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| * 2차원격자를 이루는 두 복소수 <math>\omega_1,\omega_2</math>에 대하여, <br><math>\wp(z;\omega_1,\omega_2)=\frac{1}{z^2}+ \sum_{m^2+n^2 \ne 0} \left\{ \frac{1}{(z-m\omega_1-n\omega_2)^2}- \frac{1}{\left(m\omega_1+n\omega_2\right)^2} \right\}</math><br><math>\wp(z;\omega_1,\omega_2)=\frac{1}{z^2}+ \sum_{m^2+n^2 \ne 0} \left\{ \frac{1}{(z-m\omega_1-n\omega_2)^2}- \frac{1}{\left(m\omega_1+n\omega_2\right)^2} \right\}</math><br> | | * 2차원격자를 이루는 두 복소수 <math>\omega_1,\omega_2</math>에 대하여, <br><math>\wp(z;\omega_1,\omega_2)=\frac{1}{z^2}+ \sum_{m^2+n^2 \ne 0} \left\{ \frac{1}{(z-m\omega_1-n\omega_2)^2}- \frac{1}{\left(m\omega_1+n\omega_2\right)^2} \right\}</math><br><math>\wp(z;\omega_1,\omega_2)=\frac{1}{z^2}+ \sum_{m^2+n^2 \ne 0} \left\{ \frac{1}{(z-m\omega_1-n\omega_2)^2}- \frac{1}{\left(m\omega_1+n\omega_2\right)^2} \right\}</math><br> |
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| + | 는 타원함수가 됨. |
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− | * 는 타원함수가 됨.<br>
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69번째 줄: |
67번째 줄: |
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− | * 네이버 지식인<br>
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− | ** http://kin.search.naver.com/search.naver?where=kin_qna&query=
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− | ** http://kin.search.naver.com/search.naver?where=kin_qna&query=
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− | * 도서내검색<br>
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− | ** http://books.google.com/books?q=
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− | * 도서검색<br>
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− | ** http://www.amazon.com/s/ref=nb_ss_gw?url=search-alias%3Dstripbooks&field-keywords=
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− | * http://ko.wikipedia.org/wiki/
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− | * http://en.wikipedia.org/wiki/
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− | * http://www.wolframalpha.com/input/?i=
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− | * http://front.math.ucdavis.edu/search?a=&t=&c=&n=40&s=Listings&q=
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− | * http://www.ams.org/mathscinet/search/publications.html?pg4=AUCN&s4=&co4=AND&pg5=TI&s5=&co5=AND&pg6=PC&s6=&co6=AND&pg7=ALLF&co7=AND&Submit=Search&dr=all&yrop=eq&arg3=&yearRangeFirst=&yearRangeSecond=&pg8=ET&s8=All&s7=
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− | * 다음백과사전 http://enc.daum.net/dic100/search.do?q=
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− | * [http://mathnet.kaist.ac.kr/mathnet/math_list.php?mode=list&ftype=&fstr= 대한수학회 수학 학술 용어집]
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− | * [http://navercast.naver.com/science/list 네이버 오늘의과학]
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− | * 네이버 뉴스 검색 (키워드 수정)<br>
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− | ** http://news.search.naver.com/search.naver?where=news&x=0&y=0&sm=tab_hty&query=
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− | ** http://news.search.naver.com/search.naver?where=news&x=0&y=0&sm=tab_hty&query=
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− | ** http://news.search.naver.com/search.naver?where=news&x=0&y=0&sm=tab_hty&query=
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− | ** http://news.search.naver.com/search.naver?where=news&x=0&y=0&sm=tab_hty&query=
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− | * 구글 블로그 검색 http://blogsearch.google.com/blogsearch?q=
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− | * 트렌비 블로그 검색 http://www.trenb.com/search.qst?q=
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− | * http://commons.wikimedia.org/w/index.php?title=Special%3ASearch&search=
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− | * http://images.google.com/images?q=
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− | * [http://www.artchive.com/ http://www.artchive.com]
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− | * http://www.youtube.com/results?search_type=&search_query=
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정의
- 2차원격자를 이루는 두 복소수 \(\omega_1,\omega_2\)에 대하여,
\(\wp(z;\omega_1,\omega_2)=\frac{1}{z^2}+ \sum_{m^2+n^2 \ne 0} \left\{ \frac{1}{(z-m\omega_1-n\omega_2)^2}- \frac{1}{\left(m\omega_1+n\omega_2\right)^2} \right\}\)
\(\wp(z;\omega_1,\omega_2)=\frac{1}{z^2}+ \sum_{m^2+n^2 \ne 0} \left\{ \frac{1}{(z-m\omega_1-n\omega_2)^2}- \frac{1}{\left(m\omega_1+n\omega_2\right)^2} \right\}\)
는 타원함수가 됨.
\(\wp\)의 로랑급수
- 원점에서의 로랑급수는 다음과 같이 주어짐.
\(\wp(z)=z^{-2}+\frac{g_2}{20}z^2+\frac{g_3}{28}z^4+\frac{g_2^2}{1200}z^6+O(z^8)\)
여기서 \(g_2= 60\sum{}' \Omega_{m,n}^{-4},\qquad g_3=140\sum{}' \Omega_{m,n}^{-6}\)
(증명)
\(\zeta(z)=\frac{1}{z}+\sum_{\omega \in \Omega}\frac{z^2}{\omega^2(z-\omega)}=\frac{1}{z}+\sum_{\omega \in \Omega}(\frac{1}{z-\omega}+\frac{1}{\omega}+\frac{z}{\omega}^2) \) 를 정의하자.
\(\wp(z)=-\zeta'(z)=\sum_{\omega\in \Omega} \frac{1}{(z-m)^2}- \frac{1}{\omega^2} \right\}\) 이므로 \(\zeta(z)=\frac{1}{z}+\sum_{\Omega}(\frac{1}{z-\omega}+\frac{1}{\omega}+\frac{z}{\omega}^2)\) 의 로랑급수를 구한 뒤, 미분을 하면 된다.
\(\zeta(z)=\frac{1}{z}+\sum_{\omega \in \Omega}\frac{z^2}{\omega^2(z-\omega)}=\frac{1}{z}+\sum_{\omega \in \Omega}\frac{z^2}{\omega^2}(-\frac{1}{\omega}-\frac{z}{\omega^2}-\frac{z^2}{\omega^3}-\cdots)\)
\(=\frac{1}{z}+\sum_{\omega \in \Omega}(-\frac{z^2}{\omega^3}-\frac{z^3}{\omega^4}-\frac{z^4}{\omega^5}-\cdots)=\frac{1}{z}-G_3z^2-G_4z^3-\cdots=\frac{1}{z}-\sum_{n=2}^{\infty}G_{2n}z^{2n-1}\). 여기서 \(G_{2n}=\sum_{\omega\in \Omega} \frac{1}{\omega^{2n}}\).
따라서 \(\wp(z)=\frac{1}{z^2}-\sum_{n=2}^{\infty}(2n-1)G_{2n}z^{2n-2}\).
간단한 소개
상위 주제
하위페이지
재미있는 사실
역사