aarjaneiro/planetmath huggingface dataset:
sum-product theorem

sum-product theorem

Suppose

𝔽

is a

finite field

. Then given a subset

A

of

𝔽

we define the

sum

of

A

to be the set

A

+

A

=

{

a

+

b

:

a

,

b

∈

A

}

and the product to be the set

A

⋅

A

=

{

a

⋅

b

:

a

,

b

∈

A

}

.

We concern ourselves with estimating the size of

A

+

A

and

A

⋅

A

relative to the size of

A

, and ultimately also to the size of

𝔽

.

If

A

is empty then

A

+

A

is empty as is

A

⋅

A

and so

|

A

|

=

|

A

+

A

|

=

|

A

⋅

A

|

. Now suppose

A

is non-empty then let

a

∈

A

.
Then

a

+

A

=

{

a

+

b

:

b

∈

A

}

⊂

A

+

A

so

|

A

|

≤

|

A

+

A

|

. If

A

=

{

0

}

then

A

⋅

A

=

A

so finally assume

a

∈

A

,

a

≠

0

. Then we have

a

⋅

A

=

{

a

⋅

b

:

b

∈

A

}

⊂

A

⋅

A

so in any case it always follows that

|

A

|

≤

|

A

+

A

|

,

|

A

⋅

A

|

.

(1)

Now if

𝔽

has a

proper subfield

𝔽

0

– for instance

𝔽

=

G

⁢

F

⁢

(

p

2

)

and

𝔽

0

=

G

⁢

F

⁢

(

P

)

– then setting

A

=

𝔽

0

makes

A

=

A

+

A

=

A

⋅

A

and so in this situation the bound in (

1

)
is tight, that is,

|

A

|

=

|

A

+

A

|

=

|

A

⋅

A

|

. So we insist now that

𝔽

is a

prime field

, so it has no proper subfields.

We would like to understand what size

A

must have to ensure that either

A

+

A

or

A

⋅

A

is larger than

A

. (Note this is not the same as asking
if

A

≠

A

+

A

or

A

⋅

A

as we are concerned only with growth in size not the change in the elements of the set.) Clearly

A

=

{

0

}

fails, as does

A

=

𝔽

and with some intuition as guidance it is safe to presume that

A

must be large enough to have enough elements to produce many elements as a sum or product but also small enough that these these new elements outgrow
the size of

A

. This is the content of the following important result.

Theorem 1

(Sum-Product estimate:Bourgain-Katz-Tao (2003)).

Let

F

=

Z

p

be the field of prime order

p

. Let

A

be any subset of

F

such that

|

𝔽

|

δ

<

|

A

|

<

|

𝔽

|

1

-

δ

for some

δ

>

0

. Then

max

⁡

{

|

A

+

A

|

,

|

A

⋅

A

|

}

≥

C

⁢

|

A

|

1

+

ε

for some

ε

>

0

which depends on

δ

and some constant

C

which also depends on

δ

.

The proof is non-trivial. Jean Bourgain was awarded the Fields’ medal in 1994,
Terence Tao in 2006 with the prize in part due to his various contributions in

additive number theory

.

http://www.arxiv.org/abs/math/0301343

Bourgain, Katz, and Tao,

A

Sum-Product estimate

in finite fields, and applications

, (preprint) arXiv:math/CO/0301343 v2, 2003.

Title

sum-product theorem

Canonical name

SumproductTheorem

Date of creation

2013-03-22 16:54:48

Last modified on

2013-03-22 16:54:48

Owner

Algeboy (12884)

Last modified by

Algeboy (12884)

Numerical id

8

Author

Algeboy (12884)

Entry type

Theorem

Classification

msc 11T99

Classification

msc 05B25

Synonym

Sum-product estimate

Generated on Fri Feb 9 15:47:03 2018 by

LaTeXML
echelon factoring algorithm

echelon factoring algorithm

Here’s a specific example that’s missing from everyone’s repertoire, when it comes to having a general factoring method:

Example one

Let

N

=

7477

=

61

*

127

, and

N

≈

88.01

, and compute the following:

M

=

2

(

7477

+

1

)

=

2

7478

now,

X

=

mod

(

M

,

N

-

1

)

, and

gcd

⁡

(

X

-

[

2

0

]

,

N

)

,

gcd

⁡

(

X

-

[

2

1

]

,

N

)

,…,

gcd

⁡

(

X

-

[

2

k

]

2

,

N

)

such that

[

2

k

]

2

≤

N

.

It’ll find the factor, but we would have to use George Woltman’s FFT’s method to compute the

M

’s for larger numbers. In this example,

M

mod

(

N

-

1

)

=

246

, and

gcd

⁡

(

246

-

2

,

N

)

=

61

.

Also,

calculate

l

⁢

n

⁢

(

7477

)

≈

8.955

, so, you could continue to check

gcd

⁡

(

X

-

[

3

1

]

,

N

)

,

…

,

gcd

⁡

(

X

-

[

3

k

]

,

N

)

such that

[

3

k

]

2

≤

s

⁢

q

⁢

r

⁢

t

⁢

(

N

)

, and

gcd

⁡

(

X

-

[

5

2

]

)

such that

[

5

k

]

2

≤

N

, if a factor hasn’t shown itself. Unlike primality-proving, finding the factor would be the ”proof-in-the-pudding”!

We’d have the answer; that’s for sure! I call it a step or ”echelon”-factoring algorithm.

Example two

1500450271

*

5915587277

=

8876044532898802067

You’d use this fact to get past the first

modpow

⁢

(

)

N

+

1

=

8876044532898802067

+

1

=

2

2

*

3

*

(

2

4

+

1

)

*

(

2

*

3

4

*

(

2

*

(

2

2

*

5

*

(

2

2

*

5

3

*

(

2

*

(

2

2

*

7

+

1

)

+

1

)

+

1

)

+

1

)

+

1

)

*

(

2

3

*

3

*

5

2

*

7

*

(

2

*

(

2

*

(

2

*

5

+

1

)

+

1

)

+

1

)

*

(

2

7

*

3

2

+

1

)

+

1

)

+

1

)

M

=

2

(

8876044532898802067

+

1

)

k

=

l

⁢

o

⁢

g

⁢

2

⁢

[

8876044532898802067

]

=

15

so, 1 huge step and 31 base two subtractions, and some base 3, 5, …, 43, etc!
…
find

mod

(

2

N

+

1

,

N

-

1

)

, and

gcd

⁡

(

M

-

[

2

0

]

,

N

)

,

gcd

⁡

(

M

-

[

2

1

]

,

N

)

⁢

…

,

gcd

⁡

(

M

-

[

2

30

]

,

N

)

, etc. and you’ll have a factor.

It’s the best, shortest method that you’ll ever use to check for factors, and it’s definitive, assuming we can conquer the enormous modular calculation!

In this last example, the number of steps is comparable to the 2 times 16th Root of

N

for the base 2 calculations alone. I couldn’t do the calculations by hand.

Title

echelon factoring algorithm

Canonical name

EchelonFactoringAlgorithm

Date of creation

2013-03-22 19:36:43

Last modified on

2013-03-22 19:36:43

Owner

leavemsg2 (21852)

Last modified by

leavemsg2 (21852)

Numerical id

11

Author

leavemsg2 (21852)

Entry type

Algorithm

Classification

msc 11Y05

Synonym

step

Synonym

echelon

Defines

factoring algorithm

Generated on Fri Feb 9 15:47:06 2018 by

LaTeXML
Miller-Rabin prime test

Miller-Rabin prime test

The

Miller-Rabin prime test

is a probabilistic test based on the following

Theorem 1

.

Let

N

≥

3

be an

odd number

and

N

-

1

=

2

t

⁢

n

,

n

odd. If and only if for every

a

∈

Z

with

gcd

⁡

(

a

,

N

)

=

1

we have

a

n

≡

1

mod

⁢

N

𝑜𝑟

(1a)

a

2

s

⁢

n

≡

-

1

mod

⁢

N

(1b)

for some

s

∈

{

0

,

…

,

t

-

1

}

, then

N

is prime. If

N

is not prime, then let

A

be the set of all

a

satisfying the above conditions. We have

𝐶𝑎𝑟𝑑

⁢

(

A

)

≤

1

4

⁢

φ

⁢

(

N

)

,

where

φ

is Euler’s Phi-function.

A

composite number

N

satisfying conditions (

1

) for some

a

∈

ℤ

is called a

strong

in the basis

a

. Note that this theorem states that there are no such things as

Carmichael numbers

for

strong pseudoprimes

(i.e. composite numbers that are strong pseudoprimes for all values of

a

).
From this theorem we can construct the following

algorithm

:

1.

Choose a random

2

≤

a

≤

N

-

1

.

2.

If

gcd

⁡

(

a

,

N

)

≠

1

, we have a non-trivial

divisor

of

N

, so

N

is not prime.

3.

If

gcd

⁡

(

a

,

N

)

=

1

, check if the conditions (

1

) are satisfied. If not,

N

is not prime, otherwise the probability that

N

is not prime is less than

1

4

.

In general the probability, that the test falsely reports

N

as a prime is far less than

1

4

. Of course the algorithm can be applied several times in order to reduce the probability, that

N

is not a prime but reported as one.

When comparing this test with the related

Solovay-Strassen test

one sees that this test is superior is several ways:

•

There is no need to

calculate

the

Jacobi symbol

.

•

The amount of

false

witnesses

, i.e. numbers

a

that report

N

as a prime when it is not, is at most

N

4

instead of

N

2

.

•

One can even show that

N

which passes Miller-Rabin for some

a

also passes Solovay-Strassen for that

a

, so Miller-Rabin is always better.

Note that when the test returns that

N

is composite, then this is indeed the case, so it is really a

compositeness test

rather than a test for

primality

.

Title

Miller-Rabin prime test

Canonical name

MillerRabinPrimeTest

Date of creation

2013-03-22 14:54:18

Last modified on

2013-03-22 14:54:18

Owner

mathwizard (128)

Last modified by

mathwizard (128)

Numerical id

9

Author

mathwizard (128)

Entry type

Algorithm

Classification

msc 11Y11

Synonym

Rabin prime test

Related topic

SolovayStrassenTest

Related topic

FermatCompositenessTest

Defines

strong pseudoprime

Generated on Fri Feb 9 15:47:11 2018 by

LaTeXML
Kolakoski sequence

Kolakoski sequence

A

Kolakoski sequence

is a “self-describing”

sequence

{

k

n

}

k

=

0

∞

of alternating blocks of 1’s and 2’s,
given by the following rules:

•

k

0

=

1

.

1

1

Some sources start the sequence at

k

0

=

2

, instead. This only has the effect of shifting the sequence by one position.

•

k

n

is the length of the

(

n

+

1

)

’th block.

Thus, the sequence begins 1, 2, 2, 1, 1, 2, 1, 2, 2, 1, 2, 2, 1, 1, 2, 1, …

It is conjectured that the density of 1’s in the sequence is 0.5. It is not known whether the 1’s

have

a density; however, it

is

known that were this true, that density would be 0.5. It is also not known whether the sequence is a strongly recurrent sequence; this too would imply density 0.5.

Extensive computer experiments strongly support the

conjecture

. Furthermore, if

o

n

is the number of 1’s in the first

n

elements, then it appears that

o

n

=

0.5

⁢

n

+

O

⁢

(

log

⁡

n

)

. Note for comparison that for a

random

sequence of 1’s and 2’s, the number of 1’s in the first

n

elements is with high probability

0.5

⁢

n

+

O

⁢

(

n

)

.

To generate rapidly a large number of elements of the sequence, it is most efficient to build a heirarchy of generators for the sequence.

If

the conjecture is correct, then the depth of this heirarchy is only

O

⁢

(

log

⁡

n

)

to generate the first

n

elements.

This is

http://www.research.att.com/cgi-bin/access.cgi/as/njas/sequences/eisA.cgi?Anum=A000002

sequence A000002
in

http://www.research.att.com/ njas/sequences/Seis.html

the

Online Encyclopedia of Integer Sequences

.

Title

Kolakoski sequence

Canonical

name

KolakoskiSequence

Date of creation

2013-03-22 12:47:33

Last modified on

2013-03-22 12:47:33

Owner

PrimeFan (13766)

Last modified by

PrimeFan (13766)

Numerical id

6

Author

PrimeFan (13766)

Entry type

Definition

Classification

msc 11Y55

Classification

msc 94A55

Synonym

Kolakowski’s sequence

Generated on Fri Feb 9 15:47:17 2018 by

LaTeXML
arithmetic derivative

arithmetic derivative

The

arithmetic derivative

n

′

of a

natural number

n

is defined by the following rules:

•

p

′

=

1

for any prime

p

.

•

(

a

⁢

b

)

′

=

a

′

⁢

b

+

a

⁢

b

′

for any

a

,

b

∈

ℕ

(

Leibniz rule

).

To define the arithmetic derivative of a negative number, we first note that

1

′

=

0

by the Leibniz rule (

1

′

=

(

1

⋅

1

)

′

=

1

⋅

1

′

+

1

′

⋅

1

=

2

⋅

1

′

, so

1

′

=

0

), and further that we must have

0

=

1

′

=

(

(

-

1

)

⋅

(

-

1

)

)

′

=

-

2

⋅

(

-

1

)

′

,

so

(

-

1

)

′

=

0

. The

product rule

now requires that we define

(

-

n

)

′

=

-

(

n

′

)

+

n

⁢

(

-

1

)

′

=

-

(

n

′

)

.

Further, we can extend this definition to

rational numbers

by insisting that the

quotient rule

holds, i.e. for a prime

p

we should have

0

=

1

′

=

(

p

⋅

1

p

)

′

=

(

1

p

)

′

⁢

p

+

1

p

,

giving us that

(

1

p

)

′

=

-

1

p

2

,

i.e. the usual quotient rule from

calculus

. We now

complete

the definition by extending multiplicatively (i.e. using the Leibniz rule).

The arithmetic derivatives for the first few

positive integers

are 0, 1, 1, 4, 1, 5, 1, 12, 6, 7, 1, 16, 1, 9, 8, 32, 1, 21, 1, 24, 10, 13, 1, 44, 10, etc.

As a

consequence

of

p

′

=

1

for a prime

p

, the arithmetic derivative of a

semiprime

(whether

squarefree

or not) works out to

(

p

⁢

q

)

′

=

p

′

⁢

q

+

p

⁢

q

′

=

1

⁢

p

+

q

⁢

1

=

p

+

q

. For example, the arithmetic derivative of 10 is 7, which is 2 plus 5.

The only cases of

n

′

=

n

for

-

1

<

n

<

1024

are 0, 4, 27.

n

n

′

n

n

′

n

n

′

n

n

′

n

n

′

n

n

′

n

n

′

n

n

′

n

n

′

n

n

′

0

0

10

7

20

24

30

31

40

68

50

45

60

92

70

59

80

176

90

123

1

0

11

1

21

10

31

1

41

1

51

20

61

1

71

1

81

108

91

20

2

1

12

16

22

13

32

80

42

41

52

56

62

33

72

156

82

43

92

96

3

1

13

1

23

1

33

14

43

1

53

1

63

51

73

1

83

1

93

34

4

4

14

9

24

44

34

19

44

48

54

81

64

192

74

39

84

124

94

49

5

1

15

8

25

10

35

12

45

39

55

16

65

18

75

55

85

22

95

24

6

5

16

32

26

15

36

60

46

25

56

92

66

61

76

80

86

45

96

272

7

1

17

1

27

27

37

1

47

1

57

22

67

1

77

18

87

32

97

1

8

12

18

21

28

32

38

21

48

112

58

31

68

72

78

71

88

140

98

77

9

6

19

1

29

1

39

16

49

14

59

1

69

26

79

1

89

1

99

75

References

1

EJ Barbeau, “Remark on an arithmetic derivative”.

Can. Math. Bull.

4

(1961): 117 - 122

Title

arithmetic derivative

Canonical name

ArithmeticDerivative

Date of creation

2013-03-22 13:35:09

Last modified on

2013-03-22 13:35:09

Owner

PrimeFan (13766)

Last modified by

PrimeFan (13766)

Numerical id

14

Author

PrimeFan (13766)

Entry type

Definition

Classification

msc 11Z05

Related topic

Prime

Generated on Fri Feb 9 15:47:42 2018 by

LaTeXML
determining integer contraharmonic means

determining integer contraharmonic means

For determining effectively values

c

of

integer contraharmonic means

of two

positive integers

u

and

v

(

1

<

u

<

v

), it’s convenient to start from the
(7) in the
parent entry (

http://

planetmath

.org/IntegerContraharmonicMeans

):

v

=

2

⁢

u

2

w

-

u

(1)

where

w

is any

positive

factor of

2

⁢

u

2

less than

u

.  Substituting the above

expression

of

v

to the defining expression

c

=

u

2

+

v

2

u

+

v

of

c

, this gets the form

c

=

2

⁢

u

2

w

-

2

⁢

u

+

w

.

(2)

Hence one can use the formulae (1) and (2), giving in them for each desired

u

the values

w

of the positive factors of

2

⁢

u

2

, beginning from

w

:=

1

and stopping before

w

=

u

.

The for the

integer

harmonic mean

, corresponding (2), is simply

h

=

2

⁢

u

-

w

.

(3)

Example.

In the following table one sees for

u

=

36

all possible values of the

parametre

w

and the corresponding
values of

c

and

h

; the pertinent values of

v

are given, too.

w

1

2

3

4

6

8

9

12

16

18

24

27

32

v

2556

1260

828

612

396

288

252

180

126

108

72

60

45

c

2521

1226

795

580

366

260

225

156

106

90

60

51

41

h

71

70

69

68

66

64

63

60

56

54

48

45

40

As one sees, the contraharmonic and the harmonic mean may differ considerably, but also the

difference

1 is possible.

References

1

J. Pahikkala

: “On

contraharmonic mean

and

Pythagorean triples

”.  –

Elemente der Mathematik

65

:2 (2010).

Title

determining integer contraharmonic means

Canonical

name

DeterminingIntegerContraharmonicMeans

Date of creation

2013-11-19 18:13:25

Last modified on

2013-11-19 18:13:25

Owner

pahio (2872)

Last modified by

pahio (2872)

Numerical id

15

Author

pahio (2872)

Entry type

Algorithm

Classification

msc 11Z05

Classification

msc 11A05

Classification

msc 11D09

Classification

msc 11D45

Related topic

LinearFormulasForPythagoreanTriples

Generated on Fri Feb 9 15:47:47 2018 by

LaTeXML
example of arithmetic derivative

example of arithmetic derivative

Consider the

natural number

6

. Using the rules of the

arithmetic derivative

we get:

6

′

=

(

2

⋅

3

)

′

=

2

′

⋅

3

+

2

⋅

3

′

=

1

⋅

3

+

2

⋅

1

=

5

Below is a list of the 10 first natural numbers and their first and second arithmetic derivatives:

n

1

2

3

4

5

6

7

8

9

10

n

′

0

1

1

4

1

5

1

12

6

7

n

′′

0

0

0

4

0

1

0

16

5

1

Title

example of arithmetic derivative

Canonical name

ExampleOfArithmeticDerivative

Date of creation

2013-03-22 13:35:12

Last modified on

2013-03-22 13:35:12

Owner

Johan (1032)

Last modified by

Johan (1032)

Numerical id

7

Author

Johan (1032)

Entry type

Example

Classification

msc 11Z05

Generated on Fri Feb 9 15:47:49 2018 by

LaTeXML
quotient rule for arithmetic derivative

quotient rule for arithmetic derivative

Theorem

.

If the notion of

arithmetic derivative

is extended to

rational numbers

, then we have that, for every

a

,

b

∈

Z

with

b

≠

0

:

(

a

b

)

′

=

a

′

⁢

b

-

b

′

⁢

a

b

2

Proof.

Note that

a

′

=

(

b

⋅

a

b

)

′

=

b

⋅

(

a

b

)

′

+

b

′

⋅

a

b

by the Leibniz rule.

Thus,

b

⋅

(

a

b

)

′

=

a

′

-

b

′

⋅

a

b

=

a

′

⁢

b

-

b

′

⁢

a

b

.

It follows that

(

a

b

)

′

=

a

′

⁢

b

-

b

′

⁢

a

b

2

.

∎

Title

quotient rule for arithmetic derivative

Canonical name

QuotientRuleForArithmeticDerivative

Date of creation

2013-03-22 17:04:44

Last modified on

2013-03-22 17:04:44

Owner

Wkbj79 (1863)

Last modified by

Wkbj79 (1863)

Numerical id

4

Author

Wkbj79 (1863)

Entry type

Theorem

Classification

msc 11Z05

Generated on Fri Feb 9 15:48:07 2018 by

LaTeXML
cardinality of monomials

cardinality of monomials

Theorem 1

.

If

S

is a

finite set

of

variable

symbols, then the number of

monomials

of
degree

n

constructed from these symbols is

(

n

+

m

-

1

n

)

, where

m

is the

cardinality

of

S

.

Proof.

The proof proceeds by inducion on the cardinality of

S

. If

S

has but one

element

, then there is but one monomial of degree

n

, namely the sole element
of

S

raised to the

n

-th power. Since

(

n

+

1

-

1

n

)

=

1

, the

conclusion

holds when

m

=

1

.

Suppose, then, that the result holds whenver

m

<

M

for some

M

. Let

S

be
a set with exactly

M

elements and let

x

be an element of

S

. A monomial
of degree

n

constructed from elements of

S

can be expressed as the

product

of a power of

x

and a monomial constructed from the elements of

S

∖

{

x

}

. By the

induction hypothesis

, the number of monomials of degree

k

constructed from elements of

S

∖

{

x

}

is

(

k

+

M

-

2

k

)

.

Summing

over the possible powers to which

x

may be raised, the number of
monomials of degree

n

constructed from the elements of

S

is as follows:

∑

k

=

0

n

(

k

+

M

-

2

k

)

=

(

k

+

M

-

1

k

)

∎

Theorem 2

.

If

S

is an

infinite set

of variable symbols, then the number of monomials
of degree

n

constructed from these symbols equals the cardinality of

S

.

Title

cardinality of monomials

Canonical

name

CardinalityOfMonomials

Date of creation

2013-03-22 16:34:42

Last modified on

2013-03-22 16:34:42

Owner

rspuzio (6075)

Last modified by

rspuzio (6075)

Numerical id

10

Author

rspuzio (6075)

Entry type

Theorem

Classification

msc 12-00

Generated on Fri Feb 9 16:36:47 2018 by

LaTeXML
homogeneous polynomial

homogeneous polynomial

A

polynomial

P

⁢

(

x

1

,

⋯

,

x

n

)

of degree

k

is called

homogeneous

if

P

⁢

(

c

⁢

x

1

,

⋯

,

c

⁢

x

n

)

=

c

k

⁢

P

⁢

(

x

1

,

⋯

,

x

n

)

for all

constants

c

.

An

equivalent

definition

is that all terms of the polynomial have the same degree (i.e.

k

).

Observe that a polynomial

P

is homogeneous iff

deg

⁡

P

=

ord

⁡

P

.

As an important example of

homogeneous polynomials

one can mention the

symmetric polynomials

.

Title

homogeneous polynomial

Canonical

name

HomogeneousPolynomial

Date of creation

2013-03-22 13:21:11

Last modified on

2013-03-22 13:21:11

Owner

jgade (861)

Last modified by

jgade (861)

Numerical id

11

Author

jgade (861)

Entry type

Definition

Classification

msc 12-00

Generated on Fri Feb 9 16:36:49 2018 by

LaTeXML
monomial

monomial

A

monomial

is a

product

of non-negative powers of

variables

. It may also include an optional

coefficient

(which is sometimes ignored when discussing particular properties of monomials). A

polynomial

can be thought of as a sum over a set of monomials.

For example, the following are monomials.

1

x

x

2

⁢

y

x

⁢

y

⁢

z

3

⁢

x

4

⁢

y

2

⁢

z

3

-

z

If there are

n

variables from which a monomial may be formed, then
a monomial may be represented without its coefficient as a vector of

n

naturals. Each position in this vector would correspond to a particular
variable, and the value of the element at each position would correspond
to the power of that variable in the monomial. For instance, the monomial

x

2

⁢

y

⁢

z

3

formed from the set of variables

{

w

,

x

,

y

,

z

}

would be represented as

(

0

2

1

3

)

T

. A constant would be a

zero vector

.

Given this

representation

, we may define a few more

concepts

. First, the

degree of a monomial

is the sum of the elements of its vector representation. Thus, the degree of

x

2

⁢

y

⁢

z

3

is

0

+

2

+

1

+

3

=

6

,
and the degree of a constant is 0. If a polynomial is represented as a sum
over a set of monomials, then the degree of a polynomial can be defined as the
degree of the monomial of largest degree belonging to that polynomial.

Title

monomial

Canonical name

Monomial

Date of creation

2013-03-22 12:34:32

Last modified on

2013-03-22 12:34:32

Owner

bbukh (348)

Last modified by

bbukh (348)

Numerical id

5

Author

bbukh (348)

Entry type

Definition

Classification

msc 12-00

Defines

degree of a monomial

Generated on Fri Feb 9 16:36:53 2018 by

LaTeXML
order and degree of polynomial

order and degree of polynomial

Let

f

be a

polynomial

in two

variables

, viz.

f

⁢

(

x

,

y

)

=

∑

i

,

j

a

i

⁢

j

⁢

x

i

⁢

y

j

.

1

1

In order to simplify the notation, the

definition

is given in terms of a polynomial in two variables, however the definition naturally scales to any number of variables.

Then the

total degree

of

f

is given by:

deg

⁢

f

=

sup

⁡

{

i

+

j

|

a

i

⁢

j

≠

0

}

Note the degree of the zero-polynomial is

-

∞

, since

sup

⁡

∅

(per definition) is

-

∞

, thus

deg

⁢

f

∈

ℕ

∪

{

0

}

∪

{

-

∞

}

.

Similarly the

order

of

f

is given by:

ord

⁢

f

=

inf

⁡

{

i

+

j

|

a

i

⁢

j

≠

0

}

Note the order of the zero-polynomial is

∞

(because

inf

⁡

∅

=

∞

). Thus

ord

⁢

f

∈

ℕ

∪

{

0

}

∪

{

∞

}

.

Please note that the term order is not as common as degree. In fact, it is perhaps more frequently associated with power series (a form of generalized polynomials) than with ordinary polynomials. Also be aware that the term order occasionally is used as a synonym for degree.

Title

order and degree of polynomial

Canonical

name

OrderAndDegreeOfPolynomial

Date of creation

2013-03-22 13:38:21

Last modified on

2013-03-22 13:38:21

Owner

jgade (861)

Last modified by

jgade (861)

Numerical id

8

Author

jgade (861)

Entry type

Definition

Classification

msc 12-00

Related topic

ZeroPolynomial2

Defines

order

Defines

degree

Generated on Fri Feb 9 16:36:54 2018 by

LaTeXML
weighted homogeneous polynomial

weighted homogeneous polynomial

Let

𝔽

be either the real or

complex numbers

.

Definition

.

Let

p

:

𝔽

n

→

𝔽

be a

polynomial

in

n

variables

and take

integers

d

1

,

d

2

,

…

,

d

n

.
The polynomial

p

is said to be

weighted

homogeneous of degree

k

if for all

t

>

0

we have

p

⁢

(

t

d

1

⁢

x

1

,

t

d

2

⁢

x

2

,

…

,

t

d

n

⁢

x

n

)

=

t

k

⁢

p

⁢

(

x

1

,

x

2

,

…

,

x

n

)

.

The

d

1

,

…

,

d

n

are called the

weights

of the variables

x

1

,

…

,

x

n

.

Note that if

d

1

=

d

2

=

…

=

d

n

=

1

then this definition is the standard

homogeneous polynomial

.

Title

weighted homogeneous polynomial

Canonical name

WeightedHomogeneousPolynomial

Date of creation

2013-03-22 15:21:18

Last modified on

2013-03-22 15:21:18

Owner

jirka (4157)

Last modified by

jirka (4157)

Numerical id

5

Author

jirka (4157)

Entry type

Definition

Classification

msc 12-00

Generated on Fri Feb 9 16:36:56 2018 by

LaTeXML