Did I Divide This Polynomial Correctly?

[Lancelot59]

As part of solving a DE I need to make this friendlier to integrate:

\frac{2u}{1+u^{-2}}
I figured trying to divide it couldn't hurt. I got:
u^{-1}+u^{-3}

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?

 

[Dick]

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.

 

[symbolipoint]

A more comfortable form of \frac{2u}{1+u^{-2}} may be found if you start like this:
\frac{2u}{1+u^{-2}} \frac{u^2}{u^2}

 

[SteamKing]

Hmm... I've got 1 over some function times what is almost the derivative of that function ... and I want to integrate that combination ...

 

[Lancelot59]

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):
u|1+u^{-2}

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

For 1:u*\frac{1}{u}=1
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:
\frac{1}{u}+u^{-3}+\frac{0}{1+u^{-2}}=\frac{1}{u}+u^{-3}

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

I should be solving:
1+u^{-2}|u

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.

 

[Dick]

I'm home now.

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

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

For 1:u*\frac{1}{u}=1
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:
\frac{1}{u}+u^{-3}+\frac{0}{1+u^{-2}}=\frac{1}{u}+u^{-3}

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

I should be solving:
1+u^{-2}|u


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.

 

[Lancelot59]

Sounds like a plan! Thanks for the help.