# Division by increasing power order

Euclid conducting a proof (Raphael's fresco School Of Athens, 1511)

The division of polynomials by increasing power order is similar to the usual Euclidean division but in reversed order : the terms of lower degree of the dividend are eliminated first. Some useful algorithms can be deduced from this process, such as series expansion or partial fraction decomposition.

EUCLIDEAN DIVISION

It is well known that given two polynomials $A$ and $B \neq 0$ on a commutative field K, it is always possible to perform the Euclidean division

$\displaystyle A = B Q + R , \deg(R)<\deg(B)$

The corresponding algorithm consists of dividing by decreasing power order, which means that at each step a division by B is performed on polynomials of decreasing degrees until $\deg(R)<\deg(B)$. Here is an example to recall the method :

Divide $X^3+1$ by $X+1$.

Step 1 (degree 3) : $X^3+1 = (X+1)(X^2)+(-X^2+1)$

Step 2 (degree 2) : $-X^2+1 = (X+1)(-X)+(X+1)$

Step 3 (degree 1) : $X+1 = (X+1)(1)+0$

DIVISION BY INCREASING POWER ORDER

There exists an alternative method to divide polynomials, so called by increasing power order. Here is the theory.

Theorem

Let $n \in \mathbb{N}$ and $A,B$ two polynomials in $K[X]$. We write them $A=a_0+a_1 X+\cdots+a_p X^p$ and $B=b_0+b_1 X+\cdots+c_m X^m$. We assume $b_0 \neq 0$. Then there exist a unique pair $(Q,R)$ of polynomials in $K[X]$ such that

$\displaystyle \left\{ \begin{array}{l} A = B Q + X^{n+1} R \\ \deg(Q) \leq n \end{array} \right.$

Proof

We begin to prove existence by induction. Let $\mathcal{P}(n)$ be the property for any $n \in \mathbb{N}$.

Case $\mathcal{P}(n=0)$. Let $Q = \frac{a_0}{b_0}$, which is allowed since $b_0 \neq 0$. Then $A - B Q$ has no constant term, $X$ must divide $A - B Q$, so there exists $R = \frac{A - B Q}{X}$ such that $A = B Q + X R$.

Case $\mathcal{P}(n+1)$ under hypothesis that $\mathcal{P}(k \leq n)$ is true. After division of $A$ at order $n$, we have $A = B Q + X^{n+1} R$ with $\deg(Q) \leq n$. After division of $R$ at order 0, we have $R = \lambda B + X S$ with $\deg(\lambda) = 0 \leq n$. Now we replace $R$ into $A$ to get $A = B (Q + \lambda X^{n+1}) + X^{n+2} S$ with $\deg(Q + \lambda X^{n+1}) \leq n+1$. Therefore, the property $\mathcal{P}(n+1)$ holds with quotient $Q + \lambda X^{n+1}$ and remainder $S$.

Now we prove unicity. Let $(Q,R)$ and $(Q',R')$ be two solutions. Then we have $B(Q-Q')+X^{n+1}(R-R') = 0$ with $\deg(Q) \leq n$ and $\deg(Q') \leq n$. Clearly, $X^{n+1}$ must divide $B(Q-Q')$. However, $b_0 \neq 0$ implies that $X$ is not an irreducible factor of $B$ and thus $X^{n+1}$ does not divide $B$. Hence, $X^{n+1}$ must divide $Q-Q'$. Since $\deg(Q-Q') \leq n$ it is necessary that $Q-Q' = 0$. It immediately follows $R-R' = 0$.

SERIES EXPANSION

The division by increasing power order can help to compute series expansion. Let us consider for example the following series truncated at order 6 :

$\displaystyle \tan X = \frac{\sin X}{\cos X}$

$\displaystyle \tan X = \frac{X-\frac{X^3}{6}+\frac{X^5}{120}+o(X^6)}{1-\frac{X^2}{2}+\frac{X^4}{24}+o(X^5)}$

Step 1 : $X - \frac{X^3}{6} + \frac{X^5}{120} = \left( 1-\frac{X^2}{2} + \frac{X^4}{24} \right) X + \left( \frac{X^3}{3}-\frac{X^5}{30} \right)$

Step 2 : $\frac{X^3}{3} - \frac{X^5}{30} = \left( 1 - \frac{X^2}{2} + \frac{X^4}{24} \right) \frac{X^3}{3} + \left( \frac{2 X^5}{15} - \frac{X^7}{72} \right)$

Step 3 : $\frac{2 X^5}{15} - \frac{X^7}{72} = \left( 1 - \frac{X^2}{2} + \frac{X^4}{24} \right) \frac{2 X^5}{15} + \left( \frac{19 X^7}{360} - \frac{X^9}{180} \right)$

Hence we have the following expansion

$\displaystyle \tan X = X + \frac{X^3}{3} + \frac{2 X^5}{15} + o(X^6)$

PARTIAL FRACTION DECOMPOSITION

The division by increasing power order can also help to compute partial fraction decomposition. Let us consider for example the following fraction :

$\displaystyle \frac{1+X}{X^4 (1+X^2)}$

We divide $1+X$ by $1+X^2$ as follows.

Step 1 : $1+X = (1+X^2) 1 + (X-X^2)$

Step 2 : $X-X^2 = (1+X^2) X + (-X^2-X^3)$

Step 3 : $-X^2-X^3 = (1+X^2) (-X^2) + (-X^3+X^4)$

Step 4 : $-X^3+X^4 = (1+X^2) (-X^3) + (X^4+X^5)$

Hence :

$\displaystyle 1+X = \left( 1+X^2 \right) \left( 1+X-X^2-X^3 \right) + \left( X^4+X^5 \right)$

$\displaystyle \frac{1+X}{X^4 (1+X^2)} = \frac{1+X-X^2-X^3}{X^4} + \frac{1+X}{1+X^2}$

$\displaystyle \frac{1+X}{X^4 (1+X^2)} = \frac{1}{X^4} + \frac{1}{X^3} - \frac{1}{X^2} - \frac{1}{X} + \frac{1+X}{1+X^2}$