Solve the recurrence relation with no initial conditions

Question

Solve the recurrence relation $$a_n = 6a_{n-1} − 12a_{n−2} + 8a_{n−3} + 3$$

I forgot how to these kind of questions. I remember that the condition for $n$ should be $n \ge 3$. How should I approach after? Thank you.

Answer 1

Using characteristic polynomials, you get $$(\lambda-2)^3=0$$ which results in $\lambda =2$ with multiplicity $3$

Thus the homogenous solution is $$A2^n+Bn2^n+Cn^22^n$$ and the complementary solution is $c=-3$

Adding together we get $$a_n= A2^n+Bn2^n+Cn^22^n-3$$ where the coefficients are found by the initial values.

Answer 2

Using Characteristic polynomial (a good reference!) method will be quicker. But we need to make it homogeneous first, from $$a_n=6a_{n-1}−12a_{n−2}+8a_{n−3}+3$$ $$a_{n+1}=6a_{n}−12a_{n−1}+8a_{n−2}+3$$ $$a_n-a_{n+1}=6a_{n-1}−12a_{n−2}+8a_{n−3}- 6a_{n}+12a_{n−1}-8a_{n−2}$$ leading to $$a_{n+1}-7a_{n}+18a_{n−1}−20a_{n−2}+8a_{n−3}=0$$ which is homogeneous and has the following characteristic polynomial: $$x^4-7x^3+18x^2−20x+8=0$$ $$\left(x-2\right)^3\left(x-1\right)=0$$

with the general solution $$a_n=A\cdot(1)^n+\left(B+C\cdot n + D\cdot n^2\right)\cdot2^n=\\ A+\left(B+C\cdot n + D\cdot n^2\right)\cdot2^n$$ $A,B,C,D$ are constants, typically identified from the initial conditions, but you didn't specify any.

Answer 3

It's the discrete version of the solutions to linear differential equations with constant coefficients:

• First you solve the homogeneous recurrence relation: $$a_n = 6a_{n-1} − 12a_{n−2} + 8a_{n−3}$$
• Then you find a particular solution of the complete recurrence relation. As the non-homogeneous part is a constant, you can search for a constant $a_n=K$.
• Finally, add this particular solution to the general solution of the homogeneous recurrence relation.

Solution of the homogeneous recurrence relation :

Searching for solutions of the form $a_n=\lambda^n$ leads to the characteristic equation : $$\lambda^3-6\lambda^2+12\lambda-8=(\lambda-2)^3=0.$$ This equation has a unique root 2 with multiplicity $3$, so a basis of the vector space of solutions of the homogeneous recurrence relation is $$\bigl\{ 2^n,\, n2^n, \,n^2 2^n \bigr\}.$$