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Topic: Long division of 1/(1+u)
Replies: 13   Last Post: Sep 23, 2013 12:27 PM

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Dave L. Renfro

Posts: 4,542
Registered: 12/3/04
Re: Long division of 1/(1+u)
Posted: Sep 23, 2013 12:27 PM
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Hetware wrote:

http://mathforum.org/kb/message.jspa?messageID=9281657

> This is something I should know how to do. Perhaps if I had
> bothered to show up for school, I would have learned it.
> But that's decades gone by.
>
> My calculus book has an exercise that begins with the
> following.
>
> Show, by long division or otherwise, that
>
> 1/(1+u) = 1 - u + u^2 - (u^3)/(1+u)
>
> I'm not sure what it means to perform long division
> on 1/(1+u). Is that a resource in the Internet that
> might help me figure that out?


Others have answered, but I don't believe anyone has
yet mentioned that this used to be called "division in
ascending powers" (or some variation on this phrase).

If you google in google-books the word "division" along
with the phrase "ascending powers", you'll get a lot of
19th century and early 20th century algebra textbooks that
explain and illustrate the method. Typically, the process
can continue indefinitely, but you can stop at any point
and use the current remainder. In the example you gave,
the process was continued up to and including the u^2 term,
with -u^3 being the remainder, and hence -(u^3)/(1+u) goes
at the end.

Your calculus text doesn't assume you know this (and I suspect
that virtually everyone in a beginning calculus course after
the 1950s or so would not know this), since the instructions
include "or otherwise". The obvious "or otherwise" method to
me is to simply multiply both sides by 1 + x and see that you
get an identity. If you have to write up a rigorous proof, then
start with this identity, assume that 1 + u is not equal to zero,
and then reverse the steps you did when you originally obtained
the identity.

http://www.google.com/search?q=division+%22ascending+powers%22&tbm=bks

Dave L. Renfro



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