Results 1 to 10 of 10

Math Help - partial function again . help me

  1. #1
    Junior Member
    Joined
    Jul 2008
    Posts
    52

    partial function again . help me

    x-2/(x^2+1)(x-1)^2 =


    i really cant solved. help help help me. i know the methods. but i stil couldnt find the answer
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Joined
    Aug 2007
    From
    USA
    Posts
    3,110
    Thanks
    2
    Let's see if you "know the methods". How did you set it up?
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Junior Member
    Joined
    Jul 2008
    Posts
    52
    ax+b/(x^2+1),c/(x+2),d/(x+2)^2
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor
    Joined
    Aug 2007
    From
    USA
    Posts
    3,110
    Thanks
    2
    Quote Originally Posted by qweiop90 View Post
    ax+b/(x^2+1),c/(x+2),d/(x+2)^2
    I'm sure you mean this: (ax+b)/(x^2+1) + c/(x-1) + d/(x-1)^2

    Okay, except for the bad notation and the "-1" magically changing to "+2", what's next?

    \frac{ax+b}{x^{2}+1}+\frac{c}{x-1}+\frac{d}{(x-1)^{2}}
    Last edited by TKHunny; July 29th 2008 at 09:42 AM.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Junior Member
    Joined
    Jul 2008
    Posts
    52
    by the way how do you do tht symbol.

    x-2 is identical to ax+b(x-1)^2 +c (x^2+1)(x-1) +d(x^2+1)
    when x is -1 , d = -1/2. others i have no idea
    Follow Math Help Forum on Facebook and Google+

  6. #6
    MHF Contributor
    Joined
    Aug 2007
    From
    USA
    Posts
    3,110
    Thanks
    2
    Quote Originally Posted by qweiop90 View Post
    x-2 is identical to ax+b(x-1)^2 +c (x^2+1)(x-1) +d(x^2+1)
    Again with the bad notation. This may be part of the problem you are having.

    I'm sure you mean: (x-2) = (ax+b)(x-1)^2 +c (x^2+1)(x-1) +d(x^2+1)

    When x = -1
    1) I do not understand the use of x = -1. Don't you mean +1?

    2) x = 1 is NOT in the Domain. How can you substitute it? I realize this is a popular method for this sort of thing. I never have liked it for the reason just stated.

    The solution is in Algebra. No need for a trick that works most of the time but really makes no sense.

    Expand and collect on the right-hand side to get x^{3}(a+c) + x^{2}(-2a+b-c+d) + x(a-2b+c) + (b-c+d).

    Matching corresponding powers of x in the numerators gives

    a+c = 0
    -2a+b-c+d = 0
    a-2b+c = 1
    b-c+d = -2

    You must solve this system.

    If you already have one of the values from somewhere else, you may as well use it.
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Junior Member
    Joined
    Jul 2008
    Posts
    52
    2) x = 1 is NOT in the Domain. How can you substitute it? I realize this is a popular method for this sort of thing. I never have liked it for the reason just stated.

    currently i am studying in lower6, as i said , that is the frm the example tht is written from the book.
    Follow Math Help Forum on Facebook and Google+

  8. #8
    MHF Contributor
    Joined
    Aug 2007
    From
    USA
    Posts
    3,110
    Thanks
    2
    Well, what can I say? Some examples and methods are more harm than they are help.

    Others may disagree with me. Obviously your author does. Soroban does. For me, just don't use it. Run away screaming. Trust me on this. Use algebra rather than trickery, every time.

    My views. I welcome others'.
    Last edited by TKHunny; July 29th 2008 at 01:34 PM.
    Follow Math Help Forum on Facebook and Google+

  9. #9
    Junior Member
    Joined
    Jul 2008
    Posts
    52
    ok . yah. when i applied substituting methods, it makes me more confused. that is why . anyway thanks for your help.
    Follow Math Help Forum on Facebook and Google+

  10. #10
    MHF Contributor
    Joined
    Aug 2007
    From
    USA
    Posts
    3,110
    Thanks
    2
    Quote Originally Posted by qweiop90 View Post
    ok . yah. when i applied substituting methods, it makes me more confused. that is why . anyway thanks for your help.
    What you are saying is that your choices are:

    1) Use the trick, knowing that it doesn't work all the time and that it has a shaky theoretical basis, or

    2) Fail.

    That is not a good idea.

    The solution is not difficult. Just be careful and deliberate.

    After substituting d = -1/2 and a = -c you are left with:

    b + c = 1/2 and b = -1/2

    This makes c = 1 and a = -1.

    Where is the confusing part? Careful and deliberate. You aren't trusting me, yet. {encourage}{encourage}{encourage}
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 1
    Last Post: May 4th 2010, 12:49 PM
  2. Partial derivative of a function
    Posted in the Calculus Forum
    Replies: 8
    Last Post: March 13th 2010, 04:09 AM
  3. Replies: 3
    Last Post: December 4th 2008, 12:09 AM
  4. partial function
    Posted in the Algebra Forum
    Replies: 4
    Last Post: July 21st 2008, 04:16 PM
  5. Partial differentiation with a log function
    Posted in the Calculus Forum
    Replies: 4
    Last Post: May 2nd 2008, 07:06 PM

Search Tags


/mathhelpforum @mathhelpforum