# 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:

 $\frac{m}{n}\;=\;\frac{m_{1}}{p_{1}^{\nu_{1}}}+\frac{m_{2}}{p_{2}^{\nu_{2}}}+% \cdots+\frac{m_{t}}{p_{t}^{\nu_{t}}}$

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

 $1\;=\;xu\!+\!yp^{\nu}.$

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 PartialFractions 2013-03-22 14:18:10 2013-03-22 14:18:10 pahio (2872) pahio (2872) 34 pahio (2872) Definition msc 11A41 partial fractions of fractional numbers CategoryOfAdditiveFractions fractional number