Problems expanding a function

Question

I'm trying to expand the function f(x). Can someone please tell me what I'm doing wrong? Thanks!

I started off with this function: $$ f(x) = {z^3 \over (z-{1 \over 4})(z-{3 \over 4})(z-{1 \over 2})} $$

And then said that it equals:

$$ {A \over (z-{1 \over 4})}+ {B \over (z-{3 \over 4})}+ {C \over (z-{1 \over 2})} $$

I then calculated A, B and C by doing:

$$ A(z-{3 \over 4})(z-{1 \over 2})+ B(z-{1 \over 4})(z-{1 \over 2})+ C(z-{1 \over 4})(z-{3 \over 4})=z^3 $$

$$ = A(z^2-{5 \over 4}z + {3 \over 8})+ B(z^2-{3 \over 4}z + {1 \over 8})+ C(z^2-z + {3 \over 16}) $$

I then expanded that to the following equation system, which according to Wolfram has the solutions $a = 0, b = 0, c = 0$. This is the solution that I was hoping to get

$$ {{1 \over 2} \over{1 - {1\over 4}z^{-1}}}+ {{9 \over 2} \over{1 - {3\over 4}z^{-1}}}+ {4 \over{1 - {1\over 2}z^{-1}}} $$

What am I doing wrong?

Answer 1

The original problem was to get from $$ f(x) = {1 \over (1-{1 \over 4}z^{-1})(1-{3 \over 4}z^{-1})(1-{1 \over 2}z^{-1})} $$ to $$ {{1 \over 2} \over{1 - {1\over 4}z^{-1}}}+ {{9 \over 2} \over{1 - {3\over 4}z^{-1}}}+ {4 \over{1 - {1\over 2}z^{-1}}} $$ I don't think you can do this by multiplying top and bottom by $z^3$. Instead, you want to introduce some symbol to stand for $z^{-1}$; let's let $y=z^{-1}$. Then by the usual partial fractions technique, we set up $${1 \over (1-{1 \over 4}y)(1-{3 \over 4}y)(1-{1 \over 2}y)}={A\over1-(1/4)y}+{B\over1-(3/4)y}+{C\over1-(1/2)y} $$ Clearing fractions, $$1=A(1-(3/4)y)(1-(1/2)y)+B(1-(1/4)y)(1-(1/2)y)+C(1-(1/4)y)(1-(3/4)y)$$ Let $y=4/3$ to get $B=9/2$. Let $y=2$ to get $C=-4$. Let $y=4$ to get $A=1/2$, and you're done (only notice I'm getting $C=-4$, not $C=4$).

Answer 2

This partial fraction decomposition only works if the terms on the bottom are all linear, i.e. of the form $a + bz$. Clearly $1 - \frac{1}{4}z^{-1}$ is not of the form, nor are any of the others.

To remedy this, need to get the function in terms of positive powers of $z$. We therefore multiply the top and bottom by $z^3$ to get $$f(z) = \frac{z^3}{(z - \frac{1}{4})(z - \frac{3}{4})(z - \frac{1}{2})}$$ Now we can use polynomial long division, followed by partial fractions, to get $f(z)$ in a form we can integrate.