aarjaneiro/planetmath huggingface dataset:
Lipschitz condition and differentiability result

Lipschitz condition and differentiability result

About lipschitz continuity of

differentiable functions

the following holds.

Theorem 1

.

Let

X

,

Y

be

Banach spaces

and let

A

be a convex (see convex set), open subset of

X

.
Let

f

:

A

¯

→

Y

be a

function

which is

continuous

in

A

¯

and

differentiable

in

A

. Then

f

is

lipschitz continuous

on

A

¯

if and only if the

derivative

D

⁢

f

is

bounded

on

A

i.e.

sup

x

∈

A

⁡

∥

D

⁢

f

⁢

(

x

)

∥

<

+

∞

.

Proof.

Suppose that

f

is lipschitz continuous:

∥

f

⁢

(

x

)

-

f

⁢

(

y

)

∥

≤

L

⁢

∥

x

-

y

∥

.

Then given any

x

∈

A

and any

v

∈

X

, for all small

h

∈

ℝ

we have

∥

f

⁢

(

x

+

h

⁢

v

)

-

f

⁢

(

x

)

h

∥

≤

L

.

Hence, passing to the limit

h

→

0

it must hold

∥

D

⁢

f

⁢

(

x

)

∥

≤

L

.

On the other hand suppose that

D

⁢

f

is bounded on

A

:

∥

D

⁢

f

⁢

(

x

)

∥

≤

L

,

∀

x

∈

A

.

Given any two points

x

,

y

∈

A

¯

and given any

α

∈

Y

*

consider the function

G

:

[

0

,

1

]

→

ℝ

G

⁢

(

t

)

=

⟨

α

,

f

⁢

(

(

1

-

t

)

⁢

x

+

t

⁢

y

)

⟩

.

For

t

∈

(

0

,

1

)

it holds

G

′

⁢

(

t

)

=

⟨

α

,

D

⁢

f

⁢

(

(

1

-

t

)

⁢

x

+

t

⁢

y

)

⁢

[

y

-

x

]

⟩

and hence

|

G

′

⁢

(

t

)

|

≤

L

⁢

∥

α

∥

⁢

∥

y

-

x

∥

.

Applying Lagrange

mean-value theorem

to

G

we know that there exists

ξ

∈

(

0

,

1

)

such that

|

⟨

α

,

f

⁢

(

y

)

-

f

⁢

(

x

)

⟩

|

=

|

G

⁢

(

1

)

-

G

⁢

(

0

)

|

=

|

G

′

⁢

(

ξ

)

|

≤

∥

α

∥

⁢

L

⁢

∥

y

-

x

∥

and since this is true for all

α

∈

Y

*

we get

∥

f

⁢

(

y

)

-

f

⁢

(

x

)

∥

≤

L

⁢

∥

y

-

x

∥

which is the desired claim.
∎

Title

Lipschitz condition and differentiability result

Canonical name

LipschitzConditionAndDifferentiabilityResult

Date of creation

2013-03-22 13:32:42

Last modified on

2013-03-22 13:32:42

Owner

paolini (1187)

Last modified by

paolini (1187)

Numerical id

5

Author

paolini (1187)

Entry type

Result

Classification

msc 26A16

Generated on Fri Feb 9 19:50:00 2018 by

LaTeXML
Lipschitz function

Lipschitz function

Let

W

⊆

X

⊆

ℂ

and

f

:

X

→

ℂ

. Then

f

is

on

W

if there exists an

M

∈

ℝ

such that, for all

x

,

y

∈

W

,

x

≠

y

|

f

⁢

(

x

)

-

f

⁢

(

y

)

|

≤

M

⁢

|

x

-

y

|

If

a

,

b

∈

ℝ

with

a

<

b

and

f

:

[

a

,

b

]

→

ℝ

is

Lipschitz

on

(

a

,

b

)

, then

f

is

absolutely continuous

on

[

a

,

b

]

.

Example:

Is

f

⁢

(

x

)

=

1

x

,

x

∈

[

0

,

1

]

a

Lipschitz function

.

We need to estimate the constant

M

.

|

f

⁢

(

x

)

-

f

⁢

(

y

)

|

=

|

1

x

-

1

y

|

=

|

x

-

y

x

⁢

y

|

=

|

x

-

y

x

⁢

y

⁢

(

x

+

y

)

|

=

1

|

x

⁢

y

⁢

(

x

+

y

)

|

⁢

|

x

-

y

|

.

It follows that

M

=

1

|

x

⁢

y

⁢

(

x

+

y

)

|

and

f

⁢

(

x

)

is not Lipschitz at

x

=

0

.

Title

Lipschitz function

Canonical name

LipschitzFunction

Date of creation

2013-03-22 14:01:42

Last modified on

2013-03-22 14:01:42

Owner

bwebste (988)

Last modified by

bwebste (988)

Numerical id

12

Author

bwebste (988)

Entry type

Definition

Classification

msc 26A16

Defines

Lipschitz

Generated on Fri Feb 9 19:50:02 2018 by

LaTeXML
uniform continuity of Lipschitz functions

uniform continuity of Lipschitz functions

Proposition 1

.

An Hölder

continuous mapping

is

uniformly continuous

.
In particular any

Lipschitz continuous

mapping is uniformly continuous.

Proof.

Let

f

:

X

→

Y

be a mapping such that for some

C

>

0

and

α

with

0

<

α

≤

1

one
has

d

Y

⁢

(

f

⁢

(

p

)

,

f

⁢

(

q

)

)

≤

C

⁢

d

X

⁢

(

p

,

q

)

α

.

For every given

ϵ

>

0

, choose

δ

=

(

ϵ

/

(

C

+

1

)

)

1

α

.
If

p

,

q

∈

X

are given points satisfying

d

X

⁢

(

p

,

q

)

<

δ

then

d

Y

⁢

(

f

⁢

(

p

)

,

f

⁢

(

q

)

)

≤

C

⁢

δ

α

≤

C

⁢

ϵ

C

+

1

<

ϵ

,

as desired.
∎

Title

uniform continuity of Lipschitz functions

Canonical name

UniformContinuityOfLipschitzFunctions

Date of creation

2013-03-22 15:06:16

Last modified on

2013-03-22 15:06:16

Owner

paolini (1187)

Last modified by

paolini (1187)

Numerical id

8

Author

paolini (1187)

Entry type

Theorem

Classification

msc 26A16

Generated on Fri Feb 9 19:50:04 2018 by

LaTeXML
iteration

iteration

Let

f

:

X

→

X

be a function,

X

being any set. The

n

-th iteration of a function is the function which is obtained if

f

is applied

n

times, and is denoted by

f

n

. More formally we define:

f

0

⁢

(

x

)

=

x

and

f

n

+

1

⁢

(

x

)

=

f

⁢

(

f

n

⁢

(

x

)

)

for

nonnegative integers

n

. If

f

is invertible, then by going backwards we can define the iterate also for

negative

n

.

Title

iteration

Canonical

name

Iteration

Date of creation

2013-03-22 12:43:40

Last modified on

2013-03-22 12:43:40

Owner

mathwizard (128)

Last modified by

mathwizard (128)

Numerical id

9

Author

mathwizard (128)

Entry type

Definition

Classification

msc 26A18

Synonym

iterate

Generated on Fri Feb 9 19:50:06 2018 by

LaTeXML
periodic point

periodic point

Let

f

:

X

→

X

be a

function

and

f

n

its

n

-th

iteration

. A point

x

is called a

periodic point

of period

n

of

f

if it is a

fixed point

of

f

n

. The least

n

for which

x

is a fixed point of

f

n

is called

prime period

or

least period

.

If

f

is a function

ℝ

to

ℝ

or

ℂ

to

ℂ

then a periodic point

x

of prime period

n

is called

hyperbolic

if

|

(

f

n

)

′

⁢

(

x

)

|

≠

1

,

attractive

if

|

(

f

n

)

′

⁢

(

x

)

|

<

1

and

repelling

if

|

(

f

n

)

′

⁢

(

x

)

|

>

1

.

Title

periodic point

Canonical name

PeriodicPoint

Date of creation

2013-03-22 12:43:38

Last modified on

2013-03-22 12:43:38

Owner

mathwizard (128)

Last modified by

mathwizard (128)

Numerical id

14

Author

mathwizard (128)

Entry type

Definition

Classification

msc 26A18

Defines

hyperbolic periodic point

Defines

attractive periodic point

Defines

repelling periodic point

Defines

least period

Defines

prime period

Generated on Fri Feb 9 19:50:07 2018 by

LaTeXML
0/0 is possible and has an answer

0/0 is possible and has an answer

So, I am happy to tell you that I have solved 0/0 by means of utilizing

set theory

.
Set Theory is a part of math dealing with sets of objects.
We all know that
1*0=0
2*0=0
…
30000*0=0
So that means that
Since P(R) is the whole set of real numbers
P(R)*0=0
Meaning to say, 0 divided by 0 is R, or the entire set of real numbers.
That is my solution to this mysterious

expression

that was yet to be solved.

Title

0/0 is possible and has an answer

Canonical

name

00IsPossibleAndHasAnAnswer

Date of creation

2014-08-02 19:59:17

Last modified on

2014-08-02 19:59:17

Owner

imaginary

.i (1001376)

Last modified by

imaginary.i (1001376)

Numerical id

4

Author

imaginary.i (1001376)

Entry type

Result

Generated on Fri Feb 9 19:50:09 2018 by

LaTeXML
BODMAS Rule application

BODMAS Rule application

Today I want to share what Is BODMAS Rule & How to use it?

BODMAS rule decides the

order of operations

for add, subtract, multiply, divide, etc as below shown
B-Brackets (do all operations contained in the brackets first)
O-Orders (powers and

square roots

etc)
D-Division
M-Multiplication
A-Addition
S-

Subtraction

Letâs see an example and check how BODMASS Rule works
30-(2*6+15/3) +8*3/6

Step1: Brackets
2*6+15/3 = 12+5 = 17

Step2:

Division

30-17+8*1/2

Step3:

Multiplication

30-17+4

Step 4:

Addition

and Subtraction
30-17+4=17

20+30-5/4*2+ (1+6)
(5-6/4)+9*7
1-6*(2+9)/8

BODMAS Rule is very helpful in solving various

algebra

math problems.

Title

BODMAS Rule application

Canonical

name

BODMASRuleApplication

Date of creation

2014-08-08 11:22:45

Last modified on

2014-08-08 11:22:45

Owner

burgess (1001318)

Last modified by

burgess (1001318)

Numerical id

1

Author

burgess (1001318)

Entry type

Application

Generated on Fri Feb 9 19:50:10 2018 by

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contractive sequence

contractive sequence

The

sequence

a

0

,

a

1

,

a

2

,

…

(1)

in a

metric space

(

X

,

d

)

is called

contractive

, iff there is a

real number

r

∈

(

0

,

1

)

such that for any

positive integer

n

the

inequality

d

⁢

(

a

n

,

a

n

+

1

)

≦

r

⋅

d

⁢

(

a

n

-

1

,

a

n

)

(2)

is true.

We will prove the

Theorem

.

If the sequence (1) is contractive, it is
a

Cauchy sequence

.

Proof.

Suppose that the sequence (1) is
contractive. Let

ε

be an arbitrary positive number and

m

,

n

some positive integers from which e.g.

n

is

greater

than

m

,

n

=

m

+

δ

.

Using repeatedly the

triangle inequality

we get

d

⁢

(

a

m

,

a

n

)

≦

d

⁢

(

a

m

,

a

m

+

1

)

+

d

⁢

(

a

m

+

1

,

a

m

+

δ

)

≦

d

⁢

(

a

m

,

a

m

+

1

)

+

d

⁢

(

a

m

+

1

,

a

m

+

2

)

+

d

⁢

(

a

m

+

2

,

a

m

+

δ

)

…

≦

d

⁢

(

a

m

,

a

m

+

1

)

+

d

⁢

(

a

m

+

1

,

a

m

+

2

)

+

d

⁢

(

a

m

+

2

,

a

m

+

3

)

+

…

+

d

⁢

(

a

n

-

1

,

a

n

)

.

Now the contractiveness gives the inequalities

d

⁢

(

a

1

,

a

2

)

≦

r

⁢

d

⁢

(

a

0

,

a

1

)

,

d

⁢

(

a

2

,

a

3

)

≦

r

⁢

d

⁢

(

a

1

,

a

2

)

≦

r

2

⁢

d

⁢

(

a

0

,

a

1

)

,

d

⁢

(

a

3

,

a

4

)

≦

r

⁢

d

⁢

(

a

2

,

a

3

)

≦

r

3

⁢

d

⁢

(

a

0

,

a

1

)

,

…

d

⁢

(

a

m

,

a

m

+

1

)

≦

r

m

⁢

d

⁢

(

a

0

,

a

1

)

,

…

d

⁢

(

a

n

-

1

,

a

n

)

≦

r

n

-

1

⁢

d

⁢

(

a

0

,

a

1

)

,

by which we obtain the estimation

d

⁢

(

a

m

,

a

n

)

≦

d

⁢

(

a

0

,

a

1

)

⁢

(

r

m

+

r

m

+

1

+

…

+

r

m

+

δ

-

1

)

=

d

⁢

(

a

0

,

a

1

)

⁢

r

m

⁢

(

1

+

r

+

r

2

+

…

+

r

δ

-

1

)

=

d

⁢

(

a

0

,

a

1

)

⁢

r

m

⁢

1

-

r

δ

1

-

r

<

d

⁢

(

a

0

,

a

1

)

⁢

r

m

1

-

r

.

The last expression tends to zero as

m

→

∞

.  Thus
there exists a positive number

M

such that

d

⁢

(

a

m

,

a

n

)

⁢

<

ε

⁢

for each

⁢

m

>

⁢

M

when

n

>

m

.  Consequently, (1) is a Cauchy sequence.

Remark.

The assertion of the Theorem cannot be
reversed. E.g. in the usual metric of

ℝ

, the
sequence

1

,

1

2

,

1

3

,

…

converges

to 0
and hence is Cauchy, but for it the ratio

|

a

n

-

a

n

+

1

|

:

|

a

n

-

1

-

a

n

|

=

1

-

2

n

+

1

tends to 1 as

n

→

∞

.

Cf. sequences of

bounded variation

(

http://planetmath.org/SequenceOfBoundedVariation

).

References

1

Paul Loya

:

Amazing and Aesthetic
Aspects of

Analysis

: On the incredible

infinite

.  A Course in Undergraduate Analysis, Fall 2006.
Available in http://www.math.binghamton.edu/dennis/478.f07/EleAna.pdf

Title

contractive sequence

Canonical name

ContractiveSequence

Date of creation

2014-11-30 16:45:13

Last modified on

2014-11-30 16:45:13

Owner

pahio (2872)

Last modified by

pahio (2872)

Numerical id

13

Author

pahio (2872)

Entry type

Theorem

Generated on Fri Feb 9 19:50:14 2018 by

LaTeXML
derivative of inverse function

derivative of inverse function

Theorem

.

If the

real function

f

has an

inverse function

f

←

and the

derivative

of

f

at the point

x

=

f

←

⁢

(

y

)

is distinct from zero, then

f

←

is also

differentiable

at the point

y

and

f

←

′

⁢

(

y

)

=

1

f

′

⁢

(

x

)

.

(1)

That is, the derivatives of a

function

and its inverse function are

inverse numbers

of each other, provided that they have been taken at the points which correspond to each other.

{it Proof.
Now we have

f

⁢

(

f

←

⁢

(

y

)

)

=

f

⁢

(

x

)

=

y

.

The derivatives of both sides must be equal:

d

d

⁢

y

⁢

[

f

⁢

(

f

←

⁢

(

y

)

)

]

=

d

d

⁢

y

⁢

y

Using the

chain rule

we get

f

′

⁢

(

f

←

⁢

(

y

)

)

⋅

f

←

′

⁢

(

y

)

=

1

,

whence

f

←

′

⁢

(

y

)

=

1

f

′

⁢

(

f

←

⁢

(

y

)

)

.

This is same as the asserted (1).

Examples.

For simplicity, we express here the functions by symbols

y

and the inverse functions by

x

.

1.

y

=

tan

⁡

x

,

x

=

arctan

⁡

y

;

d

⁢

x

d

⁢

y

=

1

d

⁢

y

d

⁢

x

=

1

1

+

tan

2

⁡

x

=

1

1

+

y

2

2.

y

=

sin

⁡

x

,

x

=

arcsin

⁡

y

;

d

⁢

x

d

⁢

y

=

1

d

⁢

y

d

⁢

x

=

1

cos

⁡

x

=

1

+

1

-

sin

2

⁡

x

=

+

1

1

-

y

2

3.

y

=

x

2

,

x

=

±

y

;

d

⁢

x

d

⁢

y

=

1

d

⁢

y

d

⁢

x

=

1

2

⁢

x

=

1

±

2

⁢

y

If the

variable

symbol

y

in those results is changed to

x

, the results can be written

d

d

⁢

x

⁢

arctan

⁡

x

=

1

1

+

x

2

,

d

d

⁢

x

⁢

arcsin

⁡

x

=

1

1

-

x

2

,

d

d

⁢

x

⁢

x

=

1

2

⁢

x

.

Title

derivative of inverse function

Canonical

name

DerivativeOfInverseFunction

Date of creation

2015-02-21 16:02:46

Last modified on

2015-02-21 16:02:46

Owner

pahio (2872)

Last modified by

pahio (2872)

Numerical id

12

Author

pahio (2872)

Entry type

Theorem

Classification

msc 26A24

Related topic

InverseFunctionTheorem

Related topic

Derivative2

Related topic

DerivativeOfTheNaturalLogarithmFunction

Related topic

CyclometricFunctions

Related topic

SquareRoot

Related topic

LimitExamples

Related topic

IntegrationOfSqrtx21

Generated on Fri Feb 9 19:50:17 2018 by

LaTeXML
derivative of Riemann integral

derivative of Riemann integral

Let

f

be a

continuous function

from an

open subset

A

of

ℝ

2

to

ℝ

.  Suppose that also the

partial derivative

f

t

′

⁢

(

x

,

t

)

is

continuous

in

A

which contains the

line segments

along which the integration is performed and that

a

⁢

(

t

)

and

b

⁢

(

t

)

are real functions

differentiable

in some point

t

0

.  Denote

F

⁢

(

t

)

=

∫

a

⁢

(

t

)

b

⁢

(

t

)

f

⁢

(

x

,

t

)

⁢

𝑑

x

and

G

⁢

(

t

)

=

b

′

⁢

(

t

0

)

⋅

f

⁢

(

b

⁢

(

t

)

,

t

)

-

a

′

⁢

(

t

0

)

⋅

f

⁢

(

a

⁢

(

t

)

,

t

)

+

∫

a

⁢

(

t

)

b

⁢

(

t

)

f

t

′

⁢

(

x

,

t

)

⁢

𝑑

x

.

Then one has the

derivative

F

′

⁢

(

t

0

)

=

G

⁢

(

t

0

)

in all such points

t

=

t

0

.

Title

derivative of Riemann integral

Canonical name

DerivativeOfRiemannIntegral

Date of creation

2013-03-22 14:35:30

Last modified on

2013-03-22 14:35:30

Owner

PrimeFan (13766)

Last modified by

PrimeFan (13766)

Numerical id

9

Author

PrimeFan (13766)

Entry type

Theorem

Classification

msc 26A24

Classification

msc 26A42

Related topic

DifferentiationUnderIntegralSign

Generated on Fri Feb 9 19:50:19 2018 by

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differentiable function

differentiable function

Let

f

:

V

→

W

be a

function

, where

V

and

W

are

Banach spaces

.
For

x

∈

V

, the function

f

is said to be

differentiable

at

x

if its

derivative

exists at that point. Differentiability at

x

∈

V

implies

continuity

at

x

. If

S

⊂

V

, then

f

is said to
be differentiable on

S

if

f

is differentiable at every point

x

∈

S

.

For the most common example, a real function

f

:

ℝ

→

ℝ

is differentiable
if its derivative

d

⁢

f

d

⁢

x

exists for every point in the region of

interest

. For another common case of a real function of

n

variables

f

⁢

(

x

1

,

x

2

,

…

,

x

n

)

(more formally

f

:

ℝ

n

→

ℝ

),
it is not

sufficient

that the

partial derivatives

∂

⁡

f

∂

⁡

x

i

exist for

f

to be differentiable. The
derivative of

f

must exist in the original sense
at every point in the region of interest,
where

ℝ

n

is treated as a Banach space under the usual

Euclidean vector
norm

.

If the derivative of

f

is

continuous

, then

f

is said to be

C

1

. If
the

k

th derivative of

f

is continuous, then

f

is said to be

C

k

. By convention, if

f

is only continuous but does not have a continuous derivative, then

f

is said to
be

C

0

. Note the

inclusion

property

C

k

+

1

⊂

C

k

.
And if the

k

-th derivative of

f

is continuous for all

k

,
then

f

is said to be

C

∞

. In other words

C

∞

is the

intersection

C

∞

=

⋂

k

=

0

∞

C

k

.

Differentiable functions are often referred to as

smooth

. If

f

is

C

k

, then

f

is said to be

k

-smooth. Most often a function is called
smooth (without qualifiers) if

f

is

C

∞

or

C

1

, depending on the

context

.

Title

differentiable function

Canonical

name

DifferentiableFunction

Date of creation

2013-03-22 12:39:10

Last modified on

2013-03-22 12:39:10

Owner

Koro (127)

Last modified by

Koro (127)

Numerical id

24

Author

Koro (127)

Entry type

Definition

Classification

msc 26A24

Classification

msc 57R35

Synonym

smooth function

Synonym

differentiable mapping

Synonym

differentiable map

Synonym

smooth mapping

Synonym

smooth map

Synonym

continuously differentiable

Related topic

OneSidedDerivatives

Related topic

RoundFunction

Related topic

ConverseTheorem

Related topic

WeierstrassFunction

Defines

differentiable

Defines

smooth

Generated on Fri Feb 9 19:50:20 2018 by

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example of contractive sequence

example of contractive sequence

Define the

sequence

a

1

,

a

2

,

a

3

,

…

by

a

1

:=

1

,

a

n

+

1

:=

5

-

2

⁢

a

n

(

n

=

1

,

2

,

3

,

…

)

.

(1)

We see by

induction

that the

radicand

in (1) cannot become

negative

; in fact we justify that

1

≦

a

n

≦

3

(2)

for every

n

:  It’s clear when

n

=

1

. If it is true for
an

a

n

, it implies that

1

<

5

-

2

⁢

a

n

≦

3

, i.e.

1

<

a

n

+

1

≦

3

.

As for the

convergence

of the sequence, which is not

monotonic

, one could think to show that
it is a

Cauchy sequence

. Unfortunately, it is almost impossible to directly express and
estimate the needed

absolute value

of

a

m

-

a

n

. Fortunately,
the recursive definition (1) allows quite easily to estimate

|

a

n

-

a

n

+

1

|

.  Then it turns out that it’s a question of a
contractive sequence, whence it is by the parent entry (

http://planetmath.org/ContractiveSequence

) a
Cauchy sequence.

We form the

difference

a

n

-

a

n

+

1

=

(

5

-

2

⁢

a

n

-

1

-

5

-

2

⁢

a

n

)

⁢

(

5

-

2

⁢

a

n

-

1

+

5

-

2

⁢

a

n

)

5

-

2

⁢

a

n

-

1

+

5

-

2

⁢

a

n

=

-

2

⁢

(

a

n

-

1

-

a

n

)

5

-

2

⁢

a

n

-

1

+

5

-

2

⁢

a

n

where

n

>

1

.  Thus we can estimate its absolute value, by
using (2):

|

a

n

-

a

n

+

1

|

=

2

⁢

|

a

n

-

1

-

a

n

|

5

-

2

⁢

a

n

-

1

+

5

-

2

⁢

a

n

≦

2

⁢

|

a

n

-

1

-

a

n

|

5

-

2

⁢

3

+

5

-

2

⁢

3

=

|

a

n

-

1

-

a

n

|

5

-

2

⁢

3

Since

1

5

-

2

⁢

3

<

1

, our sequence (1) is
contractive, consequently Cauchy.  Therefore it

converges

to
a limit

A

.

We have

A

2

=

(

lim

n

→

∞

⁡

a

n

+

1

)

2

=

lim

n

→

∞

⁡

a

n

+

1

2

=

lim

n

→

∞

⁡

(

5

-

2

⁢

a

n

)

=

5

-

2

⁢

A

.

From the

quadratic equation

A

2

+

2

⁢

A

-

5

=

0

we get the positive
root

A

=

6

-

1

. I.e.,

lim

n

→

∞

⁡

a

n

=

6

-

1

.

(3)

Generated on Fri Feb 9 19:50:23 2018 by

LaTeXML
example of differentiation under integral sign

example of differentiation under integral sign

Differentiation

with respect to a parameter under the integral sign may sometimes yield useful formulae.  One example is given here.

We know that the equation

∫

0

1

x

m

⁢

𝑑

x

=

1

m

+

1

is valid for all

m

>

-

1

.  If one differentiates with respect to

m

under the integral sign (

http://planetmath.org/DifferentiationUnderTheIntegralSign

) in succession, one gets

∫

0

1

∂

∂

⁡

m

⁢

e

m

⁢

ln

⁡

x

⁢

𝑑

x

=

∫

0

1

e

m

⁢

ln

⁡

x

⁢

ln

⁡

x

⁢

d

⁢

x

=

∫

0

1

x

m

⁢

ln

⁡

x

⁢

d

⁢

x

=

-

1

(

m

+

1

)

2

∫

0

1

∂

∂

⁡

m

⁢

x

m

⁢

ln

⁡

x

⁢

d

⁢

x

=

∫

0

1

x

m

⁢

(

ln

⁡

x

)

2

⁢

𝑑

x

=

+

1

⋅

2

(

m

+

1

)

3

∫

0

1

∂

∂

⁡

m

⁢

x

m

⁢

(

ln

⁡

x

)

2

⁢

𝑑

x

=

∫

0

1

x

m

⁢

(

ln

⁡

x

)

3

⁢

𝑑

x

=

-

1

⋅

2

⋅

3

(

m

+

1

)

4

⋯

It’s evident that repeating the differentiation

n

times the final result is the

∫

0

1

x

m

(

ln

x

)

n

d

x

=

(

-

1

)

n

⁢

n

!

(

m

+

1

)

n

+

1

(

m

>

-

1

)

.

Title

example of

differentiation under integral sign

Canonical name

ExampleOfDifferentiationUnderIntegralSign

Date of creation

2013-03-22 17:01:56

Last modified on

2013-03-22 17:01:56

Owner

pahio (2872)

Last modified by

pahio (2872)

Numerical id

4

Author

pahio (2872)

Entry type

Example

Classification

msc 26A24

Classification

msc 26B15

Generated on Fri Feb 9 19:50:25 2018 by

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How to find whether a given number is prime or not?

How to find whether a given number is prime or not?

What is a

prime number

A number is greater than 1 is called a prime number, if it has only two factors, namely 1 and the number itself.

Prime numbers up to 100 are:2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97

Procedure to find out the prime number

Suppose A is given number.

Step 1: Find a whole number nearly greater than the

square root

of A. K ¿ square root(A)
Step 2: Test whether A is

divisible

by any prime number less than K. If yes A is not a prime number. If not, A is prime number.

Example:

Find out whether 337 is a prime number or not?

Step 1: 19 ¿ square root (337)
Prime numbers less than 19 are 2, 3, 5, 7, 11, 13, 17
Step 2: 337 is not divisible by any of them

Therefore 337 is a prime number

These are simple and easy tricks which are helpful to solve your math homework problems .

Title

How to find whether a given number is prime or not?

Canonical name

HowToFindWhetherAGivenNumberIsPrimeOrNot

Date of creation

2014-08-12 17:53:37

Last modified on

2014-08-12 17:53:37

Owner

burgess (1001318)

Last modified by

burgess (1001318)

Numerical id

1

Author

burgess (1001318)

Entry type

Topic

Generated on Fri Feb 9 19:50:27 2018 by

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