| Version 24 |
Version 23 |
| Every {\em fractional number}, i. e. such 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: |
Every {\em fractional number}, i. e. such rational number $\frac {m}{n}$ that the integer $m$ is not divisible by $n$, can be decomposed to a sum of {\em partial fractions} as follows: |
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| $$\frac{m}{n} = \frac{m_1}{p_1^{\nu_1}}+\frac{m_2}{p_2^{\nu_2}}+...+\frac{m_t}{p_t^{\nu_t}}$$ |
$$\frac{m}{n} = \frac{m_1}{p_1^{\nu_1}}+\frac{m_2}{p_2^{\nu_2}}+...+\frac{m_t}{p_t^{\nu_t}}$$ |
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| 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. |
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. |
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| \textbf{Examples:} |
\textbf{Examples:} |
| $$\frac{6}{289} = \frac{6}{17^2}$$ |
$$\frac{6}{289} = \frac{6}{17^2}$$ |
| $$-\frac{1}{24} = -\frac{3}{2^3}+\frac{1}{3^1}$$ |
$$-\frac{1}{24} = -\frac{3}{2^3}+\frac{1}{3^1}$$ |
| $$\frac{1}{504} = -\frac{1}{2^3}+\frac{32}{3^2}-\frac{24}{7^1}$$ |
$$\frac{1}{504} = -\frac{1}{2^3}+\frac{32}{3^2}-\frac{24}{7^1}$$ |
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| 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 |
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 |
| $$1 = xu+yp^{\nu}.$$ |
$$1 = xu+yp^{\nu}.$$ |
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| Dividing this equation by $p^{\nu}u$ gives the \PMlinkescapetext{decomposition} |
Dividing this equation by $p^{\nu}u$ gives the \PMlinkescapetext{decomposition} |
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| $$\frac{1}{n} = \frac{1}{p^{\nu}u} = \frac{x}{p^{\nu}}+\frac{y}{u}.$$ |
$$\frac{1}{n} = \frac{1}{p^{\nu}u} = \frac{x}{p^{\nu}}+\frac{y}{u}.$$ |
| If $u$ has more than one distinct prime factors, a similar procedure can be made for the fraction $\frac{y}{u}$, and so on. |
If $u$ has more than one distinct prime factors, a similar procedure can be made for the fraction $\frac{y}{u}$, and so on. |
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| \textbf{Note.} \,The numerators $m_1$, $m_2$, ..., $m_t$ in the decomposition are not unique. \,E. g., we have also |
\textbf{Note:} \,The numerators $m_1$, $m_2$, ..., $m_t$ in the decomposition are not unique. \,E. g., we have also |
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| $$-\frac{1}{24} = -\frac{11}{2^3}+\frac{4}{3^1}.$$ |
$$-\frac{1}{24} = -\frac{11}{2^3}+\frac{4}{3^1}.$$ |
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| Cf. the programme ``Murto''(in Finnish) or ``Murd'' (in Estonian) or ``Fraction''(in French) or ``Bruch'' (in German) or ``Bråk'' (in Swedish) at http://www.wakkanet.fi/(tilde)pahio/ohjelmi.html. |
Cf. the programme ``Murto''(in Finnish) or ``Murd'' (in Estonian) or ``Fraction''(in French) or ``Bruch'' (in German) or ``Bråk'' (in Swedish) at http://www.wakkanet.fi/(tilde)pahio/ohjelmi.html. |
