

⁢

a

⁢

/

0

π

2

⁡

cos

⁡

2

⁢

φ

=

6

⁢

a

.

Title

perimeter of astroid

Canonical

name

PerimeterOfAstroid

Date of creation

2013-03-22 17:14:06

Last modified on

2013-03-22 17:14:06

Owner

pahio (2872)

Last modified by

pahio (2872)

Numerical id

7

Author

pahio (2872)

Entry type

Example

Classification

msc 26B15

Synonym

perimetre of astroid

Related topic

Parameter

Related topic

ClairautsEquation

Related topic

SubstitutionNotation

Generated on Fri Feb 9 19:59:17 2018 by

LaTeXML