Solving second order linear recurrences using Z-transform : problem with partial fraction decomposition.

Question

This is a question of $Z$ transformations. The original question is Solve this equation using Z transform. $y_{n+2} − 3y_{n+1} + 3y_n = 1$ subject to $y_0 = 1, y_1 = 0.$ While solving it I do not understand how this fraction was simplified. Was a property of Z transform used or how the partial fractions were formed? Please add details to your answer.

Answer 1

It looks like the partial fraction decomposition has been done incorrectly. There shouldn't be a $z$ in the numerator of the first fraction inside the brackets.

You want $$ {z\over z^2-3z+3}={A\over z-a}+{B\over z-b} $$ where $A,B$ are constants and $z^2-3z+3=(z-a)(z-b)$. Clearing fractions, $$ z=(z-b)A+(z-a)B $$ so $A=a/(a-b)$ and $B=b/(b-a)$. Now, $a,b$ are just the solutions of the quadratic equation $z^2-3z+3=0$, which you can find using the quadratic formula, and then you can calculate $A$ and $B$, and then you've got the partial fraction decomposition you were after.