# partial fractions for polynomials

The theory is used for, for example, the method of partial fraction decomposition for integrating rational functions over the reals (http://planetmath.org/ALectureOnThePartialFractionDecompositionMethod).

The proofs involve fairly elementary algebra only. Although we refer to Euclidean domains in our proofs, the reader who is not familiar with abstract algebra may simply read that as “set of polynomials” (which is one particular Euclidean domain).

Also note that the proofs themselves furnish a method for actually computing the partial fraction decomposition, as a finite-time algorithm, provided the irreducible  factorization of the denominator is known. It is not an efficient way to find the partial fraction decomposition; usually one uses instead the method of making substitutions into the polynomials, to derive linear constraints on the coefficients. But what is important is that the existence proofs here justify the substitution method. The uniqueness property proved here might also simplify some calculations: it shows that we never have to consider multiple   solutions for the coefficients in the decomposition.

###### Theorem 1.

Let $p$ and $q\neq 0$ be polynomials over a field, and $n$ be any positive integer. Then there exist unique polynomials $\alpha_{1},\ldots,\alpha_{n},\beta$ such that

 $\frac{p}{q^{n}}=\beta+\frac{\alpha_{1}}{q}+\frac{\alpha_{2}}{q^{2}}+\cdots+% \frac{\alpha_{n}}{q^{n}}\,,\quad\deg\alpha_{j}<\deg q\,.$ (1)
###### Proof.

Existence has already been proven as a special case of partial fractions in Euclidean domains; we now prove uniqueness. Suppose equation (1) has been given. Multiplying by $q^{n}$ and rearranging,

 $p=\beta q^{n}+r_{1}\,,\quad r_{1}=\alpha_{1}q^{n-1}+\cdots+\alpha_{n}\,,\quad% \deg r_{1}<\deg q^{n}\,.$

But according to the division algorithm  for polynomials (also known as long division), the quotient  and remainder polynomial after a division (by $q^{n}$ in this case) are unique. So $\beta$ must be uniquely determined. Then we can rearrange:

 $p-\beta\,q^{n}=\alpha_{1}q^{n-1}+r_{2}\,,\quad r_{2}=\alpha_{2}q^{n-2}+\cdots+% \alpha_{n}\,,\quad\deg r_{2}<\deg q^{n-1}\,.$

By uniqueness of division again (by $q^{n-1}$), $\alpha_{1}$ is determined. Repeating this process, we see that all the polynomials $\alpha_{j}$ and $\beta$ are uniquely determined. ∎

###### Theorem 2.

Let $p$ and $q\neq 0$ be polynomials over a field. Let $q=\phi_{1}^{n_{1}}\,\phi_{2}^{n_{2}}\,\cdots\,\phi_{k}^{n_{k}}$ be the factorization of $q$ to irreducible factors $\phi_{i}$ (which is unique except for the ordering and constant factors). Then there exist unique polynomials $\alpha_{ij},\beta$ such that

 $\frac{p}{q}=\beta+\sum_{i=1}^{k}\sum_{j=1}^{n_{i}}\frac{\alpha_{ij}}{\phi_{i}^% {j}}\,,\quad\deg\alpha_{ij}<\deg\phi_{i}\,.$ (2)
###### Proof.

Existence has already been proven as a special case of partial fractions in Euclidean domains; we now prove uniqueness. Suppose equation (2) has been given. First, multiply the equation by $q$:

 $p=\beta\,q+\sum_{i,j}\alpha_{ij}\,\frac{q}{\phi_{i}^{j}}\,.$

The polynomial sum on the far right of this equation has degree $, because each summand has degree $\deg(\alpha_{ij}\,q/\phi_{i}^{j})<\deg\phi_{i}+\deg q-j\cdot\deg\phi_{i}\leq\deg q$. So the polynomial sum is the remainder of a division of $p$ by $q$. Then the quotient polynomial $\beta$ is uniquely determined.

Now suppose $s_{i}$ and $s^{\prime}_{i}$ are polynomials of degree $<\phi_{i}^{n_{i}}$, such that

 $\sum_{i=1}^{k}\frac{s_{i}}{\phi_{i}^{n_{i}}}=\sum_{i=1}^{k}\frac{s^{\prime}_{i% }}{\phi_{i}^{n_{i}}}\,.$ (3)

We claim that $s_{i}=s^{\prime}_{i}$. Let $q_{1}=\phi_{1}^{n_{1}}$ and $q_{2}=q/q_{1}$, and write

 $\frac{s_{1}}{q_{1}}+\frac{u}{q_{2}}=\sum_{i=1}^{k}\frac{s_{i}}{\phi_{i}^{n_{i}% }}=\sum_{i=1}^{k}\frac{s^{\prime}_{i}}{\phi_{i}^{n_{i}}}=\frac{s^{\prime}_{1}}% {q_{1}}+\frac{u^{\prime}}{q_{2}}\,,$

for some polynomials $u$ and $u^{\prime}$. Rearranging, we get:

 $(s_{1}-s^{\prime}_{1})\,q_{2}=(u^{\prime}-u)\,q_{1}\,.$

In particular, $q_{1}$ divides the left side. Since $q_{1}=\phi_{1}^{n_{1}}$ is relatively prime from $q_{2}$, it must divide the factor $(s_{1}-s^{\prime}_{1})$. But $\deg(s_{1}-s^{\prime}_{1})<\deg q_{1}$, hence $s_{1}-s^{\prime}_{1}$ must be the zero polynomial  . That is, $s_{1}=s^{\prime}_{1}$.

So we can cancel the term $s_{1}/\phi_{1}^{n_{1}}=s^{\prime}_{1}/\phi_{1}^{n_{1}}$ on both sides of equation (3). And we could repeat the argument, and show that $s_{2}$ and $s^{\prime}_{2}$ are the same, $s_{3}$ and $s^{\prime}_{3}$ are the same, and so on. Therefore, we have shown that the polynomials $s_{i}$ in the following expression

 $\frac{p}{q}-\beta=\sum_{i=1}^{k}\frac{s_{i}}{\phi_{i}^{n_{i}}}\,,\quad\deg s_{% i}<\deg\phi_{i}^{n_{i}}$

are unique. In particular, $s_{i}$ is the following numerator that results when the fractions $\alpha_{ij}/\phi_{i}^{j}$ are put under a common denominator $\phi^{i}_{n_{i}}$:

 $s_{i}=\sum_{j=1}^{n_{i}}\alpha_{ij}\,\phi_{i}^{n_{i}-j}\,.$

But by the uniqueness part of Theorem 1, the decomposition

 $\frac{s_{i}}{\phi_{i}^{n_{i}}}=\beta_{i}+\sum_{j=1}^{n_{i}}\frac{\alpha_{ij}}{% \phi_{i}^{j}}\,,\quad\deg\alpha_{ij}<\deg\phi_{i}$

uniquely determines $\alpha_{ij}$. (Note that the proof of Theorem 1 shows that $\beta_{i}=0$, as $\deg s_{i}<\deg\phi_{i}^{n_{i}}$.) ∎

Title partial fractions for polynomials PartialFractionsForPolynomials 2013-03-22 15:40:22 2013-03-22 15:40:22 stevecheng (10074) stevecheng (10074) 6 stevecheng (10074) Result msc 12E05 partial fraction decomposition of rational functions partial fractions for rational functions PartialFractionsOfExpressions ALectureOnThePartialFractionDecompositionMethod