# Polynomial long division: different answers when reordering terms

When I use polynomial long division to divide   $\frac{1}{1-x}$,   I get $\;1 + x + x^2 +x^3 + x^4 + \cdots$
But when I just change the order of terms in the divisor:   $\frac{1}{-x+1}$,   the long division algorithm gives me a very different answer: $-\frac{1}{x} - \frac{1}{x^2} - \frac{1}{x^3} - \frac{1}{x^4} - \cdots$,   which seems somewhat strange to me, because sums are supposed to be commutative, so it looks that these two answers should be equivalent. Am I right?

But I cannot see how could these two answers be equivalent. Could it be true that $-\frac{1}{x} - \frac{1}{x^2} - \frac{1}{x^3} - \frac{1}{x^4} - \cdots \;=\; 1 + x + x^2 +x^3 + x^4 + \cdots$   ?
If that is the case, how can i "prove" it? Or how can I transform one form into the other?

I already figured out that I can express   $-\frac{1}{x} - \frac{1}{x^2} - \frac{1}{x^3} - \frac{1}{x^4} - \cdots$   as negative exponents:   $-x^{-1} - x^{-2} - x^{-3} - x^{-4} - \cdots$   and then factor the $-1$'s inside the exponents:   $-x^{-1\cdot1} - x^{-1\cdot2} - x^{-1\cdot3} - x^{-1\cdot4} - \cdots$   which can be seen as a power of a power:   $-(x^{-1})^1 - (x^{-1})^2 - (x^{-1})^3 - (x^{-1})^4 - \cdots$   and the minus sign can be factored out too:   $-\left[ (x^{-1})^1 + (x^{-1})^2 + (x^{-1})^3 + (x^{-1})^4 + \cdots \right]$   which almost gives me the form of the other series with raising positive exponents. But there's that dangling minus sign in front of it, and the 0th power term is missing :-/ so I don't know how to massage it any further to get the other form of the expansion. Any ideas?

Edit:
I know about Taylor series expansions and I can expand  $\frac{1}{1-x}$   this way into   $1 + x + x^2 + x^3 + \cdots$   (I don't know how to get the other form this way, though).

Also, I know about the convergence condition for the infinite series,   $|x| < 1$   in the case of the first answer, and   $|x| > 1$   or   $|\frac{1}{x}| < 1$   in the case of the second answer. I just didn't expect that this could matter in my case, when I just reorder the terms in the divisor, which shouldn't produce different answer.

I suspected that this could be somehow related to the fact that raising powers of a fraction is the same as falling powers of its inverse. But I cannot figure out why just one of these or the other is given, depending on the order of terms in the long division, and this is what confuses me the most.

Seeing as you tagged this sequences and series, I presume you know that you have to be careful when talking about adding up an infinite sequence of terms.

There is an infinite series hidden in your work when you write $$\frac{1}{1-x}=1+x + x^2 +x^3 + \cdots,$$ but this series only converges when $|x|<1$. This sum is the Taylor series for $\frac{1}{1-x}$.

There is an infinite series hidden in your work when you write $$\frac{1}{1-x} = -\frac{1}{x}- \frac{1}{x^2}-\frac{1}{x^3}- \cdots,$$ and this infinite series only converges when $|x|>1$ (or, equivalently, when $\left | \frac{1}{x} \right|<1$). This sum is the Laurent series for $\frac{1}{1-x}$.

So both expressions are correct (for the right values of $x$) but they are not both correct at the same time. One is valid for $|x|<1$ and the other for $|x|>1$.

Just to clarify the situation a little further:

You are correct when you say that dividing by $-x+1$ should be the same as dividing by $1-x$. There is no problem with this as long as we stop after finitely many steps:

If we stop after three steps in the polynomial long division algorithm, we get $$\frac{1}{1-x}=1+x +x^2 +\frac{x^3}{1-x}.$$ That is, we get an answer of $1+x + x^2$ with a remainder of $x^3$ that has not yet been divided by $1-x$.

Or if we do it the other way, after three steps, we get $$\frac{1}{1-x}= - \frac{1}{x}-\frac{1}{x^2}-\frac{1}{x^3} +\frac{1}{x^3}\times \frac{1}{1-x}.$$ That is, we get an answer of $- \frac{1}{x}-\frac{1}{x^2}-\frac{1}{x^3}$ with a remainder of $\frac{1}{x^3}$ that has not yet been divided by $1-x$.

Let's assume that $x \neq 0$ and $x \neq 1$ so that everything is well-defined. Then $$\frac{1}{1-x}=1+x +x^2 +\frac{x^3}{1-x} =- \frac{1}{x}-\frac{1}{x^2}-\frac{1}{x^3} +\frac{1}{x^3}\times \frac{1}{1-x}.$$ Both ways of doing it do give us the same answer.

So what's the problem? When we try continuing this to infinitely many terms, we need these infinite series to converge to the correct answer. It is not the order $-x+1$ versus $1-x$ that leads to different answers; it is the change from writing down a sum with finitely many terms to writing down a series with infinitely many terms that we have to be careful about.

• Hmm so these two forms of the answer are like two sides of the same coin, or two parts of the same answer, one for $|x| < 1$, and the other one for $|x| > 1$ ? – SasQ Sep 26 '13 at 11:44
• I know about the Taylor series and its divergence conditions. But I wanted to research how to get the same expansions another way, through polynomial long division, since I saw in Newton's book that this is possible (and this was the technique used before derivatives and Taylor series has been invented). But when I tried it myself, I noticed that I get very different answers when I just reorder the terms, which is somewhat strange to me. – SasQ Sep 26 '13 at 11:52
• Thanks for the hint with the Laurent series. I didn't know that. – SasQ Sep 26 '13 at 13:12

Note that the series $\;1 + x + x^2 +x^3 + x^4 + \cdots$ converges only for $|x|<1$. If this series is convergent thet you cannot write $\frac{1}{1-x}=-\frac{1}{x} - \frac{1}{x^2} - \frac{1}{x^3} - \frac{1}{x^4} - \cdots$ because $|\frac{1}{x}|>1$ and hence the last series diverges and therefore the equality does not hold. Also the other way around, if the last equality holds then the first one will not hold.

• I know about the convergence condition, but I didn't expected it matter here, when I just reorder the terms in my divisor, since sums are supposed to be commutative and just changing the order of terms shouldn't affect the actual meaning of the answer, I suspected, but just change its form, if anything. – SasQ Sep 26 '13 at 11:47