当前位置：网站首页>[combinatorics] Application of exponential generating function (multiple set arrangement problem | different balls in different boxes | derivation of exponential generating function of odd / even sequ

[combinatorics] Application of exponential generating function (multiple set arrangement problem | different balls in different boxes | derivation of exponential generating function of odd / even sequ

2022-07-03 03:05:00 【Programmer community】

List of articles

- - - Full Permutation formula of multiple sets
    - Exponential generating function Deal with the arrangement of multiple sets introduce
    - Exponential generating function Deal with the arrangement of multiple sets Formula derivation
    - Exponential generating function Handle Finite number string problem
    - Exponential generating function Handle n Bit string problem
    - Exponential generating function Handle n Bit string problem ( Examination questions )

Full Permutation formula of multiple sets

Given multiset , Yes

$k$ Elements , Each element

n_i

$n_{i}$ individual ;

{

⋅

⋯
 

⋅

}

S = \{n_1 \cdot a_1 , n_2 \cdot a_2 , \cdots , n_k \cdot a_k\}

$S = {n_{1} \cdot a_{1}, n_{2} \cdot a_{2}, \dots, n_{k} \cdot a_{k}}$

Full Permutation of multiple sets Formula is

⋯

\cfrac{n!}{n_1! n_2!\cdots n_k!}

$\frac{n !}{n _{1} ! n _{2} ! \dots n _{k} !}$

among

⋯

n=n_1 + n_2 + n_3 + \cdots + n_k

$n = n_{1} + n_{2} + n_{3} + \dots + n_{k}$ ;

Exponential generating function Deal with the arrangement of multiple sets introduce

Given multiset , Yes

$k$ Elements , Each element

n_i

$n_{i}$ individual ;

{

⋅

⋯
 

⋅

}

S = \{n_1 \cdot a_1 , n_2 \cdot a_2 , \cdots , n_k \cdot a_k\}

$S = {n_{1} \cdot a_{1}, n_{2} \cdot a_{2}, \dots, n_{k} \cdot a_{k}}$

But if it's not all arranged , Is to select some of these elements to arrange , You use Exponential generating function ;

Exponential generating function It can deal with the arrangement of multiple sets :

Exponential generating function Deal with the arrangement of multiple sets Formula derivation

Derivation of exponential generating function formula :

① Every element has to find its General

\cfrac{x^k}{k!}

$\frac{x ^{k}}{k !}$ ;

② For the first element

a_1

$a_{1}$ It's advisable Number Of The scope is

{

⋯
 

}

\{0, 1, 2, 3, \cdots , n_1\}

${0, 1, 2, 3, \dots, n_{1}}$ ,

Its exponential generating function is

⋯

\cfrac{x^0}{0!} + \cfrac{x^1}{1!} + \cfrac{x^2}{2!} + \cdots \cfrac{x^{n_1}}{ {n_1}!}

$\frac{x ^{0}}{0 !} + \frac{x ^{1}}{1 !} + \frac{x ^{2}}{2 !} + \dots \frac{x ^{n_{1}}}{n _{1} !}$

It is reduced to :

⋯

1 + \cfrac{x}{1!} + \cfrac{x^2}{2!} + \cdots \cfrac{x^{n_1}}{ {n_1}!}

$1 + \frac{x}{1 !} + \frac{x ^{2}}{2 !} + \dots \frac{x ^{n_{1}}}{n _{1} !}$

③ For the second element

a_2

$a_{2}$ The number of acceptable Of The scope is

{

⋯
 

}

\{0, 1, 2, 3, \cdots , n_2\}

${0, 1, 2, 3, \dots, n_{2}}$ ;

Its exponential generating function is

⋯

\cfrac{x^0}{0!} + \cfrac{x^1}{1!} + \cfrac{x^2}{2!} + \cdots \cfrac{x^{n_2}}{ {n_2}!}

$\frac{x ^{0}}{0 !} + \frac{x ^{1}}{1 !} + \frac{x ^{2}}{2 !} + \dots \frac{x ^{n_{2}}}{n _{2} !}$

It is reduced to :

⋯

1 + \cfrac{x}{1!} + \cfrac{x^2}{2!} + \cdots \cfrac{x^{n_2}}{ {n_2}!}

$1 + \frac{x}{1 !} + \frac{x ^{2}}{2 !} + \dots \frac{x ^{n_{2}}}{n _{2} !}$

④ For the first

$k$ Elements

a_k

$a_{k}$ The number of acceptable Of The scope is

{

⋯
 

}

\{0, 1, 2, 3, \cdots , n_k\}

${0, 1, 2, 3, \dots, n_{k}}$ ;

Its exponential generating function is

⋯

\cfrac{x^0}{0!} + \cfrac{x^1}{1!} + \cfrac{x^2}{2!} + \cdots \cfrac{x^{n_k}}{ {n_k}!}

$\frac{x ^{0}}{0 !} + \frac{x ^{1}}{1 !} + \frac{x ^{2}}{2 !} + \dots \frac{x ^{n_{k}}}{n _{k} !}$

It is reduced to :

⋯

1 + \cfrac{x}{1!} + \cfrac{x^2}{2!} + \cdots \cfrac{x^{n_k}}{ {n_k}!}

$1 + \frac{x}{1 !} + \frac{x ^{2}}{2 !} + \dots \frac{x ^{n_{k}}}{n _{k} !}$

⑤ The final exponential generating function is :

(

)

(

⋯

)

(

⋯

)

⋯

(

⋯

)

G_e(x) = (1 + \cfrac{x}{1!} + \cfrac{x^2}{2!} + \cdots \cfrac{x^{n_1}}{ {n_1}!}) (1 + \cfrac{x}{1!} + \cfrac{x^2}{2!} + \cdots \cfrac{x^{n_2}}{ {n_2}!}) \cdots (1 + \cfrac{x}{1!} + \cfrac{x^2}{2!} + \cdots \cfrac{x^{n_k}}{ {n_k}!})

$G_{e} (x) = (1 + \frac{x}{1 !} + \frac{x ^{2}}{2 !} + \dots \frac{x ^{n_{1}}}{n _{1} !}) (1 + \frac{x}{1 !} + \frac{x ^{2}}{2 !} + \dots \frac{x ^{n_{2}}}{n _{2} !}) \dots (1 + \frac{x}{1 !} + \frac{x ^{2}}{2 !} + \dots \frac{x ^{n_{k}}}{n _{k} !})$

Exponential generating function Handle Finite number string problem

subject :
Yes

$4$ A digital

1, 2,3,4

$1, 2, 3, 4$ constitute

$5$ digit Number of schemes ;
among

$1$ Not more than

$2$ Time , There must be ;
among

$2$ No more than

$1$ Time ;
among

$3$ The number of occurrences can reach

$3$ Time , It's also possible not to appear ;
among

$4$ The number of occurrences must be even numbers ;

analysis :

1. $1 1 1 1 Not more than 2 2 Time , There must be , therefore At least there must be 1 1 Time , The sequence of its occurrence here is { 1 , 2 } \{1, 2\} ; Its corresponding exponential generating function is : ( x 1 1 ! + x 2 2 ! ) (\cfrac{x^1}{1!} + \cfrac{x^2}{2!}) ;$
$1$ Number of occurrences :
2. $2 2 2 2 Not more than 1 1 Time , The sequence of its occurrence here is { 0 , 1 } \{0, 1\} ; Its corresponding exponential generating function is : ( x 0 0 ! + x 1 1 ! ) ( \cfrac{x^0}{0!} + \cfrac{x^1}{1!} ) ;$
$2$ Number of occurrences :
3. $3 3 3 3 The number of occurrences can reach 3 3 Time , Can not appear , The sequence of its occurrence here is { 0 , 1 , 2 , 3 } \{0, 1, 2, 3\} ; Its corresponding exponential generating function is : ( x 0 0 ! + x 1 1 ! + x 2 2 ! + x 3 3 ! ) ( \cfrac{x^0}{0!} + \cfrac{x^1}{1!} + \cfrac{x^2}{2!} + \cfrac{x^3}{3!} ) ;$
$3$ Number of occurrences :
4. $4 4 4 4 The number of occurrences is even , The sequence of its occurrence here is { 0 , 2 , 4 } \{0, 2, 4\} ; Its corresponding exponential generating function is : ( x 0 0 ! + x 2 2 ! + x 4 4 ! ) ( \cfrac{x^0}{0!} + \cfrac{x^2}{2!} + \cfrac{x^4}{4!} ) ;$
$4$ Number of occurrences :

Explain :

① Write its corresponding generating function : Here is the arrangement , Therefore, the general term of the generating function must be divided by

$k!$ ;

(

)

(

)

(

)

(

)

(

)

(

)

(

)

(

)

(

)

G_e(x) = (\cfrac{x^1}{1!} + \cfrac{x^2}{2!}) ( \cfrac{x^0}{0!} + \cfrac{x^1}{1!} ) ( \cfrac{x^0}{0!} + \cfrac{x^1}{1!} + \cfrac{x^2}{2!} + \cfrac{x^3}{3!} ) ( \cfrac{x^0}{0!} + \cfrac{x^2}{2!} + \cfrac{x^4}{4!} )\\ = ( x + \cfrac{x^2}{2!}) ( 1+ x) ( 1 + x + \cfrac{x^2}{2!} + \cfrac{x^3}{3!} ) ( 1 + \cfrac{x^2}{2!} + \cfrac{x^4}{4!} )

$G_{e} (x) = (\frac{x ^{1}}{1 !} + \frac{x ^{2}}{2 !}) (\frac{x ^{0}}{0 !} + \frac{x ^{1}}{1 !}) (\frac{x ^{0}}{0 !} + \frac{x ^{1}}{1 !} + \frac{x ^{2}}{2 !} + \frac{x ^{3}}{3 !}) (\frac{x ^{0}}{0 !} + \frac{x ^{2}}{2 !} + \frac{x ^{4}}{4 !}) = (x + \frac{x ^{2}}{2 !}) (1 + x) (1 + x + \frac{x ^{2}}{2 !} + \frac{x ^{3}}{3 !}) (1 + \frac{x ^{2}}{2 !} + \frac{x ^{4}}{4 !})$

② Expand the above formula :

(

)

(

)

(

144

)

288

G_e(x) = (x + \cfrac{3}{2} x^2 + \cfrac{1}{2} x^3) (1 + x + x^2 + \cfrac{2}{3}x^3+ \cfrac{7}{24}x^4 + \cfrac{1}{8}x^5 + \cfrac{1}{48}x^6 + \cfrac{1}{144}x^7) \\ = x + \cfrac{5}{2}x^2 + 3x^3 + \cfrac{8}{3}x^4 + \cfrac{43}{24}x^5 + \cfrac{43}{48}x^6 + \cfrac{17}{48}x^7 + \cfrac{1}{288}x^8 + \cfrac{1}{48}x^9 + \cfrac{1}{288}x^{10}

$G_{e} (x) = (x + \frac{3}{2} x^{2} + \frac{1}{2} x^{3}) (1 + x + x^{2} + \frac{2}{3} x^{3} + \frac{7}{2 4} x^{4} + \frac{1}{8} x^{5} + \frac{1}{4 8} x^{6} + \frac{1}{1 4 4} x^{7}) = x + \frac{5}{2} x^{2} + 3 x^{3} + \frac{8}{3} x^{4} + \frac{4 3}{2 4} x^{5} + \frac{4 3}{4 8} x^{6} + \frac{1 7}{4 8} x^{7} + \frac{1}{2 8 8} x^{8} + \frac{1}{4 8} x^{9} + \frac{1}{2 8 8} x^{10}$

③ Put the above formula in Of

\cfrac{43}{24}x^5

$\frac{4 3}{2 4} x^{5}$ term Convert to

\cfrac{x^k}{k!}

$\frac{x ^{k}}{k !}$ In the form of :

215

\cfrac{43 \times 5!}{24 \times 5!}x^5 =\cfrac{43 \times 5!}{24} \times \cfrac{x^5}{5!} =\cfrac{43 \times 5 \times 4 \times 3 \times 2}{24} \times \cfrac{x^5}{5!} = 215 \times \cfrac{x^5}{5!}

$\frac{4 3 \times 5 !}{2 4 \times 5 !} x^{5} = \frac{4 3 \times 5 !}{2 4} \times \frac{x ^{5}}{5 !} = \frac{4 3 \times 5 \times 4 \times 3 \times 2}{2 4} \times \frac{x ^{5}}{5 !} = 215 \times \frac{x ^{5}}{5 !}$

④ The above calculation results

\cfrac{x^5}{5!}

$\frac{x ^{5}}{5 !}$ Of The coefficient is

215

$215$ ; therefore Four Numbers constitute

$5$ The number of schemes with digits is

215

$215$ individual ;

Exponential generating function Handle n Bit string problem

subject : seek

1,3,5,7,9

$1, 3, 5, 7, 9$ Five numbers , form

$n$ Number of schemes with digits , At the same time, the following requirements should be met ;

3,7

$3, 7$ Here is even numbers ;

1,5,9

$1, 5, 9$ There is no limit to the number of occurrences ;

analysis : Is equivalent to

$n$ Put a different ball into

1,3,5,7,9

$1, 3, 5, 7, 9$ Five boxes , The number of balls in each box Number of schemes ;

3
,
7
3,7
$3, 7$ Occurrence analysis : It can only appear Even number of times , namely The number of occurrences is a sequence
{
0
,
2
,
4
,
⋯
 
}
\{0, 2, 4, \cdots\}
${0, 2, 4, \dots}$ ;
- The corresponding exponential generating function term is : $(\cfrac{x^0}{0!} + \cfrac{x^2}{2!} + \cfrac{x^4}{4!} + \dots)$
  $(\frac{x ^{0}}{0 !} + \frac{x ^{2}}{2 !} + \frac{x ^{4}}{4 !} + \dots)$ ;
1
,
5
,
9
1,5,9
$1, 5, 9$ Occurrence analysis : There is no limit to the number of times it appears , Then the sequence of occurrences is
0
,
1
,
2
,
⋯
{0, 1, 2, \cdots}
$0, 1, 2, \dots$
- The corresponding exponential generating function term is : $\cfrac{x^0}{0!} + \cfrac{x^1}{1!} + \cfrac{x^2}{2!} + \cdots ) = e^x$
  $(\frac{x ^{0}}{0 !} + \frac{x ^{1}}{1 !} + \frac{x ^{2}}{2 !} + \dots) = e^{x}$ ;

Explain :

① Write the corresponding Exponential generating function :

(

)

(

⋯
 

)

(

⋯
 

)

G_e(x) = ( \cfrac{x^0}{0!} + \cfrac{x^2}{2!} + \cfrac{x^4}{4!} + \cdots )^2 ( \cfrac{x^0}{0!} + \cfrac{x^1}{1!} + \cfrac{x^2}{2!} + \cfrac{x^3}{3!} + \cdots)^3

$G_{e} (x) = (\frac{x ^{0}}{0 !} + \frac{x ^{2}}{2 !} + \frac{x ^{4}}{4 !} + \dots)^{2} (\frac{x ^{0}}{0 !} + \frac{x ^{1}}{1 !} + \frac{x ^{2}}{2 !} + \frac{x ^{3}}{3 !} + \dots)^{3}$

② Number of occurrences Normal natural number Sequence

{

⋯
 

}

\{ 0, 1, 2, 3, 4, \cdots\}

${0, 1, 2, 3, 4, \dots}$ Calculation of exponential generating function :

⋯

∑

∞

⋅

\begin{array}{lcl} & \cfrac{x^0}{0!} + \cfrac{x^1}{1!} + \cfrac{x^2}{2!} + \cfrac{x^3}{3!} + \cdots\\ \\ = & 1+ x + \cfrac{x^2}{2!} + \cfrac{x^3}{3!} + \cdots\\ \\ =& \sum_{k=0}^{\infty} 1 \cdot \cfrac{x^k}{k!} \\ = & e^x \end{array}

$= = = \frac{x ^{0}}{0 !} + \frac{x ^{1}}{1 !} + \frac{x ^{2}}{2 !} + \frac{x ^{3}}{3 !} + \dots 1 + x + \frac{x ^{2}}{2 !} + \frac{x ^{3}}{3 !} + \dots \sum_{k = 0}^{\infty} 1 \cdot \frac{x ^{k}}{k !} e^{x}$

② appear Even number of times Sequence

{

⋯
 

}

\{0 , 2, 4, 6 , \cdots\}

${0, 2, 4, 6, \dots}$ Calculation of exponential generating function : Eliminate odd items , Leave even items ;

Two formulas are known :

⋯

e^x = 1+ x + \cfrac{x^2}{2!} + \cfrac{x^3}{3!} + \cdots

$e^{x} = 1 + x + \frac{x ^{2}}{2 !} + \frac{x ^{3}}{3 !} + \dots$
( All terms of this formula are positive )

−

⋯

e^{-x} = 1 - x + \cfrac{x^2}{2!} - \cfrac{x^3}{3!} + \cdots

$e^{- x} = 1 - x + \frac{x ^{2}}{2 !} - \frac{x ^{3}}{3 !} + \dots$
( All even terms of the formula It's all positive , All odd directions are negative )

Add the two formulas :

−

⋯

−

⋯

\begin{array}{lcl}e^x + e^{-x} & = & 1 + x + \cfrac{x^2}{2!} + \cfrac{x^3}{3!} + \cdots \\ \\ && +1 - x + \cfrac{x^2}{2!} - \cfrac{x^3}{3!} + \cdots \\ \\ & = & 1 \times 2 + \cfrac{x^2}{2!} \times 2 + \cfrac{x^4}{4!} \times 2 + \cdots \end{array}

$e^{x} + e^{- x} = = 1 + x + \frac{x ^{2}}{2 !} + \frac{x ^{3}}{3 !} + \dots + 1 - x + \frac{x ^{2}}{2 !} - \frac{x ^{3}}{3 !} + \dots 1 \times 2 + \frac{x ^{2}}{2 !} \times 2 + \frac{x ^{4}}{4 !} \times 2 + \dots$
( The result is even numbers Sequence Exponential generating function 2 times )

Even sequence generating function calculation :

⋯

(

−

)

1 + \cfrac{x^2}{2!} + \cfrac{x^4}{4!} + \cdots = \cfrac{1}{2} (e^x + e^{-x})

$1 + \frac{x ^{2}}{2 !} + \frac{x ^{4}}{4 !} + \dots = \frac{1}{2} (e^{x} + e^{- x})$

③ take ① ② The result of is substituted into the exponential generating function :

(

)

(

⋯
 

)

(

⋯
 

)

(

−

)

(

)

(

−

)

(

)

(

∑

∞

∑

∞

∑

∞

)

∑

∞

(

⋅

)

\begin{array}{lcl}G_e(x) &=& ( \cfrac{x^0}{0!} + \cfrac{x^2}{2!} + \cfrac{x^4}{4!} + \cdots )^2 ( \cfrac{x^0}{0!} + \cfrac{x^1}{1!} + \cfrac{x^2}{2!} + \cfrac{x^3}{3!} + \cdots)^3\\ \\ &=& ( \cfrac{1}{2} (e^x + e^{-x}))^2 (e^x)^3 \\ &=& \cfrac{1}{4}( 2 e^x e^{-x} + e^{2x} + e^{-2x} ) e^{3x} \\ &=& \cfrac{1}{4}( 2 e^{3x} + e^{5x} + e^{x} ) \\ &=& \cfrac{1}{4} ( \sum_{n=0}^{\infty} \cfrac{5^n}{n!} x^n + 2\sum_{n=0}^{\infty} \cfrac{3^n}{n!} x^n + \sum_{n=0}^{\infty} \cfrac{1}{n!} x^n ) \\ &=& \cfrac{1}{4} \sum_{n=0}^{\infty} ( 5^n + 2 \cdot 3^n + 1 ) \cfrac{x^n}{n!} \\ \end{array}

$G_{e} (x) = = = = = = (\frac{x ^{0}}{0 !} + \frac{x ^{2}}{2 !} + \frac{x ^{4}}{4 !} + \dots)^{2} (\frac{x ^{0}}{0 !} + \frac{x ^{1}}{1 !} + \frac{x ^{2}}{2 !} + \frac{x ^{3}}{3 !} + \dots)^{3} (\frac{1}{2} (e^{x} + e^{- x}))^{2} (e^{x})^{3} \frac{1}{4} (2 e^{x} e^{- x} + e^{2 x} + e^{- 2 x}) e^{3 x} \frac{1}{4} (2 e^{3 x} + e^{5 x} + e^{x}) \frac{1}{4} (\sum_{n = 0}^{\infty} \frac{5 ^{n}}{n !} x^{n} + 2 \sum_{n = 0}^{\infty} \frac{3 ^{n}}{n !} x^{n} + \sum_{n = 0}^{\infty} \frac{1}{n !} x^{n}) \frac{1}{4} \sum_{n = 0}^{\infty} (5^{n} + 2 \cdot 3^{n} + 1) \frac{x ^{n}}{n !}$

thus , You can get

\cfrac{x^n}{n!}

$\frac{x ^{n}}{n !}$ The coefficient of is

(

⋅

)

\cfrac{1}{4} ( 5^n + 2 \cdot 3^n + 1 )

$\frac{1}{4} (5^{n} + 2 \cdot 3^{n} + 1)$

④

$5$ The digits are composed as required

$n$ Number of digits, number of schemes yes

(

⋅

)

\cfrac{1}{4} ( 5^n + 2 \cdot 3^n + 1 )

$\frac{1}{4} (5^{n} + 2 \cdot 3^{n} + 1)$ Kind of ;

Exponential generating function Handle n Bit string problem ( Examination questions )

subject : hold

$n$ Numbered balls , Put in

$3$ In a different box , At the same time, the following requirements should be met ;
The first

$1$ Put at least one box ;
The first

$2$ Put an odd number of boxes ;
The first

$3$ Put an even number of boxes ;

analysis :

The first $\{1, 2, 3, \cdots\}$
${1, 2, 3, \dots}$
The first $\{1, 3, 5, \cdots\}$
${1, 3, 5, \dots}$
The first $\{2, 4, 6, \cdots\}$
${2, 4, 6, \dots}$

Explain :

① Write the generating function :

(

)

(

⋯
 

)

(

⋯
 

)

(

⋯
 

)

(

⋯
 

)

(

⋯
 

)

(

⋯
 

)

\begin{array}{lcl}\\ G_e(x) &=& (\cfrac{x^1}{1!} + \cfrac{x^2}{2!} + \cfrac{x^3}{3!} + \cdots) ( \cfrac{x^1}{1!} + \cfrac{x^3}{3!} + \cfrac{x^5}{5!} + \cdots ) ( \cfrac{x^0}{0!} + \cfrac{x^2}{2!} + \cfrac{x^4}{4!} + \cdots )\\ \\ &=& ( x + \cfrac{x^2}{2!} + \cfrac{x^3}{3!} + \cdots ) ( x + \cfrac{x^3}{3!} + \cfrac{x^5}{5!} + \cdots ) ( 1 + \cfrac{x^2}{2!} + \cfrac{x^4}{4!} + \cdots ) \\ \end{array}

$G_{e} (x) = = (\frac{x ^{1}}{1 !} + \frac{x ^{2}}{2 !} + \frac{x ^{3}}{3 !} + \dots) (\frac{x ^{1}}{1 !} + \frac{x ^{3}}{3 !} + \frac{x ^{5}}{5 !} + \dots) (\frac{x ^{0}}{0 !} + \frac{x ^{2}}{2 !} + \frac{x ^{4}}{4 !} + \dots) (x + \frac{x ^{2}}{2 !} + \frac{x ^{3}}{3 !} + \dots) (x + \frac{x ^{3}}{3 !} + \frac{x ^{5}}{5 !} + \dots) (1 + \frac{x ^{2}}{2 !} + \frac{x ^{4}}{4 !} + \dots)$

② Calculation The first

$1$ individual The box Of Exponential generating function term ( except

$0$ Outer sequence ) :

We know the formula :

⋯

\begin{array}{lcl}e^x &=& \cfrac{x^0}{0!} + \cfrac{x^1}{1!} + \cfrac{x^2}{2!} + \cfrac{x^3}{3!} + \cdots\\ \\ &=& 1 + x + \cfrac{x^2}{2!} + \cfrac{x^3}{3!} + \cdots \\ \end{array}

$e^{x} = = \frac{x ^{0}}{0 !} + \frac{x ^{1}}{1 !} + \frac{x ^{2}}{2 !} + \frac{x ^{3}}{3 !} + \dots 1 + x + \frac{x ^{2}}{2 !} + \frac{x ^{3}}{3 !} + \dots$

−

⋯

−

⋯

\begin{array}{lcl}e^{-x} &=& \cfrac{x^0}{0!} - \cfrac{x^1}{1!} + \cfrac{x^2}{2!} - \cfrac{x^3}{3!} + \cdots\\ \\ &=& 1 - x + \cfrac{x^2}{2!} - \cfrac{x^3}{3!} + \cdots \\ \end{array}

$e^{- x} = = \frac{x ^{0}}{0 !} - \frac{x ^{1}}{1 !} + \frac{x ^{2}}{2 !} - \frac{x ^{3}}{3 !} + \dots 1 - x + \frac{x ^{2}}{2 !} - \frac{x ^{3}}{3 !} + \dots$

The exponential generating function corresponding to the first box :

⋯

−

\begin{array}{lcl}\\ x + \cfrac{x^2}{2!} + \cfrac{x^3}{3!} + \cdots = e^x-1 \end{array}

$x + \frac{x ^{2}}{2 !} + \frac{x ^{3}}{3 !} + \dots = e^{x} - 1$

③ Calculation The first

$2$ individual The box Of Exponential generating function term ( Odd sequence ) :

−

(

⋯
 

)

−

(

−

⋯
 

)

(

⋯
 

)

\begin{array}{lcl}\\ e^x - e^{-x} &=& (1 + x + \cfrac{x^2}{2!} + \cfrac{x^3}{3!} + \cdots) - (1 - x + \cfrac{x^2}{2!} - \cfrac{x^3}{3!} + \cdots) \\ &=& 2 ( x + \cfrac{x^3}{3!} + \cfrac{x^5}{5!} + \cdots) \\ \end{array}

$e^{x} - e^{- x} = = (1 + x + \frac{x ^{2}}{2 !} + \frac{x ^{3}}{3 !} + \dots) - (1 - x + \frac{x ^{2}}{2 !} - \frac{x ^{3}}{3 !} + \dots) 2 (x + \frac{x ^{3}}{3 !} + \frac{x ^{5}}{5 !} + \dots)$

So odd sequence Corresponding exponential generating function yes :

⋯

−

x + \cfrac{x^3}{3!} + \cfrac{x^5}{5!} + \cdots = \cfrac{e^x - e^{-x}}{2}

$x + \frac{x ^{3}}{3 !} + \frac{x ^{5}}{5 !} + \dots = \frac{e ^{x} - e ^{- x}}{2}$

④ Calculation The first

$3$ individual The box Of Exponential generating function term ( Even sequence ) :

−

(

⋯
 

)

(

−

⋯
 

)

(

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\begin{array}{lcl}\\ e^x + e^{-x} &=& (1 + x + \cfrac{x^2}{2!} + \cfrac{x^3}{3!} + \cdots) + (1 - x + \cfrac{x^2}{2!} - \cfrac{x^3}{3!} + \cdots) \\ &=& 2 ( 0 + \cfrac{x^2}{2!} + \cfrac{x^4}{4!} + \cdots) \\ \end{array}

$e^{x} + e^{- x} = = (1 + x + \frac{x ^{2}}{2 !} + \frac{x ^{3}}{3 !} + \dots) + (1 - x + \frac{x ^{2}}{2 !} - \frac{x ^{3}}{3 !} + \dots) 2 (0 + \frac{x ^{2}}{2 !} + \frac{x ^{4}}{4 !} + \dots)$

So odd sequence Corresponding exponential generating function yes :

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1 + \cfrac{x^2}{2!} + \cfrac{x^4}{4!} + \cdots = \cfrac{e^x + e^{-x}}{2}

$1 + \frac{x ^{2}}{2 !} + \frac{x ^{4}}{4 !} + \dots = \frac{e ^{x} + e ^{- x}}{2}$

⑤ take ② ③ ④ result Plug in Exponential generating function :

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\begin{array}{lcl}\\ G_e(x) &=& ( x + \cfrac{x^2}{2!} + \cfrac{x^3}{3!} + \cdots ) ( x + \cfrac{x^3}{3!} + \cfrac{x^5}{5!} + \cdots ) ( 1 + \cfrac{x^2}{2!} + \cfrac{x^4}{4!} + \cdots )\\ \\ \\ &=& ( e^x - 1) ( \cfrac{e^x - e^{-x}}{2} ) ( \cfrac{e^x + e^{-x}}{2} ) \\ \\ &=& \cfrac{1}{4} (e^x - 1) ( (e^x)^2 - (e^{-x})^2 ) \\ \\ &=& \cfrac{1}{4} (e^x - 1) ( e^{2x} - e^{-2x} ) \\ \\ &=& \cfrac{1}{4} ( e^x e^{2x} - e^x e^{-2x} - e^{2x} + e^{-2x} ) \\ \\ &=& \cfrac{1}{4} (e^{3x} - e^{-x} - e^{2x} + e{-2x} ) \\ \\ &=& \cfrac{1}{4} ( \sum_{n=0}^{\infty}\cfrac{x^n}{n!} 3^n - \sum_{n=0}^{\infty}\cfrac{x^n}{n!} (-1)^n - \sum_{n=0}^{\infty}\cfrac{x^n}{n!} 2^n + \sum_{n=0}^{\infty}\cfrac{x^n}{n!} (-2)^n) \\ \\ &=& \sum_{n=0}^{\infty}\cfrac{x^n}{n!} ( \cfrac{1}{4} ( 3^n - (-1)^n - 2^n + (-2)^n) ) \\ \\ \end{array}

$G_{e} (x) = = = = = = = = (x + \frac{x ^{2}}{2 !} + \frac{x ^{3}}{3 !} + \dots) (x + \frac{x ^{3}}{3 !} + \frac{x ^{5}}{5 !} + \dots) (1 + \frac{x ^{2}}{2 !} + \frac{x ^{4}}{4 !} + \dots) (e^{x} - 1) (\frac{e ^{x} - e ^{- x}}{2}) (\frac{e ^{x} + e ^{- x}}{2}) \frac{1}{4} (e^{x} - 1) ((e^{x})^{2} - (e^{- x})^{2}) \frac{1}{4} (e^{x} - 1) (e^{2 x} - e^{- 2 x}) \frac{1}{4} (e^{x} e^{2 x} - e^{x} e^{- 2 x} - e^{2 x} + e^{- 2 x}) \frac{1}{4} (e^{3 x} - e^{- x} - e^{2 x} + e - 2 x) \frac{1}{4} (\sum_{n = 0}^{\infty} \frac{x ^{n}}{n !} 3^{n} - \sum_{n = 0}^{\infty} \frac{x ^{n}}{n !} (- 1)^{n} - \sum_{n = 0}^{\infty} \frac{x ^{n}}{n !} 2^{n} + \sum_{n = 0}^{\infty} \frac{x ^{n}}{n !} (- 2)^{n}) \sum_{n = 0}^{\infty} \frac{x ^{n}}{n !} (\frac{1}{4} (3^{n} - (- 1)^{n} - 2^{n} + (- 2)^{n}))$