Results 1 to 10 of 10

Math Help - Can anyone tell me how to expand the follwing in terms of partial fraction?

  1. #1
    Newbie
    Joined
    May 2006
    Posts
    5

    Can anyone tell me how to expand the follwing in terms of partial fraction?

    Can anyone tell me how to expand the follwing in terms of partial fraction?

    .....2a^s
    ------------
    (s^4)-(a^4)

    I do it like this:

    ......a.................b..............c
    ---------- + --------- + ------
    s^2 + a^2.........s+a..........s-a

    but after I expand it, it is tooooooooo complicated to solve...
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Grand Panjandrum
    Joined
    Nov 2005
    From
    someplace
    Posts
    14,972
    Thanks
    4
    Quote Originally Posted by orange890
    Can anyone tell me how to expand the follwing in terms of partial fraction?

    .....2a^s
    ------------
    (s^4)-(a^4)

    I do it like this:

    ......a.................b..............c
    ---------- + --------- + ------
    s^2 + a^2.........s+a..........s-a

    but after I expand it, it is tooooooooo complicated to solve...
    1. You are using "a" to mean two different things in the same expression -
    never a good idea.

    2. This is not the usual type of expression that partial fractions are used
    for. What is the background of your question (that is why do you want to
    find a partial fraction expansion for this)?

    RonL
    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 orange890
    Can anyone tell me how to expand the follwing in terms of partial fraction?

    .....2a^s
    ------------
    (s^4)-(a^4)

    I do it like this:

    ......a.................b..............c
    ---------- + --------- + ------
    s^2 + a^2.........s+a..........s-a

    but after I expand it, it is tooooooooo complicated to solve...
    You could try:

    <br />
\frac{2 a^s}{s^4-a^4}=2 a^s \left\{ \frac{1}{s^4-a^4} \right\}<br />
,

    then resolve the only the term in braces into partial fractions.

    RonL
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Super Member
    earboth's Avatar
    Joined
    Jan 2006
    From
    Germany
    Posts
    5,811
    Thanks
    116
    Quote Originally Posted by orange890
    Can anyone tell me how to expand the follwing in terms of partial fraction?
    .....2a^s
    ------------
    (s^4)-(a^4)

    I do it like this:
    ......a.................b..............c
    ---------- + --------- + ------
    s^2 + a^2.........s+a..........s-a

    but after I expand it, it is tooooooooo complicated to solve...
    Hello,

    I tried this way:

    1rst step:

    {2a^2\over{s^4-a^4}}={A\over{a^2+s^2}}+{B\over{a^2-s^2}}+{C\over{s^4-a^4}}

    Then you'll get the results: A = B = 1, C = 0

    2nd step (Take the second summand to partial fractions):

    {1\over{a^2-s^2}}={A\over{a+s}}+{B\over{a-s}}+{C\over{a^2-s^2}}

    Then you'll get the results: A = B = 0, C = 1

    Thus the only partial fraction I got is:

    {2a^2\over{s^4-a^4}}={1\over{a^2+s^2}}+{1\over{a^2-s^2}}

    Is this sufficient?

    Greetings

    EB
    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 earboth
    Hello,

    I tried this way:

    1rst step:

    {2a^2\over{s^4-a^4}}={A\over{a^2+s^2}}+{B\over{a^2-s^2}}+{C\over{s^4-a^4}}

    Then you'll get the results: A = B = 1, C = 0

    2nd step (Take the second summand to partial fractions):

    {1\over{a^2-s^2}}={A\over{a+s}}+{B\over{a-s}}+{C\over{a^2-s^2}}

    Then you'll get the results: A = B = 0, C = 1

    Thus the only partial fraction I got is:

    {2a^2\over{s^4-a^4}}={1\over{a^2+s^2}}+{1\over{a^2-s^2}}

    Is this sufficient?

    Greetings

    EB
    A'ha - you think the denominator is 2a^2 not 2a^s.

    RonL
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Super Member
    earboth's Avatar
    Joined
    Jan 2006
    From
    Germany
    Posts
    5,811
    Thanks
    116
    Quote Originally Posted by CaptainBlack
    A'ha - you think the denominator is 2a^2 not 2a^s.

    RonL
    Hello,

    I thought so - until now. But now I'm convinced that I need a new pair of glasses.

    Greetings

    EB
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Super Member
    earboth's Avatar
    Joined
    Jan 2006
    From
    Germany
    Posts
    5,811
    Thanks
    116

    last attempt

    Quote Originally Posted by orange890
    Can anyone tell me how to expand the follwing in terms of partial fraction?...
    Hello,

    I pick up the suggestion of CaptainBlack and my own incomplete solution (please, read all previous replies so that you understand what I'm talking about)

    \frac{2 a^s}{s^4-a^4}= a^{s-2} \left\{ \frac{2a^2}{s^4-a^4} \right\}

    \frac{2 a^s}{s^4-a^4}= a^{s-2} \left({1\over{a^2+s^2}}+{1\over{a^2-s^2}} \right)

    Greetings

    EB
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Grand Panjandrum
    Joined
    Nov 2005
    From
    someplace
    Posts
    14,972
    Thanks
    4
    Quote Originally Posted by earboth
    Hello,

    I thought so - until now. But now I'm convinced that I need a new pair of glasses.

    Greetings

    EB
    What I was thinking is not that you had misread the question, but
    your interpretation was a plausible explanation of what the Original
    Poster had intended

    Nice Avatar.

    RonL
    Follow Math Help Forum on Facebook and Google+

  9. #9
    Super Member
    earboth's Avatar
    Joined
    Jan 2006
    From
    Germany
    Posts
    5,811
    Thanks
    116
    Quote Originally Posted by CaptainBlack
    What I was thinking is not that you had misread the question, but
    your interpretation was a plausible explanation of what the Original
    Poster had intended

    Nice Avatar.

    RonL
    Hello,

    1. I am the worldchampion in reading what I want to read. So I'm pretty sure that you are right.

    2. The "s" and the "2" are separated on the keyboard by a rather wide space. So you can't mess them up by accident.

    3. If the Original POster had mixed up something I would presume that it is meant a^5\ \mbox{instead of} \ a^s. (And now I'd rather prefer to get a reaction from the Original Poster...)

    4. Thanks for the compliment about my avatar. I noticed that some members of the forum changed their outfit (Rebesque, ..., you too). So I decided to take to the public my canned brain, only to demonstrate why I've sometimes so much difficulties to get my precious thoughts out of this container: There are only very small holes in it.

    Best wishes

    EB
    Follow Math Help Forum on Facebook and Google+

  10. #10
    Grand Panjandrum
    Joined
    Nov 2005
    From
    someplace
    Posts
    14,972
    Thanks
    4
    Quote Originally Posted by earboth
    4. Thanks for the compliment about my avatar. I noticed that some members of the forum changed their outfit (Rebesque, ..., you too). So I decided to take to the public my canned brain, only to demonstrate why I've sometimes so much difficulties to get my precious thoughts out of this container: There are only very small holes in it.
    But pleanty of room for BIG thoughts

    RonL
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Expand in terms of binomials coefficients
    Posted in the Discrete Math Forum
    Replies: 4
    Last Post: November 17th 2011, 07:48 PM
  2. Fraction in Lowest terms
    Posted in the Pre-Calculus Forum
    Replies: 2
    Last Post: October 29th 2010, 05:44 AM
  3. please help improper fraction into partial fraction
    Posted in the Pre-Calculus Forum
    Replies: 7
    Last Post: March 1st 2010, 09:06 AM
  4. partial fraction from improper fraction
    Posted in the Calculus Forum
    Replies: 1
    Last Post: February 24th 2010, 02:28 AM
  5. Expand by Partial Fractions!
    Posted in the Calculus Forum
    Replies: 0
    Last Post: October 16th 2009, 09:25 AM

Search Tags


/mathhelpforum @mathhelpforum