当前位置:网站首页>[combinatorics] recursive equation (four cases where the non-homogeneous part of a linear non-homogeneous recursive equation with constant coefficients is the general solution of the combination of po

[combinatorics] recursive equation (four cases where the non-homogeneous part of a linear non-homogeneous recursive equation with constant coefficients is the general solution of the combination of po

2022-07-03 17:24:00 Programmer community

List of articles

  • One 、 Linear Nonhomogeneous recurrence equation with constant coefficients Of The nonhomogeneous part is polynomial And Index combination
  • Two 、 Four cases of general solution of recurrence equation

One 、 Linear Nonhomogeneous recurrence equation with constant coefficients Of The nonhomogeneous part is polynomial And Index combination


If “ Linear Nonhomogeneous recurrence equation with constant coefficients ” Nonhomogeneous part of , yes

n

n

n Of

t

t

t Sub polynomial , And

β

n

\beta^n

βn The index of , The combination of ;

Then the form of its special solution , yes

n

n

n Of

t

t

t Sub polynomial , And

P

β

n

P\beta^n

Pβn Of and ;

Recurrence equation :

a

n

2

a

n

1

=

n

+

3

n

a_n - 2a_{n-1} = n + 3^n

an2an1=n+3n

initial value :

a

0

=

0

a_0 = 0

a0=0

General form ( important ) :

① The nonhomogeneous part is

n

n

n Of

t

t

t Sub polynomial :

  • The characteristic root is not

    1

    1

    1 , The special solution is

    n

    n

    n Of

    t

    t

    t Sub polynomial ;

  • If the characteristic root is

    1

    1

    1 , And the multiplicity is

    e

    e

    e , So the special solution is

    n

    n

    n Of

    t

    +

    e

    t + e

    t+e Sub polynomial ;

② The nonhomogeneous part is

P

β

n

P\beta^n

Pβn :

  • The characteristic root cannot be the bottom

    β

    \beta

    β , The special solution is

    P

    β

    n

    P\beta^n

    Pβn ;

  • The characteristic root is the bottom

    β

    \beta

    β , The multiplicity of the characteristic root is

    e

    e

    e , The special solution is

    P

    n

    e

    β

    n

    Pn^e\beta^n

    Pneβn ;

③ Homogeneous parts have no double roots :

H

(

n

)

=

c

1

q

1

n

+

c

2

q

2

n

+

+

c

k

q

k

n

H(n) = c_1q_1^n + c_2q_2^n + \cdots + c_kq_k^n

H(n)=c1q1n+c2q2n++ckqkn

④ Homogeneous parts have double roots : General explanation

i

i

i term , Characteristic root

q

i

q_i

qi , Multiplicity

e

i

e_i

ei ,

H

i

(

n

)

=

(

c

i

1

+

c

i

2

n

+

+

c

i

e

i

n

e

i

1

)

q

i

n

H_i(n) = (c_{i1} + c_{i2}n + \cdots + c_{ie_i}n^{e_i - 1})q_i^n

Hi(n)=(ci1+ci2n++cieinei1)qin , Final general solution

H

(

n

)

=

i

=

1

t

H

i

(

n

)

H(n) = \sum\limits_{i=1}^tH_i(n)

H(n)=i=1tHi(n) ;

Reference blog : 【 Combinatorial mathematics 】 Recurrence equation ( Linear Nonhomogeneous recurrence equation with constant coefficients Of The nonhomogeneous part is polynomial And Index combination | Four cases of general solution )

Calculate the homogeneous partial general solution :

Homogeneous partial standard form of recurrence equation :

a

n

2

a

n

1

=

0

a_n - 2a_{n-1} = 0

an2an1=0

Characteristic equation :

x

2

=

0

x - 2 = 0

x2=0

Characteristic root :

x

=

2

x=2

x=2

Homogeneous partial general solution :

a

n

=

c

2

n

\overline{a_n} =c2^n

an=c2n

Calculate the non-homogeneous partial general solution :

The non-homogeneous part of the above recursive equation is

n

+

3

n

n + 3^n

n+3n , It consists of two parts :

n

n

n Of

t

t

t Sub polynomial :

n

n

n , The characteristic root is not

1

1

1, The corresponding special solution is

n

n

n Of

t

t

t Sub polynomial , In the form of

P

1

n

+

P

2

P_1n + P_2

P1n+P2 ;

β

n

\beta^n

βn Index :

3

n

3^n

3n , The characteristic root is not the bottom

3

3

3 , The corresponding special solution is

P

β

n

P\beta^n

Pβn form , In the form of

P

3

3

n

P_33^n

P33n ;

Complete special solution :

a

n

=

P

1

n

+

P

2

+

P

3

3

n

a^*_n = P_1n + P_2 + P_33^n

an=P1n+P2+P33n

Put the special solution into the recursive equation :

(

P

1

n

+

P

2

+

P

3

3

n

)

2

[

P

1

(

n

1

)

+

P

2

+

P

3

3

n

1

]

=

n

+

3

n

(P_1n + P_2 + P_33^n) - 2[P_1(n-1) + P_2 + P_33^{n-1}] = n + 3^n

(P1n+P2+P33n)2[P1(n1)+P2+P33n1]=n+3n

Expand the formula on the left :

P

1

n

+

(

2

P

1

P

2

)

+

P

3

3

n

1

=

n

+

3

n

-P_1n + (2P_1 - P_2) + P_33^{n-1}=n+3^n

P1n+(2P1P2)+P33n1=n+3n

According to the analysis of

n

n

n The power of , Constant term ,

3

n

3^n

3n Correspondence of items , The following equations can be obtained :

{

P

1

=

1

2

P

1

P

2

=

0

P

3

3

=

1

\begin{cases} -P_1 = 1 \\\\ 2P_1 - P_2 = 0 \\\\ \cfrac{P_3}{3} =1 \end{cases}

P1=12P1P2=03P3=1

Solve the above equations , The result is :

{

P

1

=

1

P

2

=

2

P

3

=

3

\begin{cases} P_1 = -1 \\\\ P_2 = -2 \\\\ P_3 =3 \end{cases}

P1=1P2=2P3=3

Special solution : Substitute the above constant into

a

n

=

P

1

n

+

P

2

+

P

3

3

n

a^*_n = P_1n + P_2 + P_33^n

an=P1n+P2+P33n in , Get the final special solution :

a

n

=

n

2

+

3

n

+

1

a^*_n = -n - 2 + 3^{n+1}

an=n2+3n+1 ;

Homogeneous partial general solution form :

a

n

=

c

2

n

\overline{a_n} =c2^n

an=c2n

Complete general solution :

a

n

=

a

n

+

a

n

=

c

2

n

n

2

+

3

n

+

1

a_n = \overline{a_n} + a^*_n = c2^n -n - 2 + 3^{n+1}

an=an+an=c2nn2+3n+1

Initial value

a

0

=

0

a_0 = 0

a0=0 Substitute the above general solution :

c

2

0

0

2

+

3

0

+

1

=

0

c2^0 - 0 - 2 + 3^{0+1} = 0

c2002+30+1=0

c

2

+

3

=

0

c - 2 + 3 = 0

c2+3=0

c

=

1

c=-1

c=1

The general solution of the final recursive equation is

a

n

=

2

n

n

2

+

3

n

+

1

a_n = 2^n -n - 2 + 3^{n+1}

an=2nn2+3n+1

Two 、 Four cases of general solution of recurrence equation


General form ( important ) :

① The nonhomogeneous part is

n

n

n Of

t

t

t Sub polynomial :

  • The characteristic root is not

    1

    1

    1 , The special solution is

    n

    n

    n Of

    t

    t

    t Sub polynomial ;

  • If the characteristic root is

    1

    1

    1 , And the multiplicity is

    e

    e

    e , So the special solution is

    n

    n

    n Of

    t

    +

    e

    t + e

    t+e Sub polynomial ;

② The nonhomogeneous part is

P

β

n

P\beta^n

Pβn :

  • The characteristic root cannot be the bottom

    β

    \beta

    β , The special solution is

    P

    β

    n

    P\beta^n

    Pβn ;

  • The characteristic root is the bottom

    β

    \beta

    β , The multiplicity of the characteristic root is

    e

    e

    e , The special solution is

    P

    n

    e

    β

    n

    Pn^e\beta^n

    Pneβn ;

③ Homogeneous parts have no double roots :

H

(

n

)

=

c

1

q

1

n

+

c

2

q

2

n

+

+

c

k

q

k

n

H(n) = c_1q_1^n + c_2q_2^n + \cdots + c_kq_k^n

H(n)=c1q1n+c2q2n++ckqkn

④ Homogeneous parts have double roots : General explanation

i

i

i term , Characteristic root

q

i

q_i

qi , Multiplicity

e

i

e_i

ei ,

H

i

(

n

)

=

(

c

i

1

+

c

i

2

n

+

+

c

i

e

i

n

e

i

1

)

q

i

n

H_i(n) = (c_{i1} + c_{i2}n + \cdots + c_{ie_i}n^{e_i - 1})q_i^n

Hi(n)=(ci1+ci2n++cieinei1)qin , Final general solution

H

(

n

)

=

i

=

1

t

H

i

(

n

)

H(n) = \sum\limits_{i=1}^tH_i(n)

H(n)=i=1tHi(n) ;

Reference blog : 【 Combinatorial mathematics 】 Recurrence equation ( Linear Nonhomogeneous recurrence equation with constant coefficients Of The nonhomogeneous part is polynomial And Index combination | Four cases of general solution )

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