partial fractions

Every fractional number, i. e. such a rational number $\frac{m}{n}$ that the integer $m$ is not divisible by the integer $n$, can be decomposed to a sum of partial fractions as follows:

$$\frac{m}{n}=\frac{m_1}{p_1^{{\nu }_1}}+\frac{m_2}{p_2^{{\nu }_2}}+\mathrm{\cdots }+\frac{m_t}{p_t^{{\nu }_t}}$$

Here, the $p_i$’s are distinct positive prime numbers, the ${\nu }_i$’s positive integers and the $m_i$’s some integers.  Cf. the partial fractions of expressions.

Examples:

$$\frac{6}{289}=\frac{6}{{17}^2}$$

$$-\frac{1}{24}=-\frac{3}{2^3}+\frac{1}{3^1}$$

$$\frac{1}{504}=-\frac{1}{2^3}+\frac{32}{3^2}-\frac{24}{7^1}$$

How to get the numerators $m_i$ for decomposing a fractional number $\frac{1}{n}$ to partial fractions?  First one can take the highest power $p^{\nu }$ of a prime $p$ which divides the denominator $n$.  Then  $n=p^{\nu }u$,  where  $\gcd (u,p^{\nu })=1$.  Euclid’s algorithm gives some integers $x$ and $y$ such that

$$1=xu+y{p}^{\nu }.$$

Dividing this equation by $p^{\nu }u$ gives the

$$\frac{1}{n}=\frac{1}{p^{\nu }u}=\frac{x}{p^{\nu }}+\frac{y}{u}.$$

If $u$ has more than one distinct prime factors, a similar procedure can be made for the fraction $\frac{y}{u}$, and so on.

Note.  The numerators  $m_1$, $m_2$, …, $m_t$  in the decomposition are not unique.  E. g., we have also

$$-\frac{1}{24}=-\frac{11}{2^3}+\frac{4}{3^1}.$$

Cf. the programme “Murto” (in Finnish) or “Murd” (in Estonian) or “Bruch” (in German) or “Bråk” (in Swedish) or “Fraction”(in French) http://www.wakkanet.fi/ pahio/ohjelmi.htmlhere.

Title	partial fractions
Canonical name	PartialFractions
Date of creation	2013-03-22 14:18:10
Last modified on	2013-03-22 14:18:10
Owner	pahio (2872)
Last modified by	pahio (2872)
Numerical id	34
Author	pahio (2872)
Entry type	Definition
Classification	msc 11A41
Synonym	partial fractions of fractional numbers
Related topic	CategoryOfAdditiveFractions
Defines	fractional number