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Math Help - Partial Fraction help

  1. #1
    Member moonnightingale's Avatar
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    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
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  2. #2
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    Quote Originally Posted by moonnightingale View Post
    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 [tex]z^3 +2z^2 - \frac{2z}{(z-1) (z-1/2)^2}[/tex].

    The code for z^3 + \frac{2z^2 -2z}{(z-1) (z-1/2)^2} is [tex]z^3 + \frac{2z^2 -2z}{(z-1) (z-1/2)^2}[/tex].
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  3. #3
    Member moonnightingale's Avatar
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    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
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  4. #4
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    Hello, moonnightingale!

    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}

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  5. #5
    Member moonnightingale's Avatar
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    Thanks a lot for prompt reply
    Kindly tell me how did u divide
    how u wrote 1+ ---- what is technique
    thanks
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  6. #6
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    Hello again, moonnightingale!

    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}<br />
&&&&&&&& 1 \\ && -- & -- & -- & -- & -- & -- & -- \\<br />
z^3-2z^2 + \frac{5}{4}z - \frac{1}{4} & \bigg) & z^3 &+& 2z^2 &-& 2z \\<br />
& & z^3 &-& 2z^2 &+& \frac{5}{4}z &-& \frac{1}{4} \\<br />
&& -- & -- & -- & -- & -- & -- & --\\<br />
&&&& 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}

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  7. #7
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    \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}
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  8. #8
    Member moonnightingale's Avatar
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    Thanks a lot for ur detailed replies
    Soroban u have really made Partial Fraction very easy for me
    Last edited by moonnightingale; July 4th 2010 at 10:56 AM.
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