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 A progressive Method for Solving recursive equations  Difference Equation Method
A progressive Method for Solving recursive equations  Difference Equation Method
Here, we only consider the following:
T(N) =C1T(N1) +C2T(N2) +... +CKT(NK) +F(N),N≥K(6.18)
. WhereCI (I = l, 2 ,...,K) Is a real constant, andCK = 0. It can be rewritten to a linear constant coefficient.KNonhomogeneous differential equations:
T(N)C1T(N1 )C2T(N2 )... CKT(NK) =F(N),N≥K(6.19)
(6.19) Linear Constant CoefficientKThe structures of nonhomogeneous ordinary differential equations are very similar, so the solutions are similar. Due to space limitations, the solution (6.19) is provided here, And the correctness proof is omitted.
Step 1, Calculate the homogeneous equation corresponding to (6.19:
T(N)C1T(N1 )C2T(N2 )... CKT(NK) = 0 (6.20)
The basic solution is to write the feature equation (6.20:
C(T) =TKC1TK1C2TK2 ... CK = 0 (6.21)
IfT=RIs (6.21) m heavy root, then (6.20) m base solutionRN,NRN,N2RN ,...,NM1RN; ifP EIθ andP eI θ is a pair of (6.21)LFor a duplicate common root, 2 of (6.20) is obtained.LBasic solutionsPNCOsNθ,PNsinNθ,N PNCOsNθ,N PNsinNθ,...,NL1PNCOsNθ,NL1PNCOsNθ. In this way, you can obtain all the roots of (6.21 ).K. BesidesK(6.20. That is, any solution of (6.20) can be expressed as thisKLinear Combination of basic solutions.
Step 2, For a special solution of (6.19. In theory, the special solution of (6.19) can be obtained by using the Laplace constant variation method. However, it is very troublesome to use explicit expressions of (6.20) general solutions, that is, linear combinations of (6.20) basic solutions. Therefore, in practice, the test method is often used, that is, based onF(N) Characteristics of the speculative form of special solutions, leave a number of adjustable constants, the speculative solution generation (6.19) and then determine. Due to the particularity of (6.19), the superposition principle can be usedF(N) Is linearly decomposed into the sum of several individual items, and corresponding special solutions of each individual item are obtained.F(N. This makes the test method more effective. For convenience, three special formsF(N), And the corresponding special solutions (6.19) are listed in Table 61, which can be applied directly. WherePI, I = 1, 2 ,..., S is a constant to be determined.
Table 61 Common special solutions for Equations (6.19)
F ( n ) format 
parts 
special solution form of equation (6.19) 
A n 
C ( A ) =0 

A is C ( T ) m of 

n S 
C (1) = 0 

1 is C ( T ) m of 

n S A n 
C ( A ) =0 

A is C ( T ) m of 

Step 3Write (6.19), that is, the general solution of (6.18 ).
(6.22)
{TI (N), I = 0, 1, 2 ,...,N} Is the basic solution of (6.20,G(N) Is a special solution for (6.19. Then, by the initial condition (6.18 ),
T(I) =TI, I = 1, 2 ,...,K1
To determine the undetermined composite constant in (6.22 {AI}, that is, Linear Equations
Or
Solve {AAnd return (6.22 ). WhereβJ =TJG(J),J= 0, 1, 2 ,...,K1.
Step 4, The approximate order of the estimated (6.22) is the requirement.
The following are two examples.
Example LRecursive equation
Its corresponding feature equation is:
C(T) =T2T1 = 0
We can solve the problem by using two single sums. The basic solution for the corresponding (6.20) is {R0n,R1n }. A special solution for (6.19) isF*(N) =8, so the corresponding (6.19) call is:
F(N) =A0R0n +A1R1n8
To meet the initial conditions, the secondorder linear equations are obtained:
Or
Or
And thus
Therefore
.
Example 2Recursive equation
T(N) = 4T(N1)4T(N2) + 2NN(6.23)
And initial conditionsT(0) = 0,T(1) = 4/3.
Its corresponding feature equation (6.21) is
C(T) =T24T+ 4 = 0
There is a dual RootR= 2. Therefore, the basic solution for (6.20) is {2n, 2nN }. BecauseF(N) = 2nN, using Table 61, a specific solution of the corresponding (6.19) is
T*(N) =N2 (P0 +P1N) 2N,
Generation (6.23)P0 = 1/2,P1 = 1/6. Therefore, the corresponding (6.19) call is:
T(N) =A02n +A1N2n +N2 (1/2 +N/6) 2N,
Make it meet the initial conditionsA0 =A1 = 0, thus
T(N) =N2 (1/2 +N/6) 2N
ThereforeT(N) =θ(N32n ).