# partial fractions for polynomials

This entry precisely states and proves
the existence and uniqueness of partial fraction decompositions
of 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 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 $q\mathrm{\ne}\mathrm{0}$ 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 polynomial
after 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 $ such that

$$ | (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 $q$:

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

The polynomial sum on the far right of this equation has degree $$, because each summand has degree $$. 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 divide
the 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 of equation (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 that the polynomials ${s}_{i}$ in the following expression

$$ |

are unique. In particular, ${s}_{i}$ is the 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 $$.) ∎

Title | partial fractions for polynomials |
---|---|

Canonical name | PartialFractionsForPolynomials |

Date of creation | 2013-03-22 15:40:22 |

Last modified on | 2013-03-22 15:40:22 |

Owner | stevecheng (10074) |

Last modified by | stevecheng (10074) |

Numerical id | 6 |

Author | stevecheng (10074) |

Entry type | Result |

Classification | msc 12E05 |

Synonym | partial fraction decomposition of rational functions |

Synonym | partial fractions for rational functions |

Related topic | PartialFractionsOfExpressions |

Related topic | ALectureOnThePartialFractionDecompositionMethod |