series expansion of gamma function & the...

$: Series Expansion of Gamma Function & the Reciprocalfractional-calculus.com/series_expansion_gamma_reciprocal.pdfThe formula of the higher derivative of the gamma function & the reciprocal$
12 Series Expansion of Gamma Function & the Reciprocal

12.1 Taylor Expansion around a

Higher Derivative of Gamma Function

The formula of the higher derivative of the gamma function & the reciprocal was discovered by Masayuki Ui

in December 2016. ( See 22 Higher Derivative of Composition Sec.3 ) I reproduce it here as follows.

Formula 12.1.0 ( Masayuki Ui )

When z is the gamma function, n z is the polygamma function and Bn,k f1 , f2 , are

Bell polynomials , the following expressions hold.

dzn

d n

( )z = ( )z Σk=1

n

Bn,k 0( )z , 1( )z , , n-1( )z (0.1+)

dzn

d n

( )z1

= ( )z1

Σk=1

n

( )-1 k Bn,k 0( )z , 1( )z , , n-1( )z (0.1-)

Proof

When f( )z = log( )z ,

f1 = dzd

log( )z = 0( )z , f2 = dzd 0( )z = 1( )z ,

fn = dzd n-2( )z = n-1( )z

Substituting these and gk = e f k =1,2,3, for the following Faà di Bruno's Formula on a composite function

g f( )x n = Σr=1

n

gk Bn,k f1 , f2 , , fn

we obtain

elogz n = e logz Σ

k=1

n

Bn,k 0( )z , 1( )z , , n-1( )z (0.1+)

When f( )z = -log( )z , in a similar way, we obtain

e-logz n = e-logz Σ

k=1

n

( )-1 kBn,k 0( )z , 1( )z , , n-1( )z (0.1-)

Using this formula, we can perform the Taylor expansion of the gamma function z and the reciprocal

1/ z . Where, we can not perform the Taylor expansion around a =0 ,-1 ,-2 ,-3 , . Because,

at these points, the differential coefficients are or 0 .

Formula 12.1.1


Bell polynomials , the following expression holds for a s.t. a 0 ,-1 ,-2 ,-3 , .

( )z = ( )a + Σn =1

n!cn( )a

( )z-a n(1.1)

- 1 -

http://fractional-calculus.com/higher_derivative_composition.pdf

where,

cn( )a = ( )a Σk=1

n

Bn,k 0( )a , 1( )a , , n-1( )a n=1,2,3,

Proof

z can be expanded to Taylor series around a 0 ,-1 ,-2 ,-3 , as follows.

( )z = ( )a + Σn =1

n! n ( )a

( )z-a n

Applying Formula 12.1.0 (0.1+) to this and replacing ( )n a with cn a , we obtain the desired formula.

Example: Taylor expansion around 2 ( symbolic calculation )

According the formula, z is expanded to Taylor series around 2. The polynomial Bn,k f1 , f2 , is

generated using the function BellY[] of formula manipulation software Mathematica . The expansion until the

3rd term is as follows.

On the other hand, when z is expanded to series around 2 using the function Series[] of Mathematica

it is as follows.

Though they seem to be different, they are the same thing. Indeed, if 0[2]=1- , 1[2]= 2/6-1

are substituted for ft z,2,3 , it is as follows.

- 2 -

Formula 12.1.2


Bell polynomials , the following expression holds for a s.t. a 0 ,-1 ,-2 ,-3 , .

( )z1

= ( )a1

+ Σn =1

n!cn( )a

( )z-a n(1.2)

where,

cn( )a = ( )a1

Σk=1

n

( )-1 k Bn,k 0( )a , 1( )a , , n-1( )a

Proof

1/ z can be expanded to Taylor series around a 0 ,-1 ,-2 ,-3 , as follows.

( )z1

= ( )a1

+ Σn =1

( )z1 n

z=a n !( )z-a n

Applying Formula 12.1.0 (0.1-) to this and replacing the differential coefficient with cn a , we obtain the

desired formula.

Example: Taylor expansion around 2 ( numeric calculation )

According the formula, 1/ z is expanded to Taylor series around 2. The polynomial Bn,k f1 , f2 , is generated using the function BellY[] of formula manipulation software Mathematica . If the right side is

expanded until 20 terms and is illustrated with the left side, it is as follows. Both sides are exactly overlapped

and the left side (blue) is invisible.

- 3 -

12.2 Laurent Expansion of Gamma Function & the Reciprocal

We can not perform the Maclaurin expansion of the gamma function z and the reciprocal 1/ z .

But, we can perform the Maclaurin expansion of the 1+z and the reciprocal 1/ 1+z . Using this,

we can perform the Laurent expansion of the z and the reciprocal 1/ z around 0.

Formula 12.2.1 ( Laurent expansion )


Bell polynomials , the following expression holds

( )z = z1

+ Σn =1

n!cn

zn-1(2.1)

where,

cn = Σk=1

n

Bn,k 0( )1 , 1( )1 , , n-1( )1 n=1,2,3,

Proof

1+z can be expanded to Maclaurin series as follows.

( )1+z = 1 + Σn =1

n! n ( )1

zn

n ( )1 = ( )1 Σk=1

n

Bn,k 0( )1 , 1( )1 , , n-1( )1

Replacing ( )n ( )1 with cn and dividing both sides by z ,

z( )1+z

= z1

+ Σn =1

n!cn

zn-1

cn = ( )1 Σk=1

n

Bn,k 0( )1 , 1( )1 , , n-1( )1

Since 1+z = z z , ( )1 =1 , we obtain the desired expression.

Numeric Calculation

According the formula, z is expanded to Laurent series around 0. The polynomial Bn,k f1 , f2 , is generated using the function BellY[] of formula manipulation software Mathematica . The expansion until

the 4th term is as follows.

- 4 -

On the other hand, z is expanded to series around 0 using the function Series[] of Mathematica as

follows. This is consistent with the above exactly.

In addition, if ft is expanded until 20 terms and is illustrated with , it is as follows. Both are almost

overlapped

Formula 12.2.2 ( reciprocal Laurent expansion )



( )z1

= z + Σn =1

n!cn

zn+1(2.2)

where,

cn = Σk=1

n

( )-1 k Bn,k 0( )1 , 1( )1 , , n-1( )1 n=1,2,3,

Proof

1+z can be expanded to Maclaurin series as follows.

( )1+z1

= 1 + Σn =1

( )1+z1 n

z=0 n !zn

( )1+z1 n

z=0 = ( )1

1Σk=1

n

( )-1 k Bn,k 0( )1 , 1( )1 , , n-1( )1

Replacing the differential coefficient with cn and multiplying both sides by z ,

( )1+zz

= z + Σn =1

cn n!zn+1

- 5 -

cn = ( )11

Σk=1

n

( )-1 k Bn,k 0( )1 , 1( )1 , , n-1( )1

Since 1+z = z z , ( )1 =1 , we obtain the desired expression.

Symbolic Calculation

According the formula, 1/ z is expanded to series around 0. The polynomial Bn,k f1 , f2 , is

generated using the function BellY[] of formula manipulation software Mathematica . The expansion until the

4th term is as follows.

On the other hand, when 1/ z is expanded to series around 0 using the function Series[] of

Mathematica, it is as follows.

Though they seem to be different, they are the same thing. Indeed, if 0[1]= - , 1[1]= 2/6 ,

3[1]= 4/15 are substituted for ft z,4 , it is as follows.

- 6 -

12.3 Maclaurin Expansion

Formula 12.3.1


Bell polynomials , the following expressions hold

( )1+z1

= 1 + Σn=1

n !cn

zn (3.1-)

( )1-z1

= 1 + Σn=1

( )-1 n

n!cn

zn (3.1+)

( )1+z/21

= 1 + Σn=1

2n n!

cnzn (3.2-)

( )1-z/21

= 1 + Σn=1

( )-1 n

2n n!

cnzn (3.1+)

where,

cn = Σk=1

n

( )-1 k Bn,k 0( )1 , 1( )1 , , n-1( )1 n=1,2,3,

Proof

1/ 1+z can be expanded to Maclaurin series as follows.

( )1+z1

= 1 + Σn=1

( )1+z1 n

z=0 n !zn

( )1+z1 n

z=0 = ( )1

1Σk=1

n

( )-1 k Bn,k 0( )1 , 1( )1 , , n-1( )1

Replacing the differential coefficient with cn we obtain (3.1-) . And reversing the sign we obtain (3.1+) .

Replacing z with z/2 in (3.1-) we obtain (3.2-) . And reversing the sign we obtain (3.2+) .

Example: 1/( )1+z/2 ( symbolic calculation )

According the formula, 1/ 1+z/2 is expanded to series around 0. The polynomial Bn,k f1 , f2 ,

is generated using the function BellY[] of formula manipulation software Mathematica . The expansion until

the 4th term is as follows.

- 7 -

On the other hand, when f z is expanded to series around 0 using the function Series[] of Mathematica,

it is as follows.

Though they seem to be different, they are the same thing. Indeed, if 0[1]= - , 1[1]= 2/6 ,

3[1]= 4/15 are substituted for fm z,4 , it is as follows.

Formula 12.3.2



( )1+z /2

= 1 + Σn=1

2n n!

cnzn (3.3-)

( )1-z /2

= 1 + Σn=1

( )-1 n

2n n!

cnzn (3.3+)

where,

cn = Σk=1

n

( )-1 k Bn,k 0 21

,1 21

, ,n-1 21

n=1,2,3,

Proof

1/ ( )1+z /2 can be expanded to Maclaurin series as follows.

( )1+z /21

= Σn=0

( )1+z /21 n

z=0 n !zn

The first term is

( )1+z /21 ( )0

z=0 = ( )1+0 /2

1 =

1

The second and the subsequent terms are obtained using Formula12.1.0

dzn

d n

( )z1

= ( )z1

Σk=1

n

( )-1 k Bn,k 0( )z , 1( )z , , n-1( )z

- 8 -

as follows.

( )1+z /21 ( )1

z=0 =

1Σk=1

1

( )-1 k B1,k 0 21+0

21 1

( )1+z /21 ( )2

z=0 =

1Σk=1

2

( )-1 k B1,k 0 21+0

, 1 21+0

21 2

( )1+z /21 n

z=0 =

1Σk=1

n

( )-1 k Bn,k 0 21

, 1 21

, ,n-1 21

2n

1

n = 1,2,3,i.e.

1+z /21

=

1 1+Σn =1

Σk=1

n

( )-1 k Bn,k 0 21

,1 21

, ,n-1 21

2n n !

z n

Multiplying by both sides and replacing the inner with cn , we obtain (3.3-) .

In a similar way, (3.3+) is also obtained. ( See the proof of Formula 12.3.3 )

Example: / ( )1-z /2 ( numeric calculation )

According the formula, / 1-z /2 is expanded to Maclaurin series. The Bell polynomial

Bn,k f1 , f2 , is generated using the function BellY[] of formula manipulation software Mathematica .

If the right side is expanded until 15 terms and is illustrated with the left side, it is as follows. Both sides

are exactly overlapped and the left side (blue) is invisible almost.

- 9 -

Formula 12.3.3



2 ( )3-z /2

= 1 + Σn =1

( )-1 n

2n n!

cnzn (3.4+)

where,

cn = Σk=1

n

( )-1 k Bn,k 0 23

, 1 23

, ,n-1 23

n =1,2,3,

Proof

1/ ( )3-z /2 can be expanded to Maclaurin series as follows.

( )3-z /21

= Σn=0

( )3-z /21 n

z=0 n !zn

The first term is

( )3-z /21 ( )0

z=0 = ( )3-0 /2

1 =

2

The second and the subsequent terms are obtained using Formula12.1.0

dzn

d n

( )z1

= ( )z1

Σk=1

n

( )-1 k Bn,k 0( )z , 1( )z , , n-1( )z

as follows.

( )3-z /21 ( )1

z=0 =

2Σk=1

1

( )-1 k B1,k 0 23-0

-21 1

( )3-z /21 ( )2

z=0 =

2Σk=1

2

( )-1 k B2,k 0 23-0

, 1 23-0

-21 2

( )3-z /21 n

z=0=

2Σk=1

n

( )-1 k Bn,k 0 23

, 1 23

, ,n-1 23

2n

( )-1 n

n = 1,2,3,i.e.

3-z /2

1=

2 1+Σ

n =1

Σk=1

n

( )-1 k Bn,k 0 23

,1 23

, ,n-1 23

2n n !

( )-1 nz n

Multiplying by /2 both sides and replacing the inner with cn , we obtain (3.4+) .


According the formula, /2 ( )3-z /2 is expanded to Maclaurin series. The Bell polynomial


The expansion until the 4th term is as follows.

- 10 -


it is as follows.

Though they seem to be different, they are the same thing. Indeed, if 1[ ]3/2 = 2/2-4 and

3[ ]3/2 = 4-96 are substituted for fm z,4 , it is as follows.

- 11 -

12.4 Taylor Expansion around 1 (Part 1)

Formula 12.4.1



( )z1

= 1 + Σn =1

n!cn

( )z-1 n(4.1-)

( )2-z1

= 1 + Σn =1

( )-1 n

n!cn

( )z-1 n(4.1+)

( )1+z /21

= 1 + Σn =1

2n n!

cn( )z-1 n

(4.2-)

( )3-z /21

= 1 + Σn =1

( )-1 n

2n n!

cn( )z-1 n

(4.2+)

( )1-z1

= -( )z-1 - Σn =1

( )-1 n

n!cn

( )z-1 n+1(4.5+)

where,

cn = Σk=1

n

( )-1 k Bn,k 0( )1 , 1( )1 , , n-1( )1 n=1,2,3,

Proof

Replacing z with z -1 in Formula 12.3.1 (3.1-) ~ (3.2+) , we obtain (4.1-) ~ (4.2+) . Multiplying both sides

of (4.1+) by 1-z , we obtain (4.5+) . Strictly, (4.5+) should be called reciprocal Laurent expansion.

Example: 1/ ( )1+z /2 ( numeric calculation )

According the formula, 1/ 1+z /2 is expanded to Taylor series around 1. The Bell polynomial


If the right side is expanded until 15 terms and is illustrated with the left side, it is as follows. Both sides

are exactly overlapped and the left side (blue) is invisible almost.

- 12 -

Formula 12.4.2



( )1+z1

= 1 + Σn =1

n!cn

( )z-1 n(4.5-)

where,

cn = Σk=1

n

( )-1 k Bn,k 0( )2 , 1( )2 , , n-1( )2 n=1,2,3,

Proof

1/ 1+z can be expanded to Taylor series as follows.

( )1+z1

= 1 + Σn =1

( )1+z1 n

z=1 n!( )z-1 n

( )1+z1 n

z=1 = ( )2

1Σk=1

n

( )-1 k Bn,k 0( )2 , 1( )2 , , n-1( )2

Replacing the differential coefficient with cn we obtain (4.5-) .


According the formula, 1/ 1+z is expanded to Taylor series. The polynomial Bn,k f1 , f2 , is

generated using the function BellY[] of formula manipulation software Mathematica . The expansion until

the 3rd term is as follows.

On the other hand, when 1/ 1+z is expanded to series around 1 using the function Series[] of

Mathematica, it is as follows.

- 13 -

Though they seem to be different, they are the same thing. Indeed, if 0[2]= 1- , 1[2]= 2/6-1are substituted for ft z,3 , it is as follows.

- 14 -

12.5 Taylor Expansion around 1 (Part 2)

Formula 12.5.1



( )z/2

= 1 + Σn =1

2n n!

cn( )z-1 n

(5.3-)

( )1-z/2

= 1 + Σn =1

( )-1 n

2n n!

cn( )z-1 n

(5.3+)

where,

cn = Σk=1

n

( )-1 k Bn,k 0 21

, 1 21

, ,n-1 21

n =1,2,3,

Proof

1/ z/2 can be expanded to Taylor series as follows.

( )z/21

= Σn =0

( )z/21 n

z=1 n!( )z-1 n

( )z/21 ( )0

z=1 = ( )1/2

1 =

1

( )z/21 n

z=1 =

2n

1Σk=1

n

( )-1 k Bn,k 0 21

, 1 21

, ,n-1 21

n = 1,2,3,i.e.

z/21

=

1 1+Σn =1

Σk=1

n

( )-1 k Bn,k 0 21

, 1 21

, ,n-1 21

2n n !

z -1 n

Multiplying both sides by and replacing the inner with cn , we obtain (5.3+) . In a similar way,

we obtain (5.3+) .

Example: 1/( )1-z/2 ( numeric calculation )

According the formula, / 1-z/2 is expanded to Taylor series around 1. The Bell polynomial


If the right side is expanded until 15 terms and is illustrated with the left side, it is as follows. Both sides are

exactly overlapped and the left side (blue) is invisible almost.

- 15 -

Formula 12.5.2



2( )1+z/2

= 1 + Σn =1

2n n!

cn( )z-1 n

(5.6-)

where,

cn = Σk=1

n

( )-1 k Bn,k 0 23

, 1 23

, ,n-1 23

n =1,2,3,

Proof

1/ 1+z/2 can be expanded to Taylor series as follows.

( )1+z/21

= Σn =0

( )1+z/21 n

z=1 n!( )z-1 n

( )1+z/21 ( )0

z=1 = ( )3/2

1 =

2

( )1+z/21 n

z=1 =

2n

2Σk=1

n

( )-1 k Bn,k 0 23

, 1 23

, , n-1 23

n = 1,2,3,i.e.

1+z/21

=

2 1+Σn =1

Σk=1

n

( )-1 k Bn,k 0 23

, 1 23

, ,n-1 23

2n n !

z -1 n

Multiplying both sides by /2 and replacing the inner with cn , we obtain (5.6-) .


According the formula, /2 1+z/2 is expanded to Taylor series. The polynomial Bn,k f1 , f2 ,

is generated using the function BellY[] of formula manipulation software Mathematica . The expansion until

the 3rd term is as follows.

- 16 -


it is as follows.

Though they seem to be different, they are the same thing. Indeed, if 1[3/2]= 2/2-4 are substituted

for ft z,3 , it is as follows.

2017.01.18

2017.08.04 Added Formula 12.3.2 & Formula 12.3.3 .

Kano. Kono

Alien's Mathematics

- 17 -

http://fractional-calculus.com

series expansion of gamma function & the...

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