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Homework Help: Polynomial Long Division

  1. Nov 2, 2011 #1
    As part of solving a DE I need to make this friendlier to integrate:

    [tex]\frac{2u}{1+u^{-2}}[/tex]
    I figured trying to divide it couldnt hurt. I got:
    [tex]u^{-1}+u^{-3}[/tex]

    I can't type out all the steps easily, I'm on a mobile device at the moment. That answer looks suspect, did I do it correctly?
     
  2. jcsd
  3. Nov 2, 2011 #2

    Dick

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    It's more than suspect, it's completely wrong. If u=1 then your first expression is 1, the second one is 2. Pretty bad, yes? Suggest you get off the mobile device and type in your steps.
     
  4. Nov 3, 2011 #3

    symbolipoint

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    A more comfortable form of [itex] \frac{2u}{1+u^{-2}} [/itex] may be found if you start like this:
    [itex]\frac{2u}{1+u^{-2}} \frac{u^2}{u^2}[/itex]
     
  5. Nov 3, 2011 #4

    SteamKing

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    Hmm... I've got 1 over some function times what is almost the derivative of that function ... and I want to integrate that combination ...
     
  6. Nov 3, 2011 #5
    I'm home now.

    I set it up like so (I left the two out):
    [tex]u|1+u^{-2}[/tex]

    from my understanding you're supposed to write the functions in descending order of powers.

    For 1:[tex]u*\frac{1}{u}=1[/tex]
    subtraction yields zero. You then need to find out what multiple of u gives you u-2. u*u-3=u-2, subtraction gives zero. Total remainder of zero.

    So I get the final answer of:
    [tex]\frac{1}{u}+u^{-3}+\frac{0}{1+u^{-2}}=\frac{1}{u}+u^{-3}[/tex]

    I think I just saw where I went wrong...I wrote it backwards.

    I should be solving:
    [tex]1+u^{-2}|u[/tex]

    I don't see how that works, it's missing an exponent.
     
  7. Nov 3, 2011 #6

    Dick

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    Right, you did it backwards. (1+u^(-2))/u=u^(-1)+u^(-3). The whole process is a lot easier to follow if you clear out the negative powers first, like symbolpoint suggested above.
     
  8. Nov 3, 2011 #7
    Sounds like a plan! Thanks for the help.
     
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