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| Title of object: |
partial fractions |
| Canonical Name: |
PartialFractions |
| Type: |
Definition |
| Created on: |
2004-04-14 07:00:21 |
| Modified on: |
2004-04-29 09:23:19 |
| Classification: |
msc:11A41 |
Preamble:
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Content:
Every {\em fractional number}, i. e. such rational number $\frac {m}{n}$ that $m$ is not divisible by $n$, can be decomposed to a sum of {\em partial fractions} as follows:
$$\frac{m}{n} = \frac{m_1}{p_1^{\nu_1}}+\frac{m_2}{p_2^{\nu_2}}+...+\frac{m_t}{p_t^{\nu_t}}$$
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:
$$\frac{6}{25} = \frac{6}{5^2}$$
$$-\frac{1}{24} = -\frac{3}{2^3}+\frac{1}{3^1}$$
$$\frac{1}{504} = -\frac{1}{2^3}+\frac{32}{3^2}-\frac{24}{7^1}$$
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 $\gcd{(u, p^{\nu})} = 1$. The Euclid's algorithm gives some integers $x$ and $y$ such that
$$1 = xu+yp^{\nu}.$$
Dividing this equation by $p^{\nu}u$ gives the 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.
Note: The numerators $m_1$, $m_2$, ..., $m_t$ in the decomposition are not unique. E. g., we have also
$$-\frac{1}{24} = -\frac{11}{2^3}+\frac{4}{3^1}.$$ |
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