- #1

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[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?

- Thread starter Lancelot59
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- #1

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[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

Dick

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symbolipoint

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[itex]\frac{2u}{1+u^{-2}} \frac{u^2}{u^2}[/itex]

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SteamKing

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- #5

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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

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.

- #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.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

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Sounds like a plan! Thanks for the help.

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