partial fractions for polynomials

This entry precisely states and provesthe existence and uniqueness of partial fraction decompositionsof ratios of polynomials of a single variable, with coefficients over a field.

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

The proofs involve fairly elementary algebra only. Although we referto Euclidean domains in our proofs, the reader who is not familiar with abstractalgebra 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 computingthe 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; usuallyone 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 herejustify the substitution method. The uniqueness property proved heremight also simplify some calculations: it shows that we never haveto consider multiple solutions for the coefficients in the decomposition.

Theorem 1.

Let $p$ and $q\mathrm{\ne }\mathrm{0}$ be polynomials over a field,and $n$ be any positive integer.Then there exist unique polynomials${\alpha }_{\mathrm{1}}\mathrm{,}\mathrm{\dots }\mathrm{,}{\alpha }_n\mathrm{,}\beta$ such that

(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,

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

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 $q\mathrm{\ne }\mathrm{0}$ be polynomials over a field.Let $q\mathrm{=}{\varphi }_{\mathrm{1}}^{n_{\mathrm{1}}}\mathit{}{\varphi }_{\mathrm{2}}^{n_{\mathrm{2}}}\mathit{}\mathrm{\cdots }\mathit{}{\varphi }_k^{n_k}$be the factorization of $q$ to irreducible factors ${\varphi }_i$(which is unique except for the ordering and constant factors).Then there exist unique polynomials ${\alpha }_{i\mathit{}j}\mathrm{,}\beta$suchthat

(2)

Proof.

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

$$p=\beta q+\sum _{i,j}{\alpha }_{ij}\frac{q}{{\varphi }_i^j}.$$

The polynomial sum on the far right of this equationhas degree , becauseeach summand has degree.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 $s_i^{\prime }$ are polynomials of degree ,such that

$$\sum _{i=1}^k\frac{s_i}{{\varphi }_i^{n_i}}=\sum _{i=1}^k\frac{s_i^{\prime }}{{\varphi }_i^{n_i}}.$$(3)

We claim that $s_i=s_i^{\prime }$. Let $q_1={\varphi }_1^{n_1}$ and $q_2=q/q_1$,and write

$$\frac{s_1}{q_1}+\frac{u}{q_2}=\sum _{i=1}^k\frac{s_i}{{\varphi }_i^{n_i}}=\sum _{i=1}^k\frac{s_i^{\prime }}{{\varphi }_i^{n_i}}=\frac{s_1^{\prime }}{q_1}+\frac{u^{\prime }}{q_2},$$

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

$$(s_1-s_1^{\prime })q_2=(u^{\prime }-u)q_1.$$

In particular, $q_1$ divides the left side.Since $q_1={\varphi }_1^{n_1}$ is relatively prime from $q_2$, it must dividethe factor $(s_1-s_1^{\prime })$. But ,hence $s_1-s_1^{\prime }$ must be the zero polynomial. That is, $s_1=s_1^{\prime }$.

So we can cancel the term $s_1/{\varphi }_1^{n_1}=s_1^{\prime }/{\varphi }_1^{n_1}$on both sides ofequation (3).And we could repeat the argument,and show that $s_2$ and $s_2^{\prime }$ are the same,$s_3$ and $s_3^{\prime }$ are the same, and so on.Therefore, we have shown thatthe polynomials $s_i$ in the following expression

are unique. In particular, $s_i$ isthe following numerator that results when the fractions ${\alpha }_{ij}/{\varphi }_i^j$are put under a common denominator ${\varphi }_{n_i}^i$:

$$s_i=\sum _{j=1}^{n_i}{\alpha }_{ij}{\varphi }_i^{n_i-j}.$$

But by the uniqueness part of Theorem 1,the decomposition

uniquely determines ${\alpha }_{ij}$.(Note that the proof of Theorem 1 shows that${\beta }_i=0$, as .)∎