Solutions of Linear Nonhomogeneous Recurrence Relations |
Example
Solution First we observe that the homogeneous problem
un+2 + un+1 -6un=0 |
C(n+2)2n+2+C(n+1)2n+1-6Cn2n = 2n , |
an=A2n+B(-3)n + 2n , n 0 . |
cman+m+ + c1an+1+c0an=g(n) , c0cm 0 , n 0. | (*) |
g(n)= n(b0+ b1n + + bknk ) , | (**) |
cm m + + c1 +c0 =0 . |
vn | = Bini n nM | ||
= n ( B0 + B1n + + Bk-1nk-1 + Bk nk ) nM , | (***) |
Note
f( )PM-1( ) { 0 } , f( )PM+k( ) Pk( ) . |
Examples
Solution Let f(n)=un+vn, with un being the general solution of the homogeneous problem and vn a particular solution.
un+2-6un+1+9un=0 , n 0 |
5 3n | = | vn+2-6vn+1+9vn |
= | C(n+2)23n+2-6C(n+1)23n+1+9Cn23n | |
= | 18C3n . |
f(n) = A + Bn + n2 3n , n 0 . |
an+4-5an+3+9an+2-7an+1 +2an=3 , n 0 |
Solution We first find the general solution un for the homogeneous problem. We then find a particular solution vn for the nonhomogeneous problem without considering the initial conditions. Then an=un+vn would be the general solution of the nonhomogeneous problem. We finally make use of the initial conditions to determine the arbitrary constants in the general solution so as to arrive at our required particular solution.
4-5 3 +9 2 -7 + 2 =0 |
1 =1 | with multiplicity | m1=3 , and |
2=2 | with multiplicity | m2=1 . |
un = (A+Bn+Cn2)1n+D2n , |
3 | = | vn+4-5vn+3+9vn+2-7vn+1+2vn |
= | E(n+4)3-5E(n+3)3+9E(n+2)3-7E(n+1)3+2En3 | |
= | E(n3 + 3n2 4+3n 42+43) -5E(n3 + 3n2 3+3n 32+33) | |
+ 9E(n3 + 3n2 2+3n 22+23) -7E(n3 + 3n2 1+3n 12+13) + 2En3 | ||
= | -6E , |
Note Should you find it very tedious to perform the expansions
in the above, you could just substitute, say, n=0 into
3 = E(n+4)3-5E(n+3)3+9E(n+2)3-7E(n+1)3+2En3 |
an = un + vn = A + Bn + Cn2 + D2n - n3/2. |
Initial Conditions | Induced Equations | Solutions | ||||||||||||||||||||||||||||||||||||
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an = 3 - 2n + n2 - n3/2 - 2n, n 0 . |
Solution
n2n+1 = vn+1-vn | = | 2n+1(B+C(n+1)) + D(n+1) - 2n(B+Cn) - Dn |
= | 2n(Cn+B+2C) + D , |
2n ( (C-1)n + (B+2C) ) + (D-1) = 0 . |
C-1=0 , B+2C=0 , D-1=0 . |
an = 2n(n-2)+n+A , n 0 . |
Solution We first observe
S(n+1)= (n+1)m + im = S(n) + (n+1)m . |
S(n+1)-S(n) | = | (n+1)m, n |
S(0) | = | 0 . |