Did I Divide This Polynomial Correctly?

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Homework Help Overview

The discussion revolves around the division of a polynomial expression, specifically focusing on the expression \(\frac{2u}{1+u^{-2}}\). Participants are exploring the correct approach to simplify this expression for integration purposes.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss various methods for dividing the polynomial, with one original poster attempting to simplify the expression by dividing it directly. Others question the correctness of the results and suggest alternative approaches, such as multiplying by \(u^2\) to eliminate negative exponents.

Discussion Status

The discussion is ongoing, with participants providing feedback on attempts and suggesting different methods for approaching the problem. There is recognition of errors in the original poster's reasoning, and some guidance has been offered regarding the order of operations and the handling of negative exponents.

Contextual Notes

Participants note the importance of writing functions in descending order of powers and express confusion over the handling of negative exponents. There is also mention of the challenges faced when using a mobile device to communicate mathematical steps.

Lancelot59
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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 couldn't 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?
 
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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.
 
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]
 
Hmm... I've got 1 over some function times what is almost the derivative of that function ... and I want to integrate that combination ...
 
Dick said:
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]

SteamKing said:
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.
 
Lancelot59 said:
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.
 
Sounds like a plan! Thanks for the help.
 

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