Decomposing $\frac{1 - 8z + 27z^2 - 35z^3 + 14z^4}{(1-z)^2(1-2z)^2(1-3z)}$

Question

I got stuck at this expression and couldn't decompose it further. The book I am referring to has the answer $$\frac{1 - 8z + 27z^2 - 35z^3 + 14z^4}{(1-z)^2(1-2z)^2(1-3z)} = \frac5{4(1-z)} + \frac1{2(1-z)^2} - \frac3{1-2z} - \frac2{(1-2z)^2} + \frac{17}{4(1-3z)}$$ But I am not able to derive the same. Can anyone please help me out on how to resolve it?

Answer 1

Was able to decompose it using following method

$let \frac{1−8z+27z^2−35z^3+14z^4}{(1−z)^2(1−2z)^2(1−3z)}=\frac{A}{(1-z)}+\frac{B}{(1-z)^2}+\frac{C}{(1-2z)}+\frac{D}{(1-2z)^2}+\frac{E}{(1-3z)} $

Multiply both sides by $(1−z)^2(1−2z)^2(1−3z)$

$1−8z+27z^2−35z^3+14z^4 = A(1-z)(1-2z)^2(1-3z)+ \\B(1-2z)^2(1-3z)+ \\C(1-z)^2(1-2z)(1-3z)+ \\D(1-z)^2(1-3z)+ \\E(1-z)^2(1-2z)^2\\$

$= A(1-8z+23z^2-28z^3+12z^4) + \\B(1-7z+16z^2-12z^3)+\\C(1-7z+17z^2-17z^3+6z^4)+\\D(1-5z+7z^2-3z^3)+\\E(1-6z+13z^2-2z^3+4z^4)\\$

$ =(A+B+C+D+E)-\\(8A+7A+7C+5D+6E)z+\\(23A+16B+17C+7D+13E)z^2-\\(28A+12B+17C+3D+12E)z^3+\\(12A+6C+4E)z^4\\$

Comparing LHS with RHS gives us 5 equations:

$A+B+C+D+E=1$

$8A+7B+7C+5D+6E=8$

$23A+16B+17C+7D+13E=27$

$28A+12B+17C+3D+12E=35$

$12A+6C+4E=14$

Solving these 5 equation in 5 variables gives { A = 5/4, B = 1/2, C = -3, D = -2, E = 17/4 }

Answer 2

Solving a n-dimensional linear system of equation is a very simple problem, but it may take a lot of time to solve it. Moreover you have to pay attention to mistakes in the calculations.

There is one more trick to find the coefficients: It is useful if the grade of the zeros is 1, that is there are no higher orders like $(z-z_0)^k$ for $k>1$.

Now, as an example, we calculate a coefficient:

$$\frac{1−8z+27z^2−35z^3+14z^4}{(1−z)^2(1−2z)^2(1−3z)}=\frac{A}{(1-z)}+\frac{B}{(1-z)^2}+\frac{C}{(1-2z)}+\frac{D}{(1-2z)^2}+\frac{E}{(1-3z)}$$

Multiply the equation by $(1-3z)$.

$$\frac{1−8z+27z^2−35z^3+14z^4}{(1−z)^2(1−2z)^2}=(1-3z)\left(\frac{A}{(1-z)}+\frac{B}{(1-z)^2}+\frac{C}{(1-2z)}+\frac{D}{(1-2z)^2}\right)+E$$

Since the equation hold for every $z\in\mathbb{R}\setminus\{\text{zeros}\}$ or $z\in\mathbb{C}\setminus\{\text{zeros}\}$ you can compute the limit $z\to1/3$. The left hand side is equal to 17/4 and the right hand side is obviously equal to E. So you have calculated the first coefficient.

You can do the same procedure for $B$ and $D$. Then you are just confronted with a linear system of 2 equations, which is much simpler.