# “Polynomial” of $\frac{\nu\sin x}{(1-\nu)+\nu\cos x}$?

I learned the following formula from a numerical analysis book, \begin{align} \frac{\nu\sin x}{(1-\nu)+\nu\cos x}&=\nu(x-\frac{1}{6}x^3+\cdots)(1-\frac{1}{2}\nu x^2+\cdots)^{-1}\\ &=\nu x-\frac{1}{6}\nu(1-3\nu)x^3+\cdots \end{align} This kind of manipulation is very common in numerical analysis. How can I get the last equality? Because the order is from low to high, I have no idea how to use the polynomial long division in this case.

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The algorithm is essentially the same as the familiar one, just backwards. $1-\frac{1}{2}\nu x^2 +\cdots$ goes into $1$ once. Write it down below the $1$, subtract. We get difference $\frac{1}{2}\nu x^2 -\cdots$. Then $1-\frac{1}{2}\nu x^2+\cdots$ goes into this $\frac{1}{2}\nu x^2$ times. Continue, if necessary. The only difference is that the thing you are dividing by should, sort of, begin with a non-zero constant. Here it does, the constant is $1$. – André Nicolas Nov 11 '11 at 23:35
I don't want to typeset an example, spacing too tricky. Let me explain another way. We want to divide $2-x$ by say $1+2x+3x^3+4x^3+\cdots$, which I will call $A(x)$. The $1$ in $A(x)$ goes into $2$ twice, so write a $2$ as part of the quotient. Multiply $A(x)$ by $1$, subtract the result from $2-x$. That gives you the remainder after the first step. It is $-5x-6x^2-8x^3+\cdots$. $A(x)$ goes into this $-5x$ times. So the next part of the quotient is $-5x$. Multiply $A(x)$ by $-5x$, subtract this from our previous remainder. The remainder starts with $4x^2$. Continue. (to be continued) – André Nicolas Nov 12 '11 at 5:59

You don't need to do any division for this sort of thing. All you ever need is $\frac1{1-x}=1+x+x^2+\dotso$

In your case, you want the result up to third order in $x$, so you expand everything up to third order:

$$\frac{\nu\sin x}{(1-\nu)+\nu\cos x}=\frac{\nu\left(x-\frac16x^3+O(x^5)\right)}{1-\frac12\nu x^2+O(x^4)}\;.$$

Now with $u = \frac12\nu x^2-O(x^4)$,

$$\frac1{1-\frac12\nu x^2+O(x^4)}=\frac1{1-u}=1+u+u^2+\dotso=1+\frac12\nu x^2+O(x^4)\;,$$

and thus

$$\begin{eqnarray} \frac{\nu\left(x-\frac16x^3+O(x^5)\right)}{1-\frac12\nu x^2+O(x^4)} &=& \nu\left(x-\frac16x^3+O(x^5)\right)\left(1+\frac12\nu x^2+O(x^4)\right) \\ &=&\nu x+\left(\frac12\nu^2-\frac16\nu\right)x^3+O(x^5) \;. \end{eqnarray}$$

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This is really helpful! I didn't notice that one should use the geometric series. – Jack Nov 13 '11 at 0:51
@Jack: It's useful for lower orders like in this case. It always works, but for higher orders it can involve multiplying out lots of powers, and then division may be easier. – joriki Nov 13 '11 at 7:26