How to calculate three constants in a linear recurrence problem.

Question

Question: Verify that $x^3 - 3x^2 + 4 = (x^2 - 4x + 4)(x+1)$

And solve linear recurrence: $f(0) = 1$, $f(1) = 0$, $f(2) = 14$, $f(n) = 3 f(n-1)- 4 f(n-3)$

The characteristic equation is already given in the verifying section and the roots can be determined from the $(x^2 - 4x + 4)(x+1)$

Roots are $x=-1,2,2$

General form: $$C_1(-1)^n + C_2(2)^n + C_3(2)^n \cdot n$$

For given, $$f(0) = 1 = C_1 + C_2$$ $$f(1) = 0 = -C_1 + 2C_2 + 2C_3$$ $$f(2) = 14 = C_1 + 4C_2 + 8C_3$$

I am having problem in calculating the three constants here. Please help.

Answer 1

As you said $$f(1)=0=-C_1+2C_2+2C_3\Rightarrow C_3=\frac{C_1-2C_2}{2}$$ So we have $$f(2)=C_1+4C_2+8(\dfrac{C_1-2C_2}{2})=5C_1-4C_2=14$$ Now just consider $$ \left\{ \begin{array}{c} C_1+C_2=1 \\ 5C_1-4C_2=14 \end{array} \right. $$ And solve it.

Answer 2

Hint: This is a system of linear equations. Look up "How to solve a system of linear equations". There are different methods like Gauss-Elimination, or stepwiese substitution,...