aarjaneiro/planetmath huggingface dataset:
bound on error of Simpson’s rule

bound on error of Simpson’s rule

If

f

is

Riemann integrable

on

[

a

,

b

]

and

|

f

(

4

)

⁢

(

x

)

|

≤

M

for all

x

∈

[

a

,

b

]

, then

|

∫

a

b

f

⁢

(

x

)

⁢

𝑑

x

-

(

b

-

a

3

⁢

n

)

⁢

(

f

⁢

(

a

)

+

f

⁢

(

b

)

+

4

⁢

∑

j

=

1

n

2

f

⁢

(

a

+

(

b

-

a

)

⁢

(

2

⁢

j

-

1

)

n

)

+

6

⁢

∑

j

=

1

n

-

2

2

f

⁢

(

a

+

(

b

-

a

)

⁢

(

2

⁢

j

)

n

)

)

|

≤

M

⁢

(

b

-

a

)

5

180

⁢

n

4

.

Title

bound on error of Simpson’s rule

Canonical name

BoundOnErrorOfSimpsonsRule

Date of creation

2013-03-22 15:57:52

Last modified on

2013-03-22 15:57:52

Owner

Wkbj79 (1863)

Last modified by

Wkbj79 (1863)

Numerical id

7

Author

Wkbj79 (1863)

Entry type

Result

Classification

msc 41A55

Classification

msc 28-00

Generated on Fri Feb 9 21:27:07 2018 by

LaTeXML
trapezoidal rule

trapezoidal rule

The

trapezoidal rule

is a method for approximating a

definite integral

by evaluating the integrand at two points. The formal rule is given by

∫

a

b

f

⁢

(

x

)

⁢

𝑑

x

≈

h

2

⁢

[

f

⁢

(

a

)

+

f

⁢

(

b

)

]

where

h

=

b

-

a

.

This rule comes from determining the area of a

right trapezoid

with bases (

http://planetmath.org/Base9

) of lengths

f

⁢

(

a

)

and

f

⁢

(

b

)

respectively and a height (

http://planetmath.org/Height6

) of length

h

. When using a graph to illustrate the trapezoidal rule, the height of the right trapezoid is actually horizontal and the bases are vertical. This may be confusing to someone who is seeing the trapezoidal rule for the first time. An example is shown below.

The figure in red need not be a right trapezoid. If either

f

⁢

(

a

)

=

0

or

f

⁢

(

b

)

=

0

, the figure will be a

right triangle

. If both

f

⁢

(

a

)

=

0

and

f

⁢

(

b

)

=

0

, the figure will be a

line segment

. In any case, the same rule for approximating the corresponding definite integral is used.

The trapezoidal rule is the first Newton-Cotes quadrature formula. It has degree of

precision

1. This means it is exact for

polynomials

of degree (

http://planetmath.org/Degree8

) less than or equal to one. We can see this with a example.

If

f

is

Riemann integrable

on

[

a

,

b

]

with

|

f

′′

⁢

(

x

)

|

≤

M

for all

x

∈

[

a

,

b

]

, then

|

∫

a

b

f

⁢

(

x

)

⁢

𝑑

x

-

h

2

⁢

[

f

⁢

(

a

)

+

f

⁢

(

b

)

]

|

≤

M

⁢

(

b

-

a

)

3

12

.

Following is an example of the trapezoidal rule.

Using the

fundamental theorem of calculus

shows

∫

0

1

x

⁢

𝑑

x

=

1

2

.

In this case, the trapezoidal rule gives the exact value,

∫

0

1

x

⁢

𝑑

x

≈

1

2

⁢

[

f

⁢

(

0

)

+

f

⁢

(

1

)

]

=

1

2

.

It is important to note that most

calculus

books give the wrong definition of the trapezoidal rule. Typically, they define it to be what is actually the

composite trapezoidal rule

, which uses the trapezoidal rule on a specified number of

subintervals

. Some examples of calculus books that define the trapezoidal rule to be what is actually the composite trapezoidal rule are:

•

Stewart, James.

Calculus.

Pacific Groves, : International Thomson Publishing Co., 1995.

•

Bittinger, Marvin L.

Calculus.

Reading, : Addison-Wesley Publishing Co., 1989.

Also note the trapezoidal rule can be derived by integrating a

linear interpolation

or by using the

method of undetermined coefficients

. The latter is probably a bit easier.

Title

trapezoidal rule

Canonical name

TrapezoidalRule

Date of creation

2013-03-22 13:39:43

Last modified on

2013-03-22 13:39:43

Owner

Wkbj79 (1863)

Last modified by

Wkbj79 (1863)

Numerical id

25

Author

Wkbj79 (1863)

Entry type

Definition

Classification

msc 41A55

Classification

msc 41A05

Synonym

trapezoid rule

Synonym

trapezium rule

Related topic

CompositeTrapezoidalRule

Generated on Fri Feb 9 21:27:10 2018 by

LaTeXML
converse to Taylor’s theorem

converse to Taylor’s theorem

Let

U

⊂

ℝ

n

be an open set.

Theorem

.

Let

f

:

U

→

R

be a

function

such that there exists a constant

C

>

0

and an integer

k

≥

0

such that for each

x

∈

U

there is a

polynomial

p

x

⁢

(

y

)

of

k

where

|

f

⁢

(

x

+

y

)

-

p

x

⁢

(

y

)

|

≤

C

⁢

|

y

|

k

+

1

for

y

near 0. Then

f

∈

C

k

⁢

(

U

)

(

f

is

k

continuously differentiable

) and the

Taylor expansion

(

http://planetmath.org/TaylorSeries

) of

k

of

f

about any

x

∈

U

is given by

p

x

.

Note that when

k

=

0

the hypothesis of the theorem is just that

f

is

Lipschitz

in

U

which certainly makes it

continuous

in

U

.

References

1

Steven G. Krantz, Harold R. Parks.
.
Birkhäuser, Boston, 2002.

Title

converse to Taylor’s theorem

Canonical name

ConverseToTaylorsTheorem

Date of creation

2013-03-22 15:05:42

Last modified on

2013-03-22 15:05:42

Owner

jirka (4157)

Last modified by

jirka (4157)

Numerical id

6

Author

jirka (4157)

Entry type

Theorem

Classification

msc 41A58

Synonym

Taylor’s theorem converse

Related topic

TaylorSeries

Related topic

BorelLemma

Generated on Fri Feb 9 21:27:16 2018 by

LaTeXML
example of Cauchy multiplication rule

example of Cauchy multiplication rule

Let us form the

Taylor expansion

of

e

x

⁢

sin

⁡

y

starting from the known Taylor expansions

e

x

=

1

+

x

+

x

2

2

!

+

x

3

3

!

+

…

,

sin

y

=

y

-

y

3

3

!

+

y

5

5

!

-

y

7

7

!

+

-

…

and multiplying these series with

Cauchy multiplication rule

. As

power series

, both series are

absolutely convergent

for all real (and

complex

) values of

x

and

y

. The rule gives immediately the series

y

+

(

-

y

3

3

!

+

x

⁢

y

)

+

(

y

5

5

!

-

x

⁢

y

3

3

!

+

x

2

⁢

y

2

!

)

+

(

-

y

7

7

!

+

x

⁢

y

5

5

!

-

x

2

⁢

y

3

2

!

⁢

3

!

+

x

3

⁢

y

3

!

)

+

(

y

9

9

!

-

x

⁢

y

7

7

!

+

x

2

⁢

y

5

2

!

⁢

5

!

-

x

3

⁢

y

3

3

!

⁢

3

!

+

x

4

⁢

y

4

!

)

+

…

(1)

The parenthesis expressions here seem a bit irregular, but we can regroup and rearrange the terms in new parentheses:

e

x

⁢

sin

⁡

y

=

y

+

x

⁢

y

1

!

⁢

1

!

+

(

x

2

⁢

y

2

!

⁢

1

!

-

y

3

3

!

)

+

(

x

3

⁢

y

3

!

⁢

1

!

-

x

⁢

y

3

1

!

⁢

3

!

)

+

(

x

4

⁢

y

4

!

⁢

1

!

-

x

2

⁢

y

3

2

!

⁢

3

!

+

y

5

5

!

)

+

…

(2)

It’s clear that the last series precisely the same terms as the preceding one. The regrouping and the rearranging of the terms is allowable, since also (1) is converges absolutely. In fact, if one would multiply the series of

e

x

with the series

y

+

y

3

3

!

+

y

5

5

!

+

…

of

sinh

⁡

y

(which converges absolutely

∀

x

∈

ℝ

), one would get the series like (1) but all signs “+”; by the Cauchy multiplication rule this series

converges

especially for each positive

x

and

y

, in which case it is a series with positive terms; hence (1) is absolutely convergent.

The form (2) can be obtained directly from the

Taylor series

formula (

http://planetmath.org/TaylorSeries

).

Title

example of Cauchy multiplication rule

Canonical name

ExampleOfCauchyMultiplicationRule

Date of creation

2013-03-22 17:29:19

Last modified on

2013-03-22 17:29:19

Owner

pahio (2872)

Last modified by

pahio (2872)

Numerical id

8

Author

pahio (2872)

Entry type

Example

Classification

msc 41A58

Classification

msc 40-00

Classification

msc 30B10

Classification

msc 26A24

Synonym

Taylor series gotten by multiplication

Related topic

HyperbolicFunctions

Generated on Fri Feb 9 21:27:19 2018 by

LaTeXML
example of Taylor polynomials for the exponential function

example of Taylor polynomials for the exponential function

Example 1

.

We construct the

n

th

Taylor polynomial

for

f

⁢

(

x

)

=

e

x

around

x

=

0

. As we know all

derivatives

of

e

x

equal

e

x

and also,

e

0

=

1

. Therefore,

f

(

n

)

⁢

(

0

)

=

1

for any

n

. Thus:

T

1

⁢

(

x

)

=

1

+

x

T

2

⁢

(

x

)

=

1

+

x

+

x

2

2

T

3

⁢

(

x

)

=

1

+

x

+

x

2

2

+

x

3

3

!

=

1

+

x

+

x

2

2

+

x

3

6

T

4

⁢

(

x

)

=

1

+

x

+

x

2

2

+

x

3

3

!

+

x

4

4

!

=

1

+

x

+

x

2

2

+

x

3

6

+

x

4

24

In fact:

T

n

⁢

(

x

)

=

1

+

x

+

x

2

2

+

x

3

3

!

+

x

4

4

!

+

…

+

x

n

n

!

Comparison of

e

x

with the Taylor pol. of deg.

1

(green),

2

(blue) and

3

(pink).

Let us use several Taylor polynomials to find

approximations

of the number

e

:

e

=

2.718281828459045

⁢

…

e

≈

T

1

⁢

(

1

)

=

1

+

1

=

2

e

≈

T

2

⁢

(

1

)

=

1

+

1

+

1

/

2

=

2.5

e

≈

T

3

⁢

(

1

)

=

1

+

1

+

1

/

2

+

1

/

6

=

8

/

3

=

2.666

⁢

6

¯

e

≈

T

4

⁢

(

1

)

=

1

+

1

+

1

/

2

+

1

/

6

+

1

/

24

=

65

/

24

=

2.708333

⁢

3

¯

e

≈

T

5

⁢

(

1

)

=

1

+

1

+

1

/

2

+

1

/

6

+

1

/

24

+

1

/

120

=

163

/

60

=

2.71666

⁢

6

¯

Title

example of Taylor polynomials for the

exponential function

Canonical name

ExampleOfTaylorPolynomialsForTheExponentialFunction

Date of creation

2013-03-22 15:04:09

Last modified on

2013-03-22 15:04:09

Owner

alozano (2414)

Last modified by

alozano (2414)

Numerical id

6

Author

alozano (2414)

Entry type

Example

Classification

msc 41A58

Related topic

LogarithmFunction

Related topic

NaturalLogBase

Related topic

EIsTranscendental

Related topic

ExponentialFunction

Generated on Fri Feb 9 21:27:22 2018 by

LaTeXML
example of Taylor polynomials for s⁢i⁢nx

example of Taylor polynomials for

s

⁢

i

⁢

n

x

In this entry we compute several

Taylor polynomials

for the

function

sin

⁡

x

around

x

=

0

and we produce graphs to compare the function with the corresponding Taylor polynomial. Recall that for a given function

y

=

f

⁢

(

x

)

(here we suppose

f

is infinitely

differentiable

) and a point

x

=

a

, the Taylor polynomial of degree

n

(

n

≥

0

) is given by:

T

n

⁢

(

x

)

=

f

⁢

(

a

)

+

f

′

⁢

(

a

)

⁢

(

x

-

a

)

+

f

′′

⁢

(

a

)

2

!

⁢

(

x

-

a

)

2

+

…

+

f

(

n

)

⁢

(

a

)

n

!

⁢

(

x

-

a

)

n

where

f

(

n

)

denotes the

n

th

derivative

of

f

⁢

(

x

)

.

From now on we assume

f

⁢

(

x

)

=

sin

⁡

x

and

a

=

0

. Notice that the derivatives of

sin

⁡

x

are cyclic:

f

′

⁢

(

x

)

=

cos

⁡

x

,

f

′′

⁢

(

x

)

=

-

sin

⁡

x

,

f

′′′

⁢

(

x

)

=

-

cos

⁡

x

,

f

(

4

)

⁢

(

x

)

=

sin

⁡

x

=

f

⁢

(

x

)

.

Therefore, the Taylor polynomials are easy to compute. In fact:

f

(

2

⁢

n

)

⁢

(

0

)

=

0

,

f

(

2

⁢

n

+

1

)

⁢

(

0

)

=

(

-

1

)

n

Thus, the first Taylor polynomial is given by:

T

1

⁢

(

x

)

=

0

+

1

⋅

x

=

x

In the following graph one can compare the function

T

1

⁢

(

x

)

=

x

and

sin

⁡

x

.

The function

y

=

sin

⁡

x

and the first Taylor polynomial.

Notice that

T

2

⁢

(

x

)

=

T

1

⁢

(

x

)

. More generally,

T

2

⁢

n

⁢

(

x

)

=

T

2

⁢

n

-

1

⁢

(

x

)

so we will not compute any other even order Taylor polynomials. However, the third degree Taylor polynomial is given by the formula:

T

3

⁢

(

x

)

=

x

-

x

3

3

!

=

x

-

x

3

6

The function

y

=

sin

⁡

x

and the third Taylor polynomial.

The Taylor polynomial of degree

5

is given by:

T

5

⁢

(

x

)

=

x

-

x

3

3

!

+

x

5

5

!

The function

y

=

sin

⁡

x

and the fifth Taylor polynomial.

Next, we compute some Taylor polynomials of higher degree. In particular, the Taylor polynomial of degree

15

has the form:

T

15

⁢

(

x

)

=

x

-

x

3

6

+

x

5

120

-

x

7

5040

+

x

9

362880

-

x

11

39916800

+

x

13

6227020800

-

x

15

1307674368000

The function

y

=

sin

⁡

x

and the Taylor polynomial of degree

15

.

Finally, we produce a detailed view of the Taylor polynomial of degree

99

. In particular, notice that the graphs are very close until

x

=

34

or so, but after that

T

99

⁢

(

x

)

behaves rather jittery and wildly.

A detail of the Taylor polynomial of degree

99

(the

interval

(

34

,

39

)

).

Title

example of Taylor polynomials for

s

⁢

i

⁢

n

x

Canonical name

ExampleOfTaylorPolynomialsForsinX

Date of creation

2013-03-22 15:03:43

Last modified on

2013-03-22 15:03:43

Owner

alozano (2414)

Last modified by

alozano (2414)

Numerical id

7

Author

alozano (2414)

Entry type

Example

Classification

msc 41A58

Related topic

ComplexSineAndCosine

Related topic

HigherOrderDerivativesOfSineAndCosine

Generated on Fri Feb 9 21:27:25 2018 by

LaTeXML
getting Taylor series from differential equation

getting Taylor series from differential equation

If a given

function

f

satisfies a

differential equation

, the

Taylor series

of

f

can sometimes be obtained easily.

Let

f

⁢

(

x

)

=

sin

⁡

(

m

⁢

arcsin

⁡

x

)

,

where

m

is a non-zero , be an example (cf. (

http://planetmath.org/Cf

) the

cyclometric functions

).  We form the

derivatives

f

′

⁢

(

x

)

=

m

1

-

x

2

⁢

cos

⁡

(

m

⁢

arcsin

⁡

x

)

,

f

′′

⁢

(

x

)

=

-

m

2

1

-

x

2

⁢

sin

⁡

(

m

⁢

arcsin

⁡

x

)

+

m

⁢

x

(

1

-

x

2

)

⁢

1

-

x

2

⁢

cos

⁡

(

m

⁢

arcsin

⁡

x

)

,

which show that

f

satisfies the differential equation

(

1

-

x

2

)

⁢

f

′′

-

x

⁢

f

′

+

m

2

⁢

f

=

0

.

Differentiating this repeatedly gives the equations

(

1

-

x

2

)

⁢

f

′′′

-

3

⁢

x

⁢

f

′′

+

(

m

2

-

1

)

⁢

f

′

=

0

,

(

1

-

x

2

)

⁢

f

(

4

)

-

5

⁢

x

⁢

f

′′′

+

(

m

2

-

4

)

⁢

f

′′

=

0

,

and so on.  Using the

sum of odd numbers

1

+

3

+

5

+

⋯

+

(

2

⁢

n

-

1

)

=

n

2

and induction on

n

yields the

recurrence relation

(

1

-

x

2

)

⁢

f

(

n

+

2

)

-

(

2

⁢

n

+

1

)

⁢

x

⁢

f

(

n

+

1

)

+

(

m

2

-

n

2

)

⁢

f

(

n

)

=

0

.

Plugging in

x

=

0

yields

f

(

n

+

2

)

(

0

)

=

(

n

2

-

m

2

)

f

(

n

)

(

0

)

(

n

=

0

,

1

,

2

,

…

)

.

Since

f

′

⁢

(

0

)

=

m

,  we have that

f

(

2

⁢

n

+

1

)

⁢

(

0

)

=

m

⁢

(

1

2

-

m

2

)

⁢

(

3

2

-

m

2

)

⁢

…

⁢

(

(

2

⁢

n

-

1

)

2

-

m

2

)

,

whereas all even derivatives of

f

vanish at

x

=

0

.  (Note that

f

is an

odd function

.)  Thus, we obtain the Taylor of

f

:

sin

⁡

(

m

⁢

arcsin

⁡

x

)

=

m

1

!

⁢

x

+

m

⁢

(

1

2

-

m

2

)

3

!

⁢

x

3

+

m

⁢

(

1

2

-

m

2

)

⁢

(

3

2

-

m

2

)

5

!

⁢

x

5

+

⋯

By the

ratio test

, this series converges for

|

x

|

<

1

.

References

1

Ernst Lindelöf:

Differentiali- ja integralilasku
ja sen sovellutukset

I.  WSOY. Helsinki (1950).

Title

getting Taylor series from differential equation

Canonical name

GettingTaylorSeriesFromDifferentialEquation

Date of creation

2013-03-22 15:06:07

Last modified on

2013-03-22 15:06:07

Owner

Wkbj79 (1863)

Last modified by

Wkbj79 (1863)

Numerical id

17

Author

Wkbj79 (1863)

Entry type

Example

Classification

msc 41A58

Related topic

ExamplesOnHowToFindTaylorSeriesFromOtherKnownSeries

Related topic

SawBladeFunction

Related topic

SpecialCasesOfHypergeometricFunction

Generated on Fri Feb 9 21:27:38 2018 by

LaTeXML
Taylor series, derivation of

Taylor series, derivation of

Let

f

⁢

(

x

)

be given by the following

power series

:

f

⁢

(

x

)

=

c

0

+

c

1

⁢

(

x

-

a

)

+

c

2

⁢

(

x

-

a

)

2

+

⋯

+

c

n

⁢

(

x

-

a

)

n

+

⋯

=

∑

k

=

0

∞

c

k

⁢

(

x

-

a

)

k

Now let’s compute a few

derivatives

at

x

=

a

.

f

⁢

(

a

)

=

c

0

;

f

′

⁢

(

a

)

=

c

1

;

f

′′

⁢

(

a

)

=

2

⁢

c

2

;

f

(

3

)

⁢

(

a

)

=

6

⁢

c

3

;

f

(

n

)

⁢

(

a

)

=

n

!

⁢

c

n

From this, it is clear that

c

n

=

f

(

n

)

⁢

(

a

)

n

!

, thus the series can be written as:

T

n

=

∑

k

=

0

n

c

k

⁢

(

x

-

a

)

k

=

∑

k

=

0

n

f

(

k

)

⁢

(

a

)

k

!

⁢

(

x

-

a

)

k

where

f

⁢

(

x

)

=

T

∞

.

Title

Taylor series

,

derivation

of

Canonical name

TaylorSeriesDerivationOf

Date of creation

2013-03-22 15:25:24

Last modified on

2013-03-22 15:25:24

Owner

apmc (9183)

Last modified by

apmc (9183)

Numerical id

6

Author

apmc (9183)

Entry type

Derivation

Classification

msc 41A58

Generated on Fri Feb 9 21:27:55 2018 by

LaTeXML
Taylor’s theorem

Taylor’s theorem

1

Taylor’s Theorem

Let

f

be a

function

which is defined on the

interval

(

a

,

b

)

and suppose the

n

th

derivative

f

(

n

)

exists on

(

a

,

b

)

. Then for all

x

and

x

0

in

(

a

,

b

)

,

R

n

⁢

(

x

)

=

f

(

n

)

⁢

(

y

)

n

!

⁢

(

x

-

x

0

)

n

with

y

strictly between

x

and

x

0

(

y

depends on the choice of

x

).

R

n

⁢

(

x

)

is the

n

th remainder of the

Taylor series

for

f

⁢

(

x

)

.

Title

Taylor’s theorem

Canonical name

TaylorsTheorem

Date of creation

2013-03-22 11:56:53

Last modified on

2013-03-22 11:56:53

Owner

Andrea Ambrosio (7332)

Last modified by

Andrea Ambrosio (7332)

Numerical id

11

Author

Andrea Ambrosio (7332)

Entry type

Theorem

Classification

msc 41A58

Related topic

TaylorSeries

Generated on Fri Feb 9 21:27:57 2018 by

LaTeXML
approximation of the log function

approximation of the log function

Because

lim

x

→

0

⁡

x

⁢

log

⁡

(

x

)

=

lim

x

→

0

⁡

x

x

-

1

we can approximate

log

⁡

(

x

)

for small

x

:

log

⁡

(

x

)

≈

x

x

-

1

x

.

A perhaps more interesting and useful result is that for

x

small we have the approximation

log

⁡

(

1

+

x

)

≈

x

.

In general, if

x

is smaller than

0.1

our approximation is practical. This occurs because for small

x

, the area under the curve (which is what

log

is a measurement of) is approximately that of a

rectangle

of height 1 and width

x

.

Now when we combine this approximation with the

formula

log

⁡

(

a

⁢

b

)

=

log

⁡

(

a

)

+

log

⁡

(

b

)

, we can now approximate the logarithm of many positive numbers. In fact,

scientific calculators

use a (somewhat more precise) version of the same procedure.

For example, suppose we wanted

log

⁡

(

1.21

)

. If we estimate

log

⁡

(

1.1

)

+

log

⁡

(

1.1

)

by taking

0.1

+

0.1

=

0.2

, we would be pretty close.

Title

approximation of the log

function

Canonical

name

ApproximationOfTheLogFunction

Date of creation

2013-03-22 15:18:38

Last modified on

2013-03-22 15:18:38

Owner

Mathprof (13753)

Last modified by

Mathprof (13753)

Numerical id

10

Author

Mathprof (13753)

Entry type

Derivation

Classification

msc 41A60

Generated on Fri Feb 9 21:27:59 2018 by

LaTeXML
asymptotic bounds for factorial

asymptotic bounds for factorial

Stirling’s

formula

furnishes the following asymptotic
bound for the factorial:

n

!

=

2

⁢

π

⁢

n

n

+

1

/

2

⁢

e

-

n

⁢

(

1

+

Θ

⁢

(

1

n

)

)

,

n

→

∞

.

The following less precise bounds are useful for counting purposes
in computer science:

n

!

=

o

⁢

(

n

n

)

(1)

n

!

=

ω

⁢

(

n

λ

⁢

n

)

for every

0

≤

λ

<

1

(2)

log

⁡

n

!

=

Θ

⁢

(

n

⁢

log

⁡

n

)

(3)

(

o

,

ω

, and

Θ

are Landau notation.)

Proof of the upper bound of equation (

1

).

We must show that for every

ε

>

0

, we have

n

!

≤

ε

⁢

n

n

for large enough

n

.

To see this, write

n

!

=

2

⁢

π

⁢

n

e

n

⁢

(

1

+

O

⁢

(

1

n

)

)

⏟

⁢

n

n

,

and observe that the middle quantity tends to

0

as

n

→

∞

,
so it can be made smaller than any fixed

ε

by taking

n

large.
∎

Proof of the lower bound of equation (

2

).

We are to show that for every

C

>

0

, we have

n

!

≥

C

⁢

n

λ

⁢

n

for large enough

n

.

We write:

n

!

=

2

⁢

π

⁢

n

⁢

(

1

+

O

⁢

(

1

n

)

)

⁢

(

n

e

1

/

(

1

-

λ

)

)

(

1

-

λ

)

⁢

n

⏟

⁢

n

λ

⁢

n

,

and the middle expression can be made to be

≥

C

by taking

n

to be large.
∎

Proof of bound for

log

⁡

n

!

, equation (

3

).

Showing

log

⁡

n

!

=

O

⁢

(

n

⁢

log

⁡

n

)

is simple:

n

!

=

n

⁢

(

n

-

1

)

⁢

⋯

⁢

(

2

)

⁢

(

1

)

≤

n

n

log

⁡

n

!

≤

n

⁢

log

⁡

n

for all

n

≥

1

.

To show

log

⁡

n

!

=

Ω

⁢

(

n

⁢

log

⁡

n

)

,
we can take equation (

2

) with the constants

C

=

1

and

λ

=

1

2

. Then

log

⁡

n

!

≥

1

2

⁢

n

⁢

log

⁡

n

for large enough

n

. In fact, it may be checked that

n

≥

1

suffices.
∎

References

1

Thomas H. Cormen, Charles E. Leiserson, Ronald L. Rivest, Clifford Stein.

Introduction

to

Algorithms

, second edition. MIT Press, 2001.

2

Michael Spivak.

Calculus

, third edition. Publish or Perish, 1994.

Title

asymptotic bounds for factorial

Canonical name

AsymptoticBoundsForFactorial

Date of creation

2013-03-22 15:54:25

Last modified on

2013-03-22 15:54:25

Owner

stevecheng (10074)

Last modified by

stevecheng (10074)

Numerical id

7

Author

stevecheng (10074)

Entry type

Result

Classification

msc 41A60

Classification

msc 68Q25

Related topic

StirlingsApproximation

Related topic

GrowthOfExponentialFunction

Generated on Fri Feb 9 21:28:02 2018 by

LaTeXML
weaker version of Stirling’s approximation

weaker version of Stirling’s approximation

One can prove a weaker version of Stirling’s

approximation

without appealing to the

gamma function

. Consider the graph of

ln

⁡

x

and note that

ln

(

n

-

1

)

!

≤

∫

1

n

ln

x

d

x

≤

ln

n

!

But

∫

ln

⁡

x

⁢

d

⁢

x

=

x

⁢

ln

⁡

x

-

x

, so

ln

(

n

-

1

)

!

≤

n

ln

n

-

n

+

1

≤

ln

n

!

and thus

n

ln

n

-

n

+

1

+

ln

n

≥

ln

(

n

-

1

)

!

+

ln

n

=

ln

n

!

≥

n

ln

n

-

n

+

1

so

ln

⁡

n

-

1

+

1

n

+

ln

⁡

n

n

≥

1

n

⁢

ln

⁡

n

!

≥

ln

⁡

n

-

1

+

1

n

As

n

gets large, the expressions on either end approach

ln

⁡

n

-

1

, so we have

1

n

⁢

ln

⁡

n

!

≈

ln

⁡

n

-

1

Multiplying through by

n

and exponentiating, we get

n

!

≈

n

n

⁢

e

-

n

Title

weaker version of Stirling’s approximation

Canonical name

WeakerVersionOfStirlingsApproximation

Date of creation

2013-03-22 16:25:21

Last modified on

2013-03-22 16:25:21

Owner

rm50 (10146)

Last modified by

rm50 (10146)

Numerical id

7

Author

rm50 (10146)

Entry type

Result

Classification

msc 41A60

Classification

msc 30E15

Classification

msc 68Q25

Generated on Fri Feb 9 21:28:04 2018 by

LaTeXML