aarjaneiro/planetmath huggingface dataset:
square root of polynomial

square root of polynomial

The

f

,  denoted by

f

, is any

polynomial

g

having the square

g

2

equal to

f

.  For example,

9

⁢

x

2

-

30

⁢

x

+

25

=

3

⁢

x

-

5

or

-

3

⁢

x

+

5

.

A polynomial needs not have a

square root

, but if it has a square root

g

, then also the opposite polynomial

-

g

is its square root.

Algorithm

.

The idea of the squaring

(

a

+

b

+

c

+

.

.

)

2

=

(

a

)

a

+

(

2

a

+

b

)

b

+

(

2

a

+

2

b

+

c

)

c

+

.

.

(see the

square of sum

) gives a method for getting the square root of a polynomial:

•

The .

•

And so on.

In the examples below, on the .

Example 1.

9

⁢

x

4

+

6

⁢

x

3

-

11

⁢

x

2

-

4

⁢

x

+

4

=

?

(

9

x

4

+

6

⁢

x

3

-

11

⁢

x

2

-

4

⁢

x

+

4

)

=

±

(

3

x

2

+

x

-

2

)

9

⁢

x

4

3

⁢

x

2

6

⁢

x

3

-

11

⁢

x

2

6

⁢

x

2

+

x

6

⁢

x

3

+

x

2

x

-

12

⁢

x

2

-

4

⁢

x

+

4

6

⁢

x

2

+

2

⁢

x

-

2

-

12

⁢

x

2

-

4

⁢

x

+

4

-

2

0

Example 2.

x

6

-

2

⁢

x

5

-

x

4

+

3

⁢

x

2

+

2

⁢

x

+

1

=

?

(

1

+

2

⁢

x

+

3

⁢

x

2

-

x

4

-

2

⁢

x

5

+

x

6

)

=

±

(

1

+

x

+

x

2

-

x

3

)

1

1

2

⁢

x

+

3

⁢

x

2

2

+

x

2

⁢

x

+

x

2

+

x

2

⁢

x

2

-

x

4

2

+

2

⁢

x

+

x

2

2

⁢

x

2

+

2

⁢

x

3

+

x

4

x

2

-

2

⁢

x

3

-

2

⁢

x

4

-

2

⁢

x

5

+

x

6

2

+

2

⁢

x

+

2

⁢

x

2

-

x

3

-

2

⁢

x

3

-

2

⁢

x

4

-

2

⁢

x

5

+

x

6

-

x

3

0

Remark.

The procedure may give a

Taylor series

expansion of the square root, if it is not a polynomial.  E.g. we get

1

+

x

=

1

+

1

2

x

-

1

8

x

2

+

1

16

x

3

-

5

128

x

4

+

-

…

References

1

Meyers Rechenduden

.  Erster verbesserter Neudruck.  Bibliographisches Institut AG, Mannheim (1960).

Title

square root of polynomial

Canonical name

SquareRootOfPolynomial

Date of creation

2013-03-22 15:32:06

Last modified on

2013-03-22 15:32:06

Owner

pahio (2872)

Last modified by

pahio (2872)

Numerical id

17

Author

pahio (2872)

Entry type

Algorithm

Classification

msc 26C99

Classification

msc 12E05

Synonym

calculation of square root of polynomial

Related topic

SquareOfSum

Related topic

BombellisMethodOfComputingSquareRoots

Generated on Fri Feb 9 20:00:53 2018 by

LaTeXML
proof of Jordan’s Inequality

proof of Jordan’s Inequality

To prove that

2

π

⁢

x

≤

sin

⁡

(

x

)

≤

x

,

∀

x

∈

[

0

,

π

2

]

consider a circle (circle with radius = 1 ). Take any point

P

on the

circumference

of the circle.

Drop the

perpendicular

from

P

to the horizontal line,

M

being the foot of the perpendicular and

Q

the

reflection

of

P

at

M

.
(refer to figure)

Let

x

=

∠

⁢

P

⁢

O

⁢

M

.

For

x

to be in

[

0

,

π

2

]

, the point

P

lies in the first

quadrant

, as shown.

The length of

line segment

P

⁢

M

is

sin

⁡

(

x

)

.
Construct a circle of radius

M

⁢

P

, with

M

as the center.

Length of line segment

P

⁢

Q

is

2

⁢

sin

⁡

(

x

)

.

Length of arc

P

⁢

A

⁢

Q

is

2

⁢

x

.

Length of arc

P

⁢

B

⁢

Q

is

π

⁢

sin

⁡

(

x

)

.

Since

P

⁢

Q

≤

length of arc

P

⁢

A

⁢

Q

(equality holds when

x

=

0

)
we have

2

⁢

sin

⁡

(

x

)

≤

2

⁢

x

.
This implies

sin

⁡

(

x

)

≤

x

Since length of arc

P

⁢

A

⁢

Q

is

≤

length of arc

P

⁢

B

⁢

Q

(equality holds true when

x

=

0

or

x

=

π

2

),
we have

2

⁢

x

≤

π

⁢

sin

⁡

(

x

)

.
This implies

2

π

⁢

x

≤

sin

⁡

(

x

)

Thus we have

2

π

⁢

x

≤

sin

⁡

(

x

)

≤

x

,

∀

x

∈

[

0

,

π

2

]

Title

proof of Jordan’s Inequality

Canonical name

ProofOfJordansInequality

Date of creation

2013-03-22 13:08:48

Last modified on

2013-03-22 13:08:48

Owner

mathcam (2727)

Last modified by

mathcam (2727)

Numerical id

17

Author

mathcam (2727)

Entry type

Proof

Classification

msc 26D05

Generated on Fri Feb 9 20:00:55 2018 by

LaTeXML
Shapiro inequality

Shapiro inequality

Let

n

be a

positive integer

and let

x

1

,

…

,

x

n

be

positive

reals.
If

n

is even and

n

≤

12

, or

n

is odd and

n

≤

23

, then

∑

i

=

1

n

x

i

x

i

+

1

+

x

i

+

2

≥

n

2

,

where the subscripts are to be understood modulo

n

.

The particular case of

n

=

3

is also known as

Nesbitts inequality

.

Title

Shapiro inequality

Canonical

name

ShapiroInequality

Date of creation

2013-03-22 13:43:15

Last modified on

2013-03-22 13:43:15

Owner

Koro (127)

Last modified by

Koro (127)

Numerical id

7

Author

Koro (127)

Entry type

Theorem

Classification

msc 26D05

Synonym

Shapiro’s

inequality

Related topic

NesbittsInequality

Generated on Fri Feb 9 20:00:57 2018 by

LaTeXML
squaring condition for square root inequality

squaring condition for square root inequality

Of the

inequalities

a

≶

b

,

•

both are undefined when

a

<

0

;

•

both can be sidewise squared when

a

≧

0

and

b

≧

0

;

•

a

>

b

is identically true if

a

≧

0

and

b

<

0

.

•

a

<

b

is identically untrue if

b

<

0

;

The above theorem may be utilised for solving inequalities involving

square roots

.

Example.

Solve the inequality

2

⁢

x

+

3

>

x

.

(1)

The reality condition

2

⁢

x

+

3

≧

0

requires that

x

≧

-

1

⁤

1

2

.  For using the theorem, we distinguish two cases according to the sign of the right hand side:

1

∘

:

-

1

⁤

1

2

≦

x

<

0

.  The inequality is identically true; we have for (1) the partial solution

-

1

⁤

1

2

≦

x

<

0

.

2

∘

:

x

≧

0

.  Now we can square both , obtaining

2

⁢

x

+

3

>

x

2

x

2

-

2

⁢

x

-

3

<

0

The zeros of

x

2

-

2

⁢

x

-

3

are

x

=

1

±

2

,  i.e.

-

1

and

3

. Since the graph of the

polynomial function

is a parabola opening upwards, the

polynomial

attains its

negative

values when

-

1

<

x

<

3

(see

quadratic inequality

).  Thus we obtain for (1) the partial solution

0

≦

x

<

3

.

Combining both partial solutions we obtain the total solution

-

1

⁤

1

2

≦

x

<

3

.

Title

squaring condition for square root inequality

Canonical name

SquaringConditionForSquareRootInequality

Date of creation

2013-03-22 17:55:56

Last modified on

2013-03-22 17:55:56

Owner

pahio (2872)

Last modified by

pahio (2872)

Numerical id

6

Author

pahio (2872)

Entry type

Theorem

Classification

msc 26D05

Classification

msc 26A09

Synonym

squaring condition

Related topic

StrangeRoot

Generated on Fri Feb 9 20:00:59 2018 by

LaTeXML
Weierstrass product inequality

Weierstrass product inequality

For any finite family

(

a

i

)

i

∈

I

of

real numbers

in the

interval

[

0

,

1

]

, we have

∏

i

(

1

-

a

i

)

≥

1

-

∑

i

a

i

.

Proof:

Write

f

=

∏

i

(

1

-

a

i

)

+

∑

i

a

i

.

For any

k

∈

I

, and any fixed values of the

a

i

for

i

≠

k

,

f

is a

polynomial

of the first degree in

a

k

.
Consequently

f

is minimal either at

a

k

=

0

or

a

k

=

1

.
That brings us down to two cases: all the

a

i

are zero, or at least
one of them is

1

. But in both cases it is clear that

f

≥

1

, QED.

Title

Weierstrass product inequality

Canonical name

WeierstrassProductInequality

Date of creation

2013-03-22 13:58:23

Last modified on

2013-03-22 13:58:23

Owner

Daume (40)

Last modified by

Daume (40)

Numerical id

5

Author

Daume (40)

Entry type

Theorem

Classification

msc 26D05

Generated on Fri Feb 9 20:01:01 2018 by

LaTeXML
numerator and denominator increased by same amount

numerator and denominator increased by same amount

Let the

positive

fraction

a

b

be altered by adding a positive number

δ

to both

a

and

b

.  Then

a

b

<

a

+

δ

b

+

δ

if

a

<

b

,

a

b

>

a

+

δ

b

+

δ

if

a

>

b

.

The asserted

inequalities

follow from the identity

a

b

-

a

+

δ

b

+

δ

=

(

a

-

b

)

⁢

δ

b

2

+

b

⁢

δ

.

Accordingly, we have for example

2

3

<

3

4

,

200

99

>

201

100

.

Title

numerator and denominator increased by same amount

Canonical name

NumeratorAndDenominatorIncreasedBySameAmount

Date of creation

2013-03-22 19:36:29

Last modified on

2013-03-22 19:36:29

Owner

pahio (2872)

Last modified by

pahio (2872)

Numerical id

9

Author

pahio (2872)

Entry type

Result

Classification

msc 26D07

Classification

msc 11A99

Related topic

AdjacentFraction

Generated on Fri Feb 9 20:01:03 2018 by

LaTeXML
Gronwall’s lemma

Gronwall’s lemma

If, for

t

0

≤

t

≤

t

1

,

ϕ

⁢

(

t

)

≥

0

and

ψ

⁢

(

t

)

≥

0

are

continuous functions

such that the

inequality

ϕ

⁢

(

t

)

≤

K

+

L

⁢

∫

t

0

t

ψ

⁢

(

s

)

⁢

ϕ

⁢

(

s

)

⁢

𝑑

s

holds on

t

0

≤

t

≤

t

1

, with

K

and

L

positive

constants

, then

ϕ

⁢

(

t

)

≤

K

⁢

exp

⁡

(

L

⁢

∫

t

0

t

ψ

⁢

(

s

)

⁢

𝑑

s

)

on

t

0

≤

t

≤

t

1

.

Title

Gronwall’s lemma

Canonical name

GronwallsLemma

Date of creation

2013-03-22 13:22:20

Last modified on

2013-03-22 13:22:20

Owner

jarino (552)

Last modified by

jarino (552)

Numerical id

4

Author

jarino (552)

Entry type

Theorem

Classification

msc 26D10

Synonym

Gronwall’s inequality

Generated on Fri Feb 9 20:01:05 2018 by

LaTeXML
proof of Gronwall’s lemma

proof of Gronwall’s lemma

The

inequality

ϕ

⁢

(

t

)

≤

K

+

L

⁢

∫

t

0

t

ψ

⁢

(

s

)

⁢

ϕ

⁢

(

s

)

⁢

𝑑

s

(1)

is

equivalent

to

ϕ

⁢

(

t

)

K

+

L

⁢

∫

t

0

t

ψ

⁢

(

s

)

⁢

ϕ

⁢

(

s

)

⁢

𝑑

s

≤

1

Multiply by

L

⁢

ψ

⁢

(

t

)

and integrate, giving

∫

t

0

t

L

⁢

ψ

⁢

(

s

)

⁢

ϕ

⁢

(

s

)

⁢

d

⁢

s

K

+

L

⁢

∫

t

0

s

ψ

⁢

(

τ

)

⁢

ϕ

⁢

(

τ

)

⁢

𝑑

τ

≤

L

⁢

∫

t

0

t

ψ

⁢

(

s

)

⁢

𝑑

s

Thus

ln

⁡

(

K

+

L

⁢

∫

t

0

t

ψ

⁢

(

s

)

⁢

ϕ

⁢

(

s

)

⁢

𝑑

s

)

-

ln

⁡

K

≤

L

⁢

∫

t

0

t

ψ

⁢

(

s

)

⁢

𝑑

s

and finally

K

+

L

⁢

∫

t

0

t

ψ

⁢

(

s

)

⁢

ϕ

⁢

(

s

)

⁢

𝑑

s

≤

K

⁢

exp

⁡

(

L

⁢

∫

t

0

t

ψ

⁢

(

s

)

⁢

𝑑

s

)

Using (

1

) in the

left hand side

of this inequality gives the result.

Title

proof of Gronwall’s lemma

Canonical name

ProofOfGronwallsLemma

Date of creation

2013-03-22 13:22:23

Last modified on

2013-03-22 13:22:23

Owner

jarino (552)

Last modified by

jarino (552)

Numerical id

5

Author

jarino (552)

Entry type

Proof

Classification

msc 26D10

Generated on Fri Feb 9 20:01:06 2018 by

LaTeXML
proof of Hermite-Hadamard integral inequality

proof of Hermite-Hadamard integral inequality

First of all, let’s recall that a

convex function

on a

open
interval

(

a

,

b

)

is

continuous

on

(

a

,

b

)

and admits left and right
derivative

f

+

⁢

(

x

)

and

f

-

⁢

(

x

)

for any

x

∈

(

a

,

b

)

. For this reason,
it’s always possible to construct at least one supporting line (

http://planetmath.org/ConvexFunctionsLieAboveTheirSupportingLines

) for

f

⁢

(

x

)

at any

x

0

∈

(

a

,

b

)

: if

f

⁢

(

x

0

)

is

differentiable

in

x

0

, one has

r

⁢

(

x

)

=

f

⁢

(

x

0

)

+

f

′

⁢

(

x

0

)

⁢

(

x

-

x

0

)

; if not, it’s obvious that all

r

⁢

(

x

)

=

f

⁢

(

x

0

)

+

c

⁢

(

x

-

x

0

)

are supporting lines for
any

c

∈

[

f

-

⁢

(

x

0

)

,

f

+

⁢

(

x

0

)

]

.

Let now

r

⁢

(

x

)

=

f

⁢

(

a

+

b

2

)

+

c

⁢

(

x

-

a

+

b

2

)

be a supporting line of

f

⁢

(

x

)

in

x

=

a

+

b

2

∈

(

a

,

b

)

.
Then,

r

⁢

(

x

)

≤

f

⁢

(

x

)

. On the other side, by
convexity definition, having defined

s

⁢

(

x

)

=

f

⁢

(

a

)

+

f

⁢

(

b

)

-

f

⁢

(

a

)

b

-

a

⁢

(

x

-

a

)

the line connecting the points

(

a

,

f

⁢

(

a

)

)

and

(

b

,

f

⁢

(

b

)

)

, one has

f

⁢

(

x

)

≤

s

⁢

(

x

)

. Shortly,

r

⁢

(

x

)

≤

f

⁢

(

x

)

≤

s

⁢

(

x

)

Integrating both

inequalities

between

a

and

b

∫

a

b

r

⁢

(

x

)

⁢

𝑑

x

≤

∫

a

b

f

⁢

(

x

)

⁢

𝑑

x

≤

∫

a

b

s

⁢

(

x

)

⁢

𝑑

x

∫

a

b

r

⁢

(

x

)

⁢

𝑑

x

=

∫

a

b

[

f

⁢

(

a

+

b

2

)

+

c

⁢

(

x

-

a

+

b

2

)

]

⁢

𝑑

x

=

f

⁢

(

a

+

b

2

)

⁢

(

b

-

a

)

+

c

⁢

∫

a

b

(

x

-

a

+

b

2

)

⁢

𝑑

x

=

f

⁢

(

a

+

b

2

)

⁢

(

b

-

a

)

∫

a

b

s

⁢

(

x

)

⁢

𝑑

x

=

∫

a

b

[

f

⁢

(

a

)

+

f

⁢

(

b

)

-

f

⁢

(

a

)

b

-

a

⁢

(

x

-

a

)

]

⁢

𝑑

x

=

f

⁢

(

a

)

⁢

(

b

-

a

)

+

f

⁢

(

b

)

-

f

⁢

(

a

)

b

-

a

⁢

∫

a

b

(

x

-

a

)

⁢

𝑑

x

=

f

⁢

(

a

)

+

f

⁢

(

b

)

2

⁢

(

b

-

a

)

and so

f

⁢

(

a

+

b

2

)

⁢

(

b

-

a

)

≤

∫

a

b

f

⁢

(

x

)

⁢

𝑑

x

≤

f

⁢

(

a

)

+

f

⁢

(

b

)

2

⁢

(

b

-

a

)

which is the thesis.

Title

proof of Hermite-Hadamard integral inequality

Canonical name

ProofOfHermiteHadamardIntegralInequality

Date of creation

2013-03-22 16:59:22

Last modified on

2013-03-22 16:59:22

Owner

Andrea Ambrosio (7332)

Last modified by

Andrea Ambrosio (7332)

Numerical id

7

Author

Andrea Ambrosio (7332)

Entry type

Proof

Classification

msc 26D10

Classification

msc 26D15

Generated on Fri Feb 9 20:01:09 2018 by

LaTeXML
relation between positive function and its gradient when its Hessian matrix is bounded

relation between positive function and its gradient when its Hessian matrix is bounded

Let

f

:

R

n

→

R

a

positive

function

, twice

differentiable

everywhere. Furthermore, let

∥

𝐇

f

⁢

(

𝐱

)

∥

2

≤

M

,

M

>

0

∀

𝐱

∈

R

n

, where

𝐇

f

⁢

(

𝐱

)

is the

Hessian matrix

of

f

⁢

(

𝐱

)

.
Then, for any

𝐱

∈

R

n

,

∥

∇

⁡

f

⁢

(

𝐱

)

∥

2

≤

2

⁢

M

⁢

f

⁢

(

𝐱

)

Proof.

Let

𝐱

,

𝐱

0

∈

R

n

be arbitrary points. By
positivity of

f

⁢

(

𝐱

)

, writing

Taylor expansion

of

f

⁢

(

𝐱

)

with Lagrange error formula around

𝐱

0

, a point

𝐜

∈

R

n

exists such that:

0

≤

f

⁢

(

𝐱

)

=

f

⁢

(

𝐱

0

)

+

∇

⁡

f

⁢

(

𝐱

0

)

⋅

(

𝐱

-

𝐱

0

)

+

1

2

⁢

(

𝐱

-

𝐱

0

)

T

⋅

𝐇

f

⁢

(

𝐜

)

⋅

(

𝐱

-

𝐱

0

)

=

|

f

⁢

(

𝐱

0

)

+

∇

⁡

f

⁢

(

𝐱

0

)

⋅

(

𝐱

-

𝐱

0

)

+

1

2

⁢

(

𝐱

-

𝐱

0

)

T

⋅

𝐇

f

⁢

(

𝐜

)

⋅

(

𝐱

-

𝐱

0

)

|

≤

f

⁢

(

𝐱

0

)

+

|

∇

⁡

f

⁢

(

𝐱

0

)

⋅

(

𝐱

-

𝐱

0

)

|

+

1

2

⁢

|

(

𝐱

-

𝐱

0

)

T

⋅

𝐇

f

⁢

(

𝐜

)

⋅

(

𝐱

-

𝐱

0

)

|

≤

f

⁢

(

𝐱

0

)

+

∥

∇

⁡

f

⁢

(

𝐱

0

)

∥

2

⁢

∥

𝐱

-

𝐱

0

∥

2

+

1

2

⁢

∥

𝐇

f

⁢

(

𝐜

)

∥

2

⁢

∥

𝐱

-

𝐱

0

∥

2

2

⁢

(by Cauchy-Schwartz
inequality)

≤

f

⁢

(

𝐱

0

)

+

∥

∇

⁡

f

⁢

(

𝐱

0

)

∥

2

⁢

∥

𝐱

-

𝐱

0

∥

2

+

1

2

⁢

M

⁢

∥

𝐱

-

𝐱

0

∥

2

2

The rightest side is a second degree

polynomial

in

variable

∥

𝐱

-

𝐱

0

∥

2

; for it to be positive for any
choice of

∥

𝐱

-

𝐱

0

∥

2

(that is,
for any choice of

𝐱

), the discriminant

∥

∇

⁡

f

⁢

(

𝐱

0

)

∥

2

2

-

4

⋅

1

2

⁢

M

⁢

f

⁢

(

𝐱

0

)

must be negative, whence the thesis.
∎

Note: The condition on the boundedness of the Hessian matrix is actually needed. In fact, in the Lagrange form

remainder

, the constant

𝐜

depends upon the point

𝐱

. Thus, if we couldn’t rely on the condition

∥

𝐇

f

⁢

(

𝐱

)

∥

2

≤

M

, we could only state

f

⁢

(

𝐱

0

)

+

∥

∇

⁡

f

⁢

(

𝐱

0

)

∥

2

⁢

∥

𝐱

-

𝐱

0

∥

2

+

1

2

⁢

∥

𝐇

f

⁢

(

𝐜

⁢

(

𝐱

)

)

∥

2

⁢

∥

𝐱

-

𝐱

0

∥

2

2

≥

0

which, not being a second degree polynomial, wouldn’t imply any particular further condition.
Moreover, in the case

n

=

1

, the lemma assumes the simpler form:
Let

f

:

ℝ

→

ℝ

a positive function, twice differentiable everywhere.
Furthermore, let

f

′′

⁢

(

x

)

≤

M

,

M

>

0

∀

x

∈

ℝ

. Then,
for any

x

∈

ℝ

,

|

f

′

⁢

(

x

)

|

≤

2

⁢

M

⁢

f

⁢

(

x

)

.

Title

relation between positive function and its gradient when its Hessian matrix is bounded

Canonical

name

RelationBetweenPositiveFunctionAndItsGradientWhenItsHessianMatrixIsBounded

Date of creation

2013-03-22 15:53:10

Last modified on

2013-03-22 15:53:10

Owner

Andrea Ambrosio (7332)

Last modified by

Andrea Ambrosio (7332)

Numerical id

16

Author

Andrea Ambrosio (7332)

Entry type

Theorem

Classification

msc 26D10

Generated on Fri Feb 9 20:01:13 2018 by

LaTeXML
another proof of Young inequality

another proof of Young inequality

Let

F

⁢

(

x

)

=

∫

0

x

ϕ

⁢

(

t

)

⁢

𝑑

t

,

and

⁢

G

⁢

(

x

)

=

∫

0

x

ϕ

-

1

⁢

(

t

)

⁢

𝑑

t

.

Since

ϕ

-

1

is

strictly increasing

,

G

is

strictly convex

, hence lies above its supporting line, i.e. for every

c

and

x

≠

c

G

⁢

(

b

)

>

G

⁢

(

c

)

+

G

′

⁢

(

c

)

⁢

(

b

-

c

)

=

G

⁢

(

c

)

+

ϕ

-

1

⁢

(

c

)

⁢

(

b

-

c

)

.

In particular, for

c

=

ϕ

⁢

(

a

)

we have

F

⁢

(

a

)

+

G

⁢

(

b

)

>

F

⁢

(

a

)

+

G

⁢

(

ϕ

⁢

(

a

)

)

+

a

⁢

(

b

-

ϕ

⁢

(

a

)

)

=

a

⁢

b

,

because

F

⁢

(

a

)

+

G

⁢

(

ϕ

⁢

(

a

)

)

=

a

⁢

ϕ

⁢

(

a

)

.

Title

another

proof of Young inequality

Canonical name

AnotherProofOfYoungInequality

Date of creation

2013-03-22 15:45:38

Last modified on

2013-03-22 15:45:38

Owner

a4karo (12322)

Last modified by

a4karo (12322)

Numerical id

10

Author

a4karo (12322)

Entry type

Proof

Classification

msc 26D15

Generated on Fri Feb 9 20:01:19 2018 by

LaTeXML
arithmetic mean

arithmetic mean

If

a

1

,

a

2

,

…

,

a

n

are

real numbers

, their

arithmetic mean

is defined as

A

.

M

.

=

a

1

+

a

2

+

…

+

a

n

n

.

The arithmetic mean is what is commonly called the

average

of the numbers.  The value of

A

.

M

.

is always between the least and the greatest of the numbers (

http://planetmath.org/MinimalAndMaximalNumber

)

a

j

.  If the numbers

a

j

are all

positive

, then

A

.

M

.

>

a

j

n

for all

j

.

A

generalization

of this

concept

is that of

weighted mean

, also known as

weighted average

.  Let

w

1

,

…

,

w

n

be numbers whose sum is not zero,
which will be known as

weights

. (Typically, these will be

strictly

positive numbers, so their sum will automatically differ from zero.) Then the
weighted mean of

a

1

,

a

2

,

…

,

a

n

is defined to be

W

.

M

.

=

w

1

⁢

a

1

+

w

2

⁢

a

2

+

…

+

w

n

⁢

a

n

w

1

+

w

2

+

…

+

w

n

.

In the special case where all the weights are equal to each other, the weighted mean equals the arithmetic mean.

Title

arithmetic mean

Canonical

name

ArithmeticMean

Date of creation

2013-03-22 11:50:42

Last modified on

2013-03-22 11:50:42

Owner

drini (3)

Last modified by

drini (3)

Numerical id

14

Author

drini (3)

Entry type

Definition

Classification

msc 26D15

Classification

msc 11-00

Synonym

average

Synonym

mean

Related topic

GeometricMean

Related topic

HarmonicMean

Related topic

ArithmeticGeometricMeansInequality

Related topic

GeneralMeansInequality

Related topic

WeightedPowerMean

Related topic

PowerMean

Related topic

GeometricDistribution2

Related topic

RootMeanSquare3

Related topic

ProofOfGeneralMeansInequality

Related topic

ProofOfArithmeticGeometricHarmonicMeansInequality

Related topic

DerivationOfHarm

Defines

weighted mean

Defines

weighted average

Generated on Fri Feb 9 20:01:21 2018 by

LaTeXML
average value of function

average value of function

The set of the values of a real function

f

defined on an

interval

[

a

,

b

]

is usually

uncountable

, and therefore for being able to speak of an

average value

of

f

in the sense of the average value

A

.

V

.

=

a

1

+

a

2

+

…

+

a

n

n

=

∑

j

=

1

n

a

j

∑

j

=

1

n

1

(1)

of a finite list

a

1

,

a

2

,

…

,

a

n

of numbers, one has to replace the sums with

integrals

.  Thus one could define

A

.

V

.

(

f

)

:=

∫

a

b

f

⁢

(

x

)

⁢

𝑑

x

∫

a

b

1

⁢

𝑑

x

,

i.e.

A

.

V

.

(

f

)

:=

1

b

-

a

⁢

∫

a

b

f

⁢

(

x

)

⁢

𝑑

x

.

(2)

For example, the average value of

x

2

on the interval

[

0

,

1

]

is

1

3

and the average value of

sin

⁡

x

on the interval

[

0

,

π

]

is

2

π

.

The definition (2) may be extended to

complex functions

f

on an arc

γ

of a

rectifiable curve

via the

contour integral

A

.

V

.

(

f

)

:=

1

l

⁢

(

γ

)

⁢

∫

γ

f

⁢

(

z

)

⁢

𝑑

z

(3)

where

l

⁢

(

γ

)

is the length (

http://planetmath.org/ArcLength

) of the arc.  If especially

γ

is a

closed curve

in a simply connected domain where

f

is

analytic

, then the average value of

f

on

γ

is always 0, as the

Cauchy integral theorem

implies.

Title

average value of function

Canonical name

AverageValueOfFunction

Date of creation

2013-03-22 19:01:54

Last modified on

2013-03-22 19:01:54

Owner

pahio (2872)

Last modified by

pahio (2872)

Numerical id

12

Author

pahio (2872)

Entry type

Definition

Classification

msc 26D15

Classification

msc 11-00

Related topic

ArithmeticMean

Related topic

Mean3

Related topic

Countable

Related topic

GaussMeanValueTheorem

Related topic

Expectation

Related topic

MeanSquareDeviation

Defines

average value

Generated on Fri Feb 9 20:01:23 2018 by

LaTeXML
Carleman’s inequality

Carleman’s inequality

Theorem

(

[

1

]

, pp. 24)
For positive real numbers

{

a

n

}

n

=

1

∞

,

Carleman’s

inequality

states that

∑

n

=

1

∞

(

a

1

⁢

a

2

⁢

⋯

⁢

a

n

)

1

/

n

≤

e

⁢

∑

n

=

1

∞

a

n

.

Although the constant

e

(the

natural log base

)
is optimal, it is possible to refine
Carleman’s inequality by

decreasing

the weight

coefficients

on the right hand side

[

2

]

.

References

1

L. Hörmander,

The

Analysis

of Linear Partial Differential Operators I,
(

Distribution

theory and

Fourier Analysis

)

, 2nd ed, Springer-Verlag, 1990.

2

B.Q. Yuan,

Refinements

of Carleman’s inequality

,
Journal of Inequalities in Pure and Applied Mathematics, Vol. 2, Issue 2, 2001,
Article 21.

http://jipam.vu.edu.au/v2n2/029_00.html

online

Title

Carleman’s inequality

Canonical name

CarlemansInequality

Date of creation

2013-03-22 13:43:17

Last modified on

2013-03-22 13:43:17

Owner

Koro (127)

Last modified by

Koro (127)

Numerical id

5

Author

Koro (127)

Entry type

Definition

Classification

msc 26D15

Generated on Fri Feb 9 20:01:25 2018 by

LaTeXML
Chebyshev’s inequality

Chebyshev’s inequality

If

x

1

,

x

2

,

…

,

x

n

and

y

1

,

y

2

,

…

,

y

n

are two sequences (at least one of them consisting of positive numbers):

•

if

x

1

<

x

2

<

⋯

<

x

n

and

y

1

<

y

2

<

⋯

<

y

n

then

(

x

1

+

x

2

+

⋯

+

x

n

n

)

⁢

(

y

1

+

y

2

+

⋯

+

y

n

n

)

≤

x

1

⁢

y

1

+

x

2

⁢

y

2

+

⋯

+

x

n

⁢

y

n

n

.

•

if

x

1

<

x

2

<

⋯

<

x

n

and

y

1

>

y

2

>

⋯

>

y

n

then

(

x

1

+

x

2

+

⋯

+

x

n

n

)

⁢

(

y

1

+

y

2

+

⋯

+

y

n

n

)

≥

x

1

⁢

y

1

+

x

2

⁢

y

2

+

⋯

+

x

n

⁢

y

n

n

.

Title

Chebyshev’s

inequality

Canonical name

ChebyshevsInequality

Date of creation

2013-03-22 11:47:36

Last modified on

2013-03-22 11:47:36

Owner

drini (3)

Last modified by

drini (3)

Numerical id

7

Author

drini (3)

Entry type

Theorem

Classification

msc 26D15

Classification

msc 18F99

Classification

msc 58Z05

Related topic

RearrangementInequality

Related topic

ProofOfRearrangementInequality

Related topic

KolmogorovsInequality

Related topic

ChebyshevsInequality2

Generated on Fri Feb 9 20:01:27 2018 by

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