# Partial Fraction help

• July 4th 2010, 04:39 AM
moonnightingale
Partial Fraction help
I have the term

z^3 +2z^2 -2z / (z-1) (z-1/2)^2

i am doing it like this way

A/(z-1) + B/(z-1/2) + C/(z-1/2)^2

I got the answer of A and C which comes to be 4 and 3/4

How to calculate B and are above answers correct
Can i check partial fraction in MATLAB
• July 4th 2010, 05:46 AM
mr fantastic
Quote:

Originally Posted by moonnightingale
I have the term

z^3 +2z^2 -2z / (z-1) (z-1/2)^2

i am doing it like this way

A/(z-1) + B/(z-1/2) + C/(z-1/2)^2

I got the answer of A and C which comes to be 4 and 3/4

How to calculate B and are above answers correct
Can i check partial fraction in MATLAB

Do you mean z^3 +2z^2 - (2z / (z-1) (z-1/2)^2) ? Or z^3 + ((2z^2 -2z) / (z-1) (z-1/2)^2) ? etc.

Please learn some basic latex so that you can better typeset your mathematical expressions:

The code for $z^3 +2z^2 - \frac{2z}{(z-1) (z-1/2)^2}$ is $$z^3 +2z^2 - \frac{2z}{(z-1) (z-1/2)^2}$$.

The code for $z^3 + \frac{2z^2 -2z}{(z-1) (z-1/2)^2}$ is $$z^3 + \frac{2z^2 -2z}{(z-1) (z-1/2)^2}$$.
• July 4th 2010, 06:27 AM
moonnightingale
Sorry i forgot to put brackets

[ z^3 +2z^2 -2z] / [(z-1) (z-1/2)^2]

whole red is numerator
ok i will try to learn latex
can u kindly explain partial fraction to me as i am always stuck when there is power in denominator
• July 4th 2010, 06:55 AM
Soroban
Hello, moonnightingale!

Quote:

I have the term: . $\dfrac{z^3 +2z^2 -2z}{(z-1) (z-\frac{1}{2})^2}$

i am doing it like this way: . $\dfrac{A}{z-1} + \dfrac{B}{z-\frac{1}{2}}+ \dfrac{C}{(z-\frac{1}{2})^2}$

I got: . $A = 4,\;C = \frac{3}{4}$

How to calculate $B$?

I have no idea how you got those answers . . . but they are correct!

First, note that the fraction is "improper" . . . we must divide it out.

. . $\dfrac{x^2 + 2z^2- 2z}{(z-1)(z-\frac{1}{2})^2} \;=\; 1 + \dfrac{4z^2 - \frac{13}{4}z + \frac{1}{4}}{(z-1)(z-\frac{1}{2})^2}$

We will decompose the fraction:

. . $\dfrac{4z^2 - \frac{13}{4}z + \frac{1}{4}}{(z-1)(z-\frac{1}{2})^2} \;=\;\dfrac{A}{z-1} + \dfrac{B}{z-\frac{1}{2}} + \dfrac{C}{(z-\frac{1}{2})^2}$

We have: . $4z^2 - \frac{13}{4}z + \frac{1}{4} \;=\;A(z-\frac{1}{2})^2 + B(z-1)(z-\frac{1}{2}) + C(z-1)$

Let $z = 1\!:\;\;4-\frac{13}{4} + \frac{1}{4} \:=\:A(\frac{1}{4}) + B(0) + C(0) \quad\Rightarrow\quad \boxed{A = 4}$ .[1]

Let $z = \frac{1}{2}\!:\;\;4(\frac{1}{4}) - \frac{13}{4}(\frac{1}{2}) + \frac{1}{4} \:=\:A(0) + B(0) + C(-\frac{1}{2}) \quad\Rightarrow\quad \boxed{C = \tfrac{3}{4}}$ .[2]

Let $z = 0\!:\;\;4(0) - \frac{13}{4}(0) + \frac{1}{4} \:=\:A(\frac{1}{4}) + B(-1)(-\frac{1}{2}) + C(-1)$

We have: . $\frac{1}{4}A + \frac{1}{2}B - C \:=\:\frac{1}{4}$

Substitute [1] and [2]: . $\frac{1}{4}(4) + \frac{1}{2}B - \frac{3}{4} \:=\:\frac{1}{4} \quad\Rightarrow\quad \boxed{B = 0}$

Therefore: . $\dfrac{z^3+2z^2-2z}{(z-1)(z-\frac{1}{2})^2} \;=\;1 + \dfrac{4}{z-1} + \dfrac{\frac{3}{4}}{(z-\frac{1}{2})^2}$

• July 4th 2010, 07:14 AM
moonnightingale
Thanks a lot for prompt reply
Kindly tell me how did u divide
how u wrote 1+ ---- what is technique
thanks
• July 4th 2010, 07:47 AM
Soroban
Hello again, moonnightingale!

Quote:

Kindly tell me how did u divide?
how u wrote 1+ ---- ?
what is technique?
thanks

It's long division . . .

We have: . $\dfrac{z^3+2z^2 - 2z}{(z-1)(z-\frac{1}{2})^2} \;=\;\dfrac{z^3 + 2z^2 - 2z}{z^3 - 2x^2 + \frac{5}{4}z - \frac{1}{4}}$

. . $\begin{array}{cccccccccc}
&&&&&&&& 1 \\ && -- & -- & -- & -- & -- & -- & -- \\
z^3-2z^2 + \frac{5}{4}z - \frac{1}{4} & \bigg) & z^3 &+& 2z^2 &-& 2z \\
& & z^3 &-& 2z^2 &+& \frac{5}{4}z &-& \frac{1}{4} \\
&& -- & -- & -- & -- & -- & -- & --\\
&&&& 4z^2 &-& \frac{13}{4}z &+& \frac{1}{4} \end{array}$

Therefore: . $\dfrac{z^3 + 2z^2 - 2z}{(z-1)(z-\frac{1}{2})^2} \;=\;1 + \dfrac{4z^2 - \frac{13}{4}z + \frac{1}{4}}{(z-1)(z-\frac{1}{2})^2}$

• July 4th 2010, 07:53 AM
Prove It
$\frac{z^3 + 2z^2 - 2z}{(z - 1)\left(z - \frac{1}{2}\right)^2} = \frac{z^3 + 2z^2 - 2z}{(z - 1)\left(z^2 - z + \frac{1}{4}\right)}$

$= \frac{z^3 + 2z^2 - 2z}{z^3 - z^2 +\frac{1}{4}z - z^2 + z - \frac{1}{4}}$

$= \frac{z^3 + 2z^2 - 2z}{z^3 - 2z^2 + \frac{5}{4}z - \frac{1}{4}}$

$= \frac{z^3 - 2z^2 + \frac{5}{4}z - \frac{1}{4} + 4z^2 - \frac{13}{4}z + \frac{1}{4}}{z^3 - 2z^2 + \frac{5}{4}z - \frac{1}{4}}$

$= \frac{z^3 - 2z^2 + \frac{5}{4}z - \frac{1}{4}}{z^3 - 2z^2 + \frac{5}{4}z - \frac{1}{4}} + \frac{4z^2 - \frac{13}{4}z + \frac{1}{4}}{z^3 - 2z^2 + \frac{5}{4}z - \frac{1}{4}}$

$= 1 + \frac{4z^2 - \frac{13}{4}z + \frac{1}{4}}{z^3 - 2z^2 + \frac{5}{4}z - \frac{1}{4}}$

$= 1 + \frac{4z^2 - \frac{13}{4}z + \frac{1}{4}}{(z - 1)\left(z - \frac{1}{2}\right)^2}$
• July 4th 2010, 10:36 AM
moonnightingale
Thanks a lot for ur detailed replies
Soroban u have really made Partial Fraction very easy for me