1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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

    User Avatar
    Science Advisor
    Homework Helper

    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

    User Avatar
    Homework Helper
    Education Advisor
    Gold Member

    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

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper

    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

    User Avatar
    Science Advisor
    Homework Helper

    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.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Polynomial Long Division
Loading...