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		<title>부분분수분해 - 편집 역사</title>
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		<updated>2026-05-04T15:11:34Z</updated>
		<subtitle>이 문서의 편집 역사</subtitle>
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		<id>https://tcatmon.com/w/index.php?title=%EB%B6%80%EB%B6%84%EB%B6%84%EC%88%98%EB%B6%84%ED%95%B4&amp;diff=818607&amp;oldid=prev</id>
		<title>2017년 2월 7일 (화) 10:41에 Maintenance script님의 편집</title>
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				<updated>2017-02-07T10:41:15Z</updated>
		
		<summary type="html">&lt;p&gt;&lt;/p&gt;
&lt;p&gt;&lt;b&gt;새 문서&lt;/b&gt;&lt;/p&gt;&lt;div&gt;Partial fraction decomposition&lt;br /&gt;
[목차]&lt;br /&gt;
== 개요 ==&lt;br /&gt;
부분분수분해는 유리식에서 분자나 분모의 차수를 낮추거나, [[급수]]를 [[망원급수]]의 형태로 바꾸고자 할 때 쓰이는 방법이다.&lt;br /&gt;
&lt;br /&gt;
예를 들어 &amp;lt;math&amp;gt;\displaystyle {1\over {AB}}={1\over{B-A}}\left({1\over A}-{1\over B}\right)&amp;lt;/math&amp;gt;에서 &amp;lt;math&amp;gt;B-A&amp;lt;/math&amp;gt;의 값이 단순하다면 합을 구하기 용이해진다.&lt;br /&gt;
&lt;br /&gt;
== 예시 ==&lt;br /&gt;
&amp;lt;math&amp;gt;\displaystyle \frac{1}{x^2-1} =\frac{1}{2}\left(\frac{1}{x-1}-\frac{1}{x+1}\right)&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
음이 아닌 정수 &amp;lt;math&amp;gt;m&amp;lt;/math&amp;gt;에 대하여&lt;br /&gt;
&amp;lt;math&amp;gt;\displaystyle \sum_{k=1}^{n}k(k+1)(k+2)\cdots(k+m)=\sum_{k=1}^{n}\frac{1}{m+2}\left\{k(k+1)\cdots(k+m)(k+m+1)-(k-1)k\cdots (k+m)\right\}=\frac{1}{m+2}{n(n+1)(n+2)\cdots(n+m+1)}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
자연수 &amp;lt;math&amp;gt;m&amp;lt;/math&amp;gt;에 대하여&lt;br /&gt;
&amp;lt;math&amp;gt;\displaystyle \sum_{k=1}^{n}{1\over k(k+1)(k+2)\cdots(k+m)}={1\over m}\left({{1\over m!}-{1\over (n+1)(n+2)\cdots(n+m)}}\right)&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
[[분류:대수학]][[분류:해석학]]&lt;/div&gt;</summary>
		<author><name>Maintenance script</name></author>	</entry>

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