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Math Help - Partial Fraction

  1. #1
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    Partial Fraction

    havin trouble solving this partial fraction:

    1/s^2-2s+2

    i think i can break it up into:

    1/(s-1)^2+1
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  2. #2
    MHF Contributor Quick's Avatar
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    Quote Originally Posted by Chadly724 View Post
    havin trouble solving this partial fraction:

    1/s^2-2s+2

    i think i can break it up into:

    1/(s-1)^2+1
    is it: \frac{1}{s^2}-2s+2 or \frac{1}{s^2-2s+2}?
    Last edited by Quick; November 16th 2006 at 06:12 PM.
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  3. #3
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    correction

    sorry there should be partheses around the entire bottom function
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    Quote Originally Posted by Chadly724 View Post
    havin trouble solving this partial fraction:

    1/s^2-2s+2

    i think i can break it up into:

    1/(s-1)^2+1
    You have,
    \frac{1}{s^2-2s+2}

    Express as,
    \frac{1}{s^2-2s+1+1}

    Thus,
    \frac{1}{(s-1)^2+1}

    Now, I presume you have an inverse Laplace transform for s>1.

    In that case if you have,
    \frac{1}{s^2+1} s>0
    Then, that is the function,
    f(x)=\sin x
    Since you have a substitution you use the shift rule.
    Thus,
    f(x)=e^x\sin x
    If the unique inverse LaPlace Transform.
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  5. #5
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    Quote Originally Posted by Chadly724 View Post
    havin trouble solving this partial fraction:

    \frac{1}{s^2-2s+2}

    i think i can break it up into:

    \frac{1}{(s-1)^2+1}
    Do you want to break this up into partial fractions or just transform into a
    standard form of LT that can be inverted by table look-up?

    My table of LT's has:

    <br />
\frac{1}{(s-b)^2+a^2}=\mathcal{L}\ \frac{e^{bt}\sin(at)}{a}<br />

    which should take case of your last form.

    However we can do the same thing with partial fractions if you will:

    <br />
\frac{1}{s^2-2s+2}=\frac{1}{ (s-(1+\bold{i})) (s-(1-\bold{i}))}=<br />
\frac{\bold{i}}{2(s-(1-\bold{i})) } - \frac{\bold{i}}{2(s-(1+\bold{i})) }<br />

    (please check the arithmetic)

    RonL
    Last edited by CaptainBlack; November 17th 2006 at 08:25 AM.
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  6. #6
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    Hello, Chadly724!

    Havin trouble solving this partial fraction: \frac{1}{s^2-2s+2}

    i think i can break it up into: \frac{1}{(s-1)^2+1}. . Yes, you can . . . but why?

    Partial Fractions are used when the denominator can be factored.
    . . Like: . \frac{1}{x^2+2x-15}\:=\:\frac{1}{(x-3)(x+5)} or: . \frac{1}{x^3+4x} \:=\:\frac{1}{x(x^2 +4)}


    As Captain Black pointed out, it can be factored: . (s - 1 - i)(s - 1 + i)
    . . but I doubt that you want to introduce complex numbers.


    If this is an integration problem:

    . . \int\frac{ds}{s^2 - 2x + 2} \;=\;\int\frac{ds}{(s-1)^2+1} \;=\;\tan^{-1}(s-1) + C

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  7. #7
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    Quote Originally Posted by Soroban View Post
    Hello, Chadly724!


    Partial Fractions are used when the denominator can be factored.
    . . Like: . \frac{1}{x^2+2x-15}\:=\:\frac{1}{(x-3)(x+5)} or: . \frac{1}{x^3+4x} \:=\:\frac{1}{x(x^2 +4)}


    As Captain Black pointed out, it can be factored: . (s - 1 - i)(s - 1 + i)
    . . but I doubt that you want to introduce complex numbers.


    If this is an integration problem:

    . . \int\frac{ds}{s^2 - 2x + 2} \;=\;\int\frac{ds}{(s-1)^2+1} \;=\;\tan^{-1}(s-1) + C
    The variable s points in the direction of a Laplace transform problem, not an intrgration.

    RonL
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