Find out the answer to "Show that−4z−1(1−14z−1)(1−4z−1)=16151(1−14z−1)−16151(1−4z−1)"

Jordin Olsen

Answered question

2022-04-30

Show that

\frac{- 4 z^{- 1}}{(1 - \frac{1}{4} z^{- 1}) (1 - 4 z^{- 1})} = \frac{16}{15} \frac{1}{(1 - \frac{1}{4} z^{- 1})} - \frac{16}{15} \frac{1}{(1 - 4 z^{- 1})}

Answer & Explanation

Draidayerabauu

Beginner2022-05-01Added 9 answers

Let $x = z^{- 1}$
Then $\frac{- 4 z^{- 1}}{(1 - \frac{1}{4} z^{- 1}) (1 - 4 z^{- 1})} = \frac{- 4 x}{(1 - \frac{1}{4} x) (1 - 4 x)}$
Suppose
$\frac{- 4 x}{(1 - \frac{1}{4} x) (1 - 4 x)} = \frac{A}{(1 - \frac{1}{4} x)} + \frac{B}{(1 - 4 x)} (1)$
$\frac{- 4 x}{(1 - \frac{1}{4} x) (1 - 4 x)} = \frac{A (1 - 4 x) + B (1 - \frac{1}{4} x)}{(1 - \frac{1}{4} x) (1 - 4 x)}$
$- 4 x = A (1 - 4 x) + B (1 - \frac{1}{4} x)$
$- 4 x = x (- 4 A - \frac{B}{4}) + A + B$
You'll have two equations involving A and B when you compare coefficients.
$- 4 A - \frac{B}{4} = - 4 (2)$
$A + B = 0 (3)$
Solve $(2)$ and $(3)$ to get $A = \frac{16}{15}$ and $B = - \frac{16}{15}$ then replace them in (1) and also revert x back into $z^{- 1}$ to get your final result.

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