Polynomial Long Division

  • Thread starter Lancelot59
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  • #1
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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?
 

Answers and Replies

  • #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.
 
  • #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]
 
  • #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 ...
 
  • #5
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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.
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]

Hmm... I've got 1 over some function times what is almost the derivative of that function ... and I want to integrate that combination ...
I don't see how that works, it's missing an exponent.
 
  • #6
Dick
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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.
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.
 
  • #7
634
1
Sounds like a plan! Thanks for the help.
 

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