Results 1 to 7 of 7

Math Help - partial fraction question

  1. #1
    Junior Member
    Joined
    Nov 2006
    Posts
    26

    partial fraction question

    using the method of partial fractions obtain the expansion of

    1-s/ s^2(s^2+1)

    Im not sure how to go about it as you cant factorise the s^2+1
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Global Moderator

    Joined
    Nov 2005
    From
    New York City
    Posts
    10,616
    Thanks
    9
    Quote Originally Posted by feage7 View Post
    using the method of partial fractions obtain the expansion of

    1-s/ s^2(s^2+1)

    Im not sure how to go about it as you cant factorise the s^2+1
    You have,
    \frac{1-s}{s^2(s^2+1)}=\frac{A}{s}+\frac{B}{s^2}+\frac{Cs+  D}{s^2+1}

    Note Cs+D because it belongs to an irreducible quadradic.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Grand Panjandrum
    Joined
    Nov 2005
    From
    someplace
    Posts
    14,972
    Thanks
    4
    Quote Originally Posted by feage7 View Post
    using the method of partial fractions obtain the expansion of

    1-s/ s^2(s^2+1)

    Im not sure how to go about it as you cant factorise the s^2+1
    What ImPerfectHacker says.

    But you can factorise s^2+1 if complex factors are used, and
    the tables of LT still work with complex terms.

    So:

    s^2+1=(s+\bold{i})(s-\bold{i})

    RonL
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Junior Member
    Joined
    Nov 2006
    Posts
    26
    Quote Originally Posted by ThePerfectHacker View Post
    You have,
    \frac{1-s}{s^2(s^2+1)}=\frac{A}{s}+\frac{B}{s^2}+\frac{Cs+  D}{s^2+1}

    Note Cs+D because it belongs to an irreducible quadradic.

    Ive tried that and spent ages trying to find away to then equate the terms on top but i still end up finding the need to use i and every time ive involved i i seem to get it wrong when i put in values at the end to check my answer
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Grand Panjandrum
    Joined
    Nov 2005
    From
    someplace
    Posts
    14,972
    Thanks
    4
    Quote Originally Posted by feage7 View Post
    Ive tried that and spent ages trying to find away to then equate the terms on top but i still end up finding the need to use i and every time ive involved i i seem to get it wrong when i put in values at the end to check my answer
    When I do it I get:

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

    RonL
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Junior Member
    Joined
    Nov 2006
    Posts
    26
    Im just going wrong everywehre

    When i put i in i get

    1-i = (Ci+D)(i)(-1)

    1-i = (-c+Di)(-1)

    1-i = C-Di

    so i get C=1 and D = 1 which isnt right

    aslo when i do A and B i end up with

    1-s= As^4 + As^2 = Bs^3 +Bs + Cs^4 + Ds^3

    equating terms in s gives me B=-1 and equating either s^4 or s^2 gives me A as 0
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Forum Admin topsquark's Avatar
    Joined
    Jan 2006
    From
    Wellsville, NY
    Posts
    9,939
    Thanks
    338
    Awards
    1
    <br />
\frac{1-s}{s^2(s^2+1)}=\frac{A}{s}+\frac{B}{s^2}+\frac{Cs+  D}{s^2+1}<br />

    \frac{A}{s}+\frac{B}{s^2}+\frac{Cs+D}{s^2+1} = \frac{As(s^2+1) + B(s^2+1) + (Cs + D)s^2}{s^2(s^2+1)}

    = \frac{(A+C)s^3 + (B+D)s^2 + As + B}{s^2(s^2+1)}

    So:
    \frac{1-s}{s^2(s^2+1)} = \frac{(A+C)s^3 + (B+D)s^2 + As + B}{s^2(s^2+1)}

    Thus
    A + C = 0
    B + D = 0
    A = -1
    B = 1

    So
    C = 1
    D = -1

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

    ================================================== ==

    In terms of the complex denominators:
    \frac{1-s}{s^2(s^2+1)} = \frac{A}{s} + \frac{B}{s^2} + \frac{C}{s+i} + \frac{D}{s-i}

    \frac{A}{s} + \frac{B}{s^2} + \frac{C}{s+i} + \frac{D}{s-i} = \frac{As(s^2+1) + B(s^2+1) + Cs^2(s-i) + Ds^2(s+i)}{s^2(s^2+1)}

    = \frac{(A+ C + D)s^3+ (B - iC + iD)s^2 + As + B}{s^2(s^2+1)}

    So:
    \frac{1-s}{s^2(s^2+1)} = \frac{(A + C + D)s^3+ (B - iC + iD)s^2 + As + B}{s^2(s^2+1)}

    Thus:
    A + C + D = 0
    B - iC + iD = 0
    A = -1
    B = 1

    So
    C + D = 1
    -iC + iD = -1

    So
    C = (1/2)(1-i)
    D = (1/2)(1+i)

    Thus:
    \frac{1-s}{s^2(s^2+1)} = -\frac{1}{s} + \frac{1}{s^2} + \frac{\frac{1}{2}(1-i)}{s+i} + \frac{\frac{1}{2}(1+i)}{s-i}

    -Dan
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. partial-fraction-decomp question
    Posted in the Pre-Calculus Forum
    Replies: 8
    Last Post: July 20th 2010, 06:39 PM
  2. A little minor question on Partial Fraction
    Posted in the Algebra Forum
    Replies: 22
    Last Post: January 16th 2010, 03:26 AM
  3. Partial fraction question
    Posted in the Calculus Forum
    Replies: 7
    Last Post: November 21st 2009, 07:58 AM
  4. Partial fraction question
    Posted in the Calculus Forum
    Replies: 1
    Last Post: November 18th 2009, 09:12 PM
  5. A partial fraction question..
    Posted in the Calculus Forum
    Replies: 2
    Last Post: October 22nd 2008, 09:27 AM

Search Tags


/mathhelpforum @mathhelpforum