partial fractionsPartialFractions2004-04-14 07:00:212008-10-17 15:22:52Definitionfractionfractional number% this is the default PlanetMath preamble. as your knowledge
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% define commands hereEvery {\em 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 {\em 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.
\textbf{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 \PMlinkescapetext{decomposition}
$$\frac{1}{n} = \frac{1}{p^{\nu}u} = \frac{x}{p^{\nu}}+\frac{y}{u}.$$
If $u$ has more than one distinct prime factors, a similar procedure can be made for the fraction $\frac{y}{u}$, and so on.
\textbf{Note.}\, The numerators\, $m_1$, $m_2$, \ldots, $m_t$\, 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) \PMlinkexternal{here}{http://www.wakkanet.fi/~pahio/ohjelmi.html}.