aarjaneiro/planetmath huggingface dataset:
Kenosymplirostic numbers

Kenosymplirostic numbers

So, basically. I am now introducing a new

number system

called kenosymplirostic numbers.
It is a type of number that you get when dividing by 0.
So, what does kenosymplirostic mean?
The name comes from two Greek words, ”

keno

” meaning ”gap” and ”

sympliroste

” meaning ”fill”.
These numbers fill the gap that was left open, the gap people thought will never be closed.

Here is how it works:

1

0

=

1

⁢

k

The value of the

numerator

is the

coefficient

of ”kenosym unit” k.

So, what about 0/0?

The thing is, it will be 0. Why? Because 0*k(the kenosym unit) is still 0 by

definition

.

And now, we talk about its place in the

complex plane

, or with the

addition

of
kenosyymplirostic numbers, the complex space.
It will go through the complex plane where it will meet in 0, the origin of real,

imaginary

and
kenosymplirostic numbers.

ADDITION AND

SUBTRACTION

So, let’s say we have:

1

⁢

k

+

3

⁢

k

It will be equal to:

1

0

+

3

0

Which will be

4

0

⁢

o

⁢

r

⁢

4

⁢

k

This means that subtraction is just the same, as we treat 0 as a

normal number

.

MULTIPLICATION

So, let’s say we have:

3

⁢

k

⋅

5

⁢

k

That would mean 3/0 * 5/0 which will be 15/0 or 15k.
(K IS NOT TREATED AS A

VARIABLE

)
Kenosymplirostic numbers cannot be divided since it will always equal to 0 per the

properties

of dividing fractions.

EXPONENT

The kenosym unit k acts like the

real number

1, where k to the power of any number n is k.
kenosym numbers with coefficients will just raise the coefficient to that power and just
copy the kenosym unit k.

RULES OF COEFFICIENTS

Only

integers

coefficients are allowed since there cant be a fourth dimension in
the complex space and to avoiid many numbers having the same kenosym.

Axioms:
Addition:

a+b=b+a (

Commutativity

)

a+(b+c)=(a+b)+c (

Associativity

)

a+0=a (

Identity element

exists)

a+(âa)=0 (

Inverse

exists)

Multiplication:

ab=ba (Commutativity)

a(bc)=(ab)c (Associativity)

a*1=a (Identity element exists)

Distributive Property:

a(b+c)=ab+ac

Shoutouts:
Herve Arki from the Mathematics G+ community for bringing up an issue. :)

As of now, there are some paradoxes that might arise, but I will resolve all once I gain
more knowledge.
Also, if someone would help suggest to me a better kenosym unit, You can comment below
this article.

Title

Kenosymplirostic numbers

Canonical

name

KenosymplirosticNumbers

Date of creation

2014-10-11 21:53:52

Last modified on

2014-10-11 21:53:52

Owner

imaginary.i (1001376)

Last modified by

imaginary.i (1001376)

Numerical id

6

Author

imaginary.i (1001376)

Entry type

Definition

Synonym

numbers

Synonym

pi

Related topic

pi

Related topic

tau

Related topic

phi

Defines

Kenosymplirostic numbers

Generated on Fri Feb 9 19:50:29 2018 by

LaTeXML
l’Hôpital’s rule

l’Hôpital’s rule

L’Hôpital’s rule states that given an unresolvable limit of the form

0

0

or

∞

∞

, the ratio of functions

f

⁢

(

x

)

g

⁢

(

x

)

will have the same limit at

c

as the ratio

f

′

⁢

(

x

)

g

′

⁢

(

x

)

. In short, if the limit of a ratio of functions approaches an

indeterminate form

, then

lim

x

→

c

⁡

f

⁢

(

x

)

g

⁢

(

x

)

=

lim

x

→

c

⁡

f

′

⁢

(

x

)

g

′

⁢

(

x

)

provided this last limit exists. L’Hôpital’s rule may be applied indefinitely as long as the conditions are satisfied. However it is important to note, that the nonexistance of

lim

⁡

f

′

⁢

(

x

)

g

′

⁢

(

x

)

does not prove the nonexistance of

lim

⁡

f

⁢

(

x

)

g

⁢

(

x

)

.

Example:

We try to determine the value of

lim

x

→

∞

⁡

x

2

e

x

.

As

x

approaches

∞

the

expression

becomes an indeterminate form

∞

∞

. By applying L’Hôpital’s rule twice we get

lim

x

→

∞

⁡

x

2

e

x

=

lim

x

→

∞

⁡

2

⁢

x

e

x

=

lim

x

→

∞

⁡

2

e

x

=

0

.

Another example of the usage of L’Hôpital’s rule can be found

http://planetmath.org/node/5741

here.

Title

l’Hôpital’s rule

Canonical

name

LHopitalsRule

Date of creation

2013-03-22 12:28:15

Last modified on

2013-03-22 12:28:15

Owner

mathwizard (128)

Last modified by

mathwizard (128)

Numerical id

13

Author

mathwizard (128)

Entry type

Theorem

Classification

msc 26A24

Classification

msc 26C15

Synonym

l’Hospital’s rule

Related topic

IndeterminateForm

Related topic

DerivationOfHarmonicMeanAsTheLimitOfThePowerMean

Related topic

ImproperLimits

Related topic

ExampleUsingStolzCesaroTheorem

Generated on Fri Feb 9 19:50:31 2018 by

LaTeXML
proof of De l’Hôpital’s rule

proof of De l’Hôpital’s rule

Let

x

0

∈

ℝ

,

I

be an

interval

containing

x

0

and let

f

and

g

be two

differentiable functions

defined on

I

∖

{

x

0

}

with

g

′

⁢

(

x

)

≠

0

for all

x

∈

I

. Suppose that

lim

x

→

x

0

⁡

f

⁢

(

x

)

=

0

,

lim

x

→

x

0

⁡

g

⁢

(

x

)

=

0

and that

lim

x

→

x

0

⁡

f

′

⁢

(

x

)

g

′

⁢

(

x

)

=

m

.

We want to prove that hence

g

⁢

(

x

)

≠

0

for all

x

∈

I

∖

{

x

0

}

and

lim

x

→

x

0

⁡

f

⁢

(

x

)

g

⁢

(

x

)

=

m

.

First of all (with little abuse of notation) we suppose that

f

and

g

are defined also in the point

x

0

by

f

⁢

(

x

0

)

=

0

and

g

⁢

(

x

0

)

=

0

. The resulting

functions

are

continuous

in

x

0

and hence in the whole interval

I

.

Let us first prove that

g

⁢

(

x

)

≠

0

for all

x

∈

I

∖

{

x

0

}

. If by

contradiction

g

⁢

(

x

¯

)

=

0

since we also have

g

⁢

(

x

0

)

=

0

, by Rolle’s

Theorem

we get that

g

′

⁢

(

ξ

)

=

0

for some

ξ

∈

(

x

0

,

x

¯

)

which is against our hypotheses.

Consider now any

sequence

x

n

→

x

0

with

x

n

∈

I

∖

{

x

0

}

.
By Cauchy’s mean value Theorem there exists a sequence

x

n

′

such that

f

⁢

(

x

n

)

g

⁢

(

x

n

)

=

f

⁢

(

x

n

)

-

f

⁢

(

x

0

)

g

⁢

(

x

n

)

-

g

⁢

(

x

0

)

=

f

′

⁢

(

x

n

′

)

g

′

⁢

(

x

n

′

)

.

But as

x

n

→

x

0

and since

x

n

′

∈

(

x

0

,

x

n

)

we get that

x

n

′

→

x

0

and hence

lim

n

→

∞

⁡

f

⁢

(

x

n

)

g

⁢

(

x

n

)

=

lim

n

→

∞

⁡

f

′

⁢

(

x

n

)

g

′

⁢

(

x

n

)

=

lim

x

→

x

0

⁡

f

′

⁢

(

x

)

g

′

⁢

(

x

)

=

m

.

Since this is true for any given sequence

x

n

→

x

0

we conclude that

lim

x

→

x

0

⁡

f

⁢

(

x

)

g

⁢

(

x

)

=

m

.

Title

proof of De l’Hôpital’s rule

Canonical name

ProofOfDeLHopitalsRule

Date of creation

2013-03-22 13:23:31

Last modified on

2013-03-22 13:23:31

Owner

paolini (1187)

Last modified by

paolini (1187)

Numerical id

10

Author

paolini (1187)

Entry type

Proof

Classification

msc 26A24

Classification

msc 26C15

Generated on Fri Feb 9 19:50:36 2018 by

LaTeXML
proof of product rule

proof of product rule

We begin with two

differentiable functions

f

⁢

(

x

)

and

g

⁢

(

x

)

and show that their product is

differentiable

, and that the

derivative

of the product has the desired form.

By simply calculating, we have for all values of

x

in the domain of

f

and

g

that

d

d

⁢

x

⁢

[

f

⁢

(

x

)

⁢

g

⁢

(

x

)

]

=

lim

h

→

0

⁡

f

⁢

(

x

+

h

)

⁢

g

⁢

(

x

+

h

)

-

f

⁢

(

x

)

⁢

g

⁢

(

x

)

h

=

lim

h

→

0

⁡

f

⁢

(

x

+

h

)

⁢

g

⁢

(

x

+

h

)

+

f

⁢

(

x

+

h

)

⁢

g

⁢

(

x

)

-

f

⁢

(

x

+

h

)

⁢

g

⁢

(

x

)

-

f

⁢

(

x

)

⁢

g

⁢

(

x

)

h

=

lim

h

→

0

⁡

[

f

⁢

(

x

+

h

)

⁢

g

⁢

(

x

+

h

)

-

g

⁢

(

x

)

h

+

g

⁢

(

x

)

⁢

f

⁢

(

x

+

h

)

-

f

⁢

(

x

)

h

]

=

lim

h

→

0

⁡

[

f

⁢

(

x

+

h

)

⁢

g

⁢

(

x

+

h

)

-

g

⁢

(

x

)

h

]

+

lim

h

→

0

⁡

[

g

⁢

(

x

)

⁢

f

⁢

(

x

+

h

)

-

f

⁢

(

x

)

h

]

=

f

⁢

(

x

)

⁢

g

′

⁢

(

x

)

+

f

′

⁢

(

x

)

⁢

g

⁢

(

x

)

.

The key

argument

here is the next to last line, where we have used the fact that both

f

and

g

are differentiable, hence the limit can be distributed across the sum to give the desired equality.

Title

proof of product rule

Canonical name

ProofOfProductRule

Date of creation

2013-03-22 12:28:00

Last modified on

2013-03-22 12:28:00

Owner

mathcam (2727)

Last modified by

mathcam (2727)

Numerical id

6

Author

mathcam (2727)

Entry type

Proof

Classification

msc 26A24

Related topic

Derivative

Related topic

ProductRule

Generated on Fri Feb 9 19:50:38 2018 by

LaTeXML
proof of sum rule

proof of sum rule

d

d

⁢

x

⁢

[

f

⁢

(

x

)

+

g

⁢

(

x

)

]

=

lim

h

→

0

⁡

f

⁢

(

x

+

h

)

+

g

⁢

(

x

+

h

)

-

f

⁢

(

x

)

-

g

⁢

(

x

)

h

=

lim

h

→

0

⁡

[

f

⁢

(

x

+

h

)

-

f

⁢

(

x

)

h

+

g

⁢

(

x

+

h

)

-

g

⁢

(

x

)

h

]

=

f

′

⁢

(

x

)

+

g

′

⁢

(

x

)

Title

proof of sum rule

Canonical name

ProofOfSumRule

Date of creation

2013-03-22 12:28:13

Last modified on

2013-03-22 12:28:13

Owner

mathcam (2727)

Last modified by

mathcam (2727)

Numerical id

5

Author

mathcam (2727)

Entry type

Proof

Classification

msc 26A24

Related topic

Derivative

Related topic

FixedPointsOfNormalFunctions

Generated on Fri Feb 9 19:50:40 2018 by

LaTeXML
proof of the power rule

proof of the power rule

The

power rule

can be derived by repeated application of the

product rule

.

Proof for all positive integers

n

The power rule has been shown to hold for

n

=

0

and

n

=

1

.
If the power rule is known to hold for some

k

>

0

, then we have

d

d

⁢

x

⁢

x

k

+

1

=

d

d

⁢

x

⁢

(

x

⋅

x

k

)

=

x

⁢

(

d

d

⁢

x

⁢

x

k

)

+

x

k

=

x

⋅

(

k

⁢

x

k

-

1

)

+

x

k

=

k

⁢

x

k

+

x

k

=

(

k

+

1

)

⁢

x

k

Thus the power rule holds for all

positive integers

n

.

Proof for all positive rationals

n

Let

y

=

x

p

/

q

. We need to show

d

⁢

y

d

⁢

x

⁢

(

x

p

/

q

)

=

p

q

⁢

x

p

/

q

-

1

(1)

The proof of this comes from

implicit differentiation

.

By definition, we have

y

q

=

x

p

. We now take the

derivative

with respect to

x

on both sides of the equality.

d

d

⁢

x

⁢

y

q

=

d

d

⁢

x

⁢

x

p

d

d

⁢

y

⁢

(

y

q

)

⁢

d

⁢

y

d

⁢

x

=

p

⁢

x

p

-

1

q

⁢

y

q

-

1

⁢

d

⁢

y

d

⁢

x

=

p

⁢

x

p

-

1

d

⁢

y

d

⁢

x

=

p

q

⁢

x

p

-

1

y

q

-

1

=

p

q

⁢

x

p

-

1

⁢

y

1

-

q

=

p

q

⁢

x

p

-

1

⁢

x

p

⁢

(

1

-

q

)

/

q

=

p

q

⁢

x

p

-

1

+

p

/

q

-

p

=

p

q

⁢

x

p

/

q

-

1

Proof for all positive irrationals

n

For

positive

irrationals

we claim

continuity

due to the fact that (

1

)
holds for all positive

rationals

, and there are positive rationals that approach any positive irrational.

Proof for negative powers

n

We again employ implicit differentiation. Let

u

=

x

, and

differentiate

u

n

with respect to

x

for some non-negative

n

. We must show

d

⁢

u

-

n

d

⁢

x

=

-

n

⁢

u

-

n

-

1

(2)

By definition we have

u

n

⁢

u

-

n

=

1

.
We begin by taking the derivative with respect to

x

on both sides of the equality. By application of the product rule we get

d

d

⁢

x

⁢

(

u

n

⁢

u

-

n

)

=

1

u

n

⁢

d

⁢

u

-

n

d

⁢

x

+

u

-

n

⁢

d

⁢

u

n

d

⁢

x

=

0

u

n

⁢

d

⁢

u

-

n

d

⁢

x

+

u

-

n

⁢

(

n

⁢

u

n

-

1

)

=

0

u

n

⁢

d

⁢

u

-

n

d

⁢

x

=

-

n

⁢

u

-

1

d

⁢

u

-

n

d

⁢

x

=

-

n

⁢

u

-

n

-

1

Title

proof of the power rule

Canonical name

ProofOfThePowerRule

Date of creation

2013-03-22 12:28:06

Last modified on

2013-03-22 12:28:06

Owner

mathcam (2727)

Last modified by

mathcam (2727)

Numerical id

9

Author

mathcam (2727)

Entry type

Proof

Classification

msc 26A24

Related topic

ProductRule

Related topic

Derivative2

Generated on Fri Feb 9 19:50:44 2018 by

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second derivative as simple limit

second derivative as simple limit

If the real function

f

is twice

differentiable

in a neighbourhood of

x

=

x

0

,  then

f

′′

⁢

(

x

0

)

=

lim

h

→

0

⁡

f

⁢

(

x

0

+

2

⁢

h

)

-

2

⁢

f

⁢

(

x

0

+

h

)

+

f

⁢

(

x

0

)

h

2

.

(1)

Proof.

The right hand side of the asserted equation is of the

indeterminate form

0

0

.  Using

http://

planetmath

.org/node/2657

l’Hôpital’s rule, we obtain

lim

h

→

0

⁡

f

⁢

(

x

0

+

2

⁢

h

)

-

2

⁢

f

⁢

(

x

0

+

h

)

+

f

⁢

(

x

0

)

h

2

=

lim

h

→

0

⁡

f

′

⁢

(

x

0

+

2

⁢

h

)

⋅

2

-

2

⁢

f

′

⁢

(

x

0

+

h

)

2

⁢

h

-

f

′

⁢

(

x

0

)

h

+

f

′

⁢

(

x

0

)

h

=

2

⁢

lim

2

⁢

h

→

0

⁡

f

′

⁢

(

x

0

+

2

⁢

h

)

-

f

′

⁢

(

x

0

)

2

⁢

h

-

lim

h

→

0

⁡

f

′

⁢

(

x

0

+

h

)

-

f

′

⁢

(

x

0

)

h

=

2

⁢

f

′′

⁢

(

x

0

)

-

f

′′

⁢

(

x

0

)

=

f

′′

⁢

(

x

0

)

.

Title

second derivative as simple limit

Canonical name

SecondDerivativeAsSimpleLimit

Date of creation

2013-03-22 19:00:00

Last modified on

2013-03-22 19:00:00

Owner

pahio (2872)

Last modified by

pahio (2872)

Numerical id

9

Author

pahio (2872)

Entry type

Result

Classification

msc 26A24

Synonym

second derivative as limit

Related topic

DifferenceQuotient

Related topic

ImproperLimits

Generated on Fri Feb 9 19:50:46 2018 by

LaTeXML
Some formulas of partnership

Some formulas of partnership

1. Partnership : When two or more than two persons run a business jointly, they are called partners and the deal is known as partnership
2. Ratio of

Division

of Gains:
(i)When investments of all the partners are for the same time, the gain or loss is distributed among the partners in the ratio of their investments.
Suppose A and B invest Rs. x and Rs. y respectively for a year in a business, then at end of the year
(A’s share of profit) : (B’s share of profit) = x:y
(ii) When investments are for different time periods, then

equivalent

capitals are calculated for a unit of time by taking (capital x number of units of time).Now, gain or loss is divided in the ratio of these capitals.
Suppose A invests Rs. x for p months and B invests Rs. y for q months, then
(A’s share of profit) : (B’s share of profit) = px : qy
3. Working and Sleeping Partners:
A partner who manages the business is known as a working partner and the one who simply invests the money is a sleeping partner
I hope these

concepts

will be helpful for you to solve some of your math problems.

Title

Some

formulas

of partnership

Canonical

name

SomeFormulasOfPartnership

Date of creation

2014-08-26 11:08:16

Last modified on

2014-08-26 11:08:16

Owner

burgess (1001318)

Last modified by

burgess (1001318)

Numerical id

1

Author

burgess (1001318)

Entry type

Definition

Generated on Fri Feb 9 19:50:48 2018 by

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Sophomore’s dream

Sophomore’s dream

The

integral

I

:=

∫

0

1

x

x

⁢

𝑑

x

(1)

may be

expanded

to a rapidly converging series as follows.

Changing the integrand to a power of

e

and using the

power series

expansion of the

exponential function

gives us

I

=

∫

0

1

e

x

⁢

ln

⁡

x

⁢

𝑑

x

=

∫

0

1

∑

n

=

0

∞

(

x

⁢

ln

⁡

x

)

n

n

!

⁢

d

⁢

x

.

(2)

Here the series is

uniformly convergent

on

[

0

,

1

]

and may be
integrated termwise:

I

=

∑

n

=

0

∞

∫

0

1

x

n

⁢

(

ln

⁡

x

)

n

n

!

⁢

𝑑

x

.

(3)

The last equation of the parent entry (

http://planetmath.org/ExampleOfDifferentiationUnderIntegralSign

)
then gives in the case

m

=

n

from (3) the result

I

=

∑

n

=

0

∞

(

-

1

)

n

(

n

+

1

)

n

+

1

,

(4)

i.e.,

∫

0

1

x

x

d

x

=

1

-

1

2

2

+

1

3

3

-

1

4

4

+

-

…

(5)

Cf. the

function

x

x

(

http://planetmath.org/FunctionXX

).

Since the series (5) satisfies the conditions of
Leibniz’ theorem for

alternating series

(

http://planetmath.org/LeibnizEstimateForAlternatingSeries

),
one may easily estimate the error made when a

partial sum

of (5) is used for the exact value of the integral

I

.  If one for example takes for

I

the sum of nine first
terms, the first omitted term is

-

1

10

10

; thus the
error is

negative

and its

absolute value

less than

10

-

10

.

Title

Sophomore’s dream

Canonical name

SophomoresDream

Date of creation

2014-07-20 10:46:23

Last modified on

2014-07-20 10:46:23

Owner

pahio (2872)

Last modified by

pahio (2872)

Numerical id

11

Author

pahio (2872)

Entry type

Derivation

Classification

msc 26A24

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

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sum rule

sum rule

The

sum rule

states that

d

d

⁢

x

⁢

[

f

⁢

(

x

)

+

g

⁢

(

x

)

]

=

f

′

⁢

(

x

)

+

g

′

⁢

(

x

)

Proof

See the proof of the sum rule (

http://

planetmath

.org/ProofOfSumRule

).

Examples

d

d

⁢

x

⁢

(

x

+

1

)

=

d

d

⁢

x

⁢

x

+

d

d

⁢

x

⁢

1

=

1

d

d

⁢

x

⁢

(

x

2

-

3

⁢

x

+

2

)

=

d

d

⁢

x

⁢

x

2

+

d

d

⁢

x

⁢

(

-

3

⁢

x

)

+

d

d

⁢

x

⁢

(

2

)

=

2

⁢

x

-

3

d

d

⁢

x

⁢

(

sin

⁡

x

+

cos

⁡

x

)

=

d

d

⁢

x

⁢

sin

⁡

x

+

d

d

⁢

x

⁢

cos

⁡

x

=

cos

⁡

x

-

sin

⁡

x

Title

sum rule

Canonical name

SumRule

Date of creation

2013-03-22 12:28:09

Last modified on

2013-03-22 12:28:09

Owner

mathcam (2727)

Last modified by

mathcam (2727)

Numerical id

8

Author

mathcam (2727)

Entry type

Theorem

Classification

msc 26A24

Related topic

Derivative

Related topic

ProductRule

Related topic

FixedPointsOfNormalFunctions

Generated on Fri Feb 9 19:50:52 2018 by

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Tests of Divisibility- Simple tricks

Tests of Divisibility- Simple tricks

With this simple short cuts you can find out a number is

divisible

by a given number
Divisible by 2: A number is divisible by 2, if its unitâs digit is any of 0, 2, 4, 6, 8.
Example: 6798512

Divisible by 3: A number is divisible by 3, if sum of its digits divisible by 3.
Example : 123456
1+2+3+4+5+6 = 21
21 is divisible by 3 so 123456 is also divisible by 3

Divisible by 4: if the last two digits of a given are divisible 4, so the number can be divisible by 4.
Example : 749232
Last two digits are 32 which are divisible by 4 so the given number is also divisible by 4

Divisible by 5: If unitâs digit of a number is either â0â or â5â it is divisible 5.
Example : 749230

Divisible by 6: If a given number is divisible by 2 and 3 (which are factors of 6), then the number is divisible by 6.
Example : 35256
Unitâs digit is 6 so divisible by 2
3+5+2+5+6 = 21 so divisible by 3
So 35256 divisible by 6

Divisible by 8: if last 3 digits of a given number is divisible 8, then the given number is divisible 8.
Example: 953360
360 is divisible by 8, so 953360 is divisible by 8

Divisible by 9: A number is divisible by 9, if sum of its digits divisible by 9.
Example : 50832
5+0+8+3+2 = 18 divisible by 9 so 50832 divisible by 9

Divisible by 10: A number is divisible 10, if it ends with 0.
Example : 508320

Divisible by 11: A number is divisible by 11,if the

difference

of sum of its digits at odd places and sum of its digits at even places , is either 0 or a number divisible by 11.
Example : 4832718
(sum of digits at odd places ) â (sum of digits at even places)
=(8+7+3+4)-(1+2+8) = 11 which is divisible by 11.
So 4832718 is divisible by 11.

I hope this simple tricks, will be very helpful to solve mathâs homework problems easily.

Title

Tests of Divisibility- Simple tricks

Canonical

name

TestsOfDivisibilitySimpleTricks

Date of creation

2014-08-07 17:35:51

Last modified on

2014-08-07 17:35:51

Owner

burgess (1001318)

Last modified by

burgess (1001318)

Numerical id

1

Author

burgess (1001318)

Entry type

Topic

Generated on Fri Feb 9 19:51:03 2018 by

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another example of Dirac sequence

another example of Dirac sequence

Let

A

n

=

[

-

1

2

n

,

1

2

n

]

and

δ

n

=

2

n

-

1

⁢

χ

A

n

for every

positive integer

n

, where

χ

S

denotes the

characteristic function

of the set

S

. Then

{

δ

n

}

is a

Dirac sequence

:

1.

δ

n

⁢

(

t

)

≥

0

for every positive integer

n

and every

t

∈

ℝ

.

2.

Let

n

be a positive integer. Then

∫

-

∞

∞

δ

n

⁢

(

t

)

⁢

𝑑

t

=

∫

-

1

2

n

1

2

n

2

n

-

1

⁢

𝑑

t

=

1

.

3.

Let

r

>

0

. Then there exists a positive integer

N

such that, for every

integer

k

>

N

, we have

1

2

k

<

r

. Thus, for every integer

k

>

N

, we have

∫

ℝ

∖

[

-

r

,

r

]

d

k

⁢

(

t

)

⁢

𝑑

t

=

0

.

Title

another example of Dirac sequence

Canonical name

AnotherExampleOfDiracSequence

Date of creation

2013-03-22 17:19:50

Last modified on

2013-03-22 17:19:50

Owner

Wkbj79 (1863)

Last modified by

Wkbj79 (1863)

Numerical id

8

Author

Wkbj79 (1863)

Entry type

Example

Classification

msc 26A30

Generated on Fri Feb 9 19:51:06 2018 by

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Cantor function

Cantor function

The

Cantor function

is a

canonical

example of a

singular function

. It is based on the

Cantor set

, and is usually defined as follows. Let

x

be a

real number

in

[

0

,

1

]

with the ternary expansion

0

.

a

1

⁢

a

2

⁢

a

3

⁢

…

, then let

N

be

∞

if no

a

n

=

1

and
otherwise let

N

be the smallest value such that

a

n

=

1

. Next let

b

n

=

1

2

⁢

a

n

for all

n

<

N

and let

b

N

=

1

. We define the Cantor function (or the Cantor ternary function) as

f

⁢

(

x

)

=

∑

n

=

1

N

b

n

2

n

.

This

function

can be easily checked to be

continuous

and

monotonic

on

[

0

,

1

]

and also

f

′

⁢

(

x

)

=

0

almost everywhere (it is

constant

on the

complement

of the Cantor set), with

f

⁢

(

0

)

=

0

and

f

⁢

(

1

)

=

1

. Another
interesting fact about this function is that the arclength of the graph is 2, hence the

calculus

arclength

formula

does not work in this case.

Figure 1: Graph of the cantor function using 20

iterations

.

This function, and functions

similar

to it are frequently called the Devil’s staircase. Such functions sometimes occur naturally in various areas of mathematics and

mathematical physics

and are not just a

pathological

oddity.

References

1

H. L. Royden.

. Prentice-Hall, Englewood Cliffs, New Jersey, 1988

Title

Cantor function

Canonical name

CantorFunction

Date of creation

2013-03-22 14:08:23

Last modified on

2013-03-22 14:08:23

Owner

jirka (4157)

Last modified by

jirka (4157)

Numerical id

9

Author

jirka (4157)

Entry type

Definition

Classification

msc 26A30

Synonym

Cantor ternary function

Synonym

Cantor-Lebesgue function

Synonym

Devil’s staircase

Related topic

CantorSet

Related topic

SingularFunction

Defines

Cantor function

Defines

Cantor ternary function

Generated on Fri Feb 9 19:51:07 2018 by

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Dirac sequence

Dirac sequence

A

Dirac sequence

is a sequence

(

δ

k

)

of functions

δ

k

, which satisfies the following conditions:

1.

δ

k

≥

0

for all

k

.

2.

∫

-

∞

∞

δ

k

⁢

(

t

)

⁢

𝑑

t

=

1

for all

k

.

3.

For every

r

>

0

and

ε

>

0

there is an

N

∈

ℕ

, such that for all

k

>

N

we have

∫

ℝ

\

[

-

r

,

r

]

δ

k

⁢

(

t

)

⁢

𝑑

t

<

ε

.

These functions “converge” to the

Dirac delta function

.

Title

Dirac sequence

Canonical

name

DiracSequence

Date of creation

2013-03-22 14:11:35

Last modified on

2013-03-22 14:11:35

Owner

mathwizard (128)

Last modified by

mathwizard (128)

Numerical id

5

Author

mathwizard (128)

Entry type

Definition

Classification

msc 26A30

Synonym

delta sequence

Related topic

DiracDeltaFunction

Related topic

FejerKernel

Generated on Fri Feb 9 19:51:09 2018 by

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