"분할수의 근사 공식 (하디-라마누잔-라데마커 공식)"의 두 판 사이의 차이

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<h5>이 항목의 스프링노트 원문주소</h5>
 
<h5>이 항목의 스프링노트 원문주소</h5>
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* [[분할수의 근사 공식 (하디-라마누잔-라데마커 공식)|하디-라마누잔-라데마커 분할수 공식]]<br>
  
 
 
 
 
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<h5 style="line-height: 3.428em; margin-top: 0px; margin-right: 0px; margin-bottom: 0px; margin-left: 0px; color: rgb(34, 61, 103); font-family: 'malgun gothic', dotum, gulim, sans-serif; font-size: 1.166em; background-image: ; background-color: initial; background-position: 0px 100%;">개요</h5>
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<h5 style="margin: 0px; line-height: 3.428em; color: rgb(34, 61, 103); font-family: 'malgun gothic',dotum,gulim,sans-serif; font-size: 1.166em; background-position: 0px 100%;">개요</h5>
  
 
 
 
 
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<h5 style="line-height: 2em; margin-top: 0px; margin-right: 0px; margin-bottom: 0px; margin-left: 0px;">첫번째 항의 크기</h5>
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<h5 style="margin: 0px; line-height: 2em;">첫번째 항의 크기</h5>
  
 
<math>K=\pi\sqrt{\frac{2}{3}</math> 로 두자
 
<math>K=\pi\sqrt{\frac{2}{3}</math> 로 두자
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<math>A_1(n)=1</math>
 
<math>A_1(n)=1</math>
  
<math>\frac{\sinh\left(\pi\sqrt{\frac{2}{3}\left(n-\frac{1}{24}\right)}\right)}{\sqrt{n-\frac{1}{24}}} \approx \frac{e^{K\sqrt{n}}}{2\sqrt{n}}</math>
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<math>\frac{\sinh\left(\pi\sqrt{\frac{2}{3}\left(n-\frac{1}{24}\right)}\right)}{\sqrt{n-\frac{1}{24}}} \approx \frac{e^{K\sqrt{n}}}{2\sqrt{n}}</math> 로부터 <math>\frac{d}{dn}\left(\frac{\sinh\left(\frac{\pi}{k}\sqrt{\frac{2}{3}\left(n-\frac{1}{24}\right)}\right)}{\sqrt{n-\frac{1}{24}}}\right)\approx \frac{Ke^{K\sqrt{n}}}{4n}</math>
 
 
 
 
 
 
 
 
  
 
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따라서 
 
 
 
 
  
<math>p(n) \approx \frac {e^{\pi\sqrt{\frac{2n}{3}}}} {4\sqrt{3}n}</math>의 유도
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<math>p(n) \approx \frac{1}{\pi\sqrt{2}}\frac{Ke^{K\sqrt{n}}}{4n}=\frac {e^{\pi\sqrt{\frac{2n}{3}}}} {4\sqrt{3}n}</math>
  
 
 
 
 
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<h5 style="margin: 0px; line-height: 3.428em; color: rgb(34, 61, 103); font-family: 'malgun gothic',dotum,gulim,sans-serif; font-size: 1.166em; background-position: 0px 100%;">재미있는 사실</h5>
  
 
 
 
 
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* [[수학사연표 (역사)|수학사연표]]
 
* [[수학사연표 (역사)|수학사연표]]
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* http://www.google.com/dictionary?langpair=en|ko&q=
 
* http://www.google.com/dictionary?langpair=en|ko&q=
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<h5 style="line-height: 3.428em; margin-top: 0px; margin-right: 0px; margin-bottom: 0px; margin-left: 0px; color: rgb(34, 61, 103); font-family: 'malgun gothic', dotum, gulim, sans-serif; font-size: 1.166em; background-image: ; background-color: initial; background-position: 0px 100%;">사전 형태의 자료</h5>
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<h5 style="margin: 0px; line-height: 3.428em; color: rgb(34, 61, 103); font-family: 'malgun gothic',dotum,gulim,sans-serif; font-size: 1.166em; background-position: 0px 100%;">사전 형태의 자료</h5>
  
 
* http://ko.wikipedia.org/wiki/
 
* http://ko.wikipedia.org/wiki/
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* http://www.wolframalpha.com/input/?i=
 
* http://www.wolframalpha.com/input/?i=
 
* [http://dlmf.nist.gov/ NIST Digital Library of Mathematical Functions]
 
* [http://dlmf.nist.gov/ NIST Digital Library of Mathematical Functions]
* [http://www.research.att.com/~njas/sequences/index.html The On-Line Encyclopedia of Integer Sequences]<br>
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* [http://www.research.att.com/%7Enjas/sequences/index.html The On-Line Encyclopedia of Integer Sequences]<br>
 
** http://www.research.att.com/~njas/sequences/?q=
 
** http://www.research.att.com/~njas/sequences/?q=
  
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* http://www.jstor.org/action/doBasicSearch?Query=
 
* http://www.jstor.org/action/doBasicSearch?Query=
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*  도서내검색<br>
 
*  도서내검색<br>
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*  네이버 뉴스 검색 (키워드 수정)<br>
 
*  네이버 뉴스 검색 (키워드 수정)<br>
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*  구글 블로그 검색<br>
 
*  구글 블로그 검색<br>

2010년 3월 13일 (토) 18:20 판

이 항목의 스프링노트 원문주소

 

 

개요

 

  • 분할수의 근사공식
    \(p(n)=\frac{1}{\pi\sqrt{2}}\sum_{k=1}^\infty A_k(n) \sqrt{k}\frac{d}{dn}\left(\frac{\sinh\left(\frac{\pi}{k}\sqrt{\frac{2}{3}\left(n-\frac{1}{24}\right)}\right)}{\sqrt{n-\frac{1}{24}}}\right)\)
    여기서 \(A_k(n)=\sum_{0 \leq h < k,(h,k)=1}e^{\pi i s(h,k)-2\pi i n \frac{h}{k}\)이고 \(s(h,k)\)는 데데킨트 합

 

 

 

첫번째 항의 크기

\(K=\pi\sqrt{\frac{2}{3}\) 로 두자

\(A_1(n)=1\)

\(\frac{\sinh\left(\pi\sqrt{\frac{2}{3}\left(n-\frac{1}{24}\right)}\right)}{\sqrt{n-\frac{1}{24}}} \approx \frac{e^{K\sqrt{n}}}{2\sqrt{n}}\) 로부터 \(\frac{d}{dn}\left(\frac{\sinh\left(\frac{\pi}{k}\sqrt{\frac{2}{3}\left(n-\frac{1}{24}\right)}\right)}{\sqrt{n-\frac{1}{24}}}\right)\approx \frac{Ke^{K\sqrt{n}}}{4n}\)

따라서 

\(p(n) \approx \frac{1}{\pi\sqrt{2}}\frac{Ke^{K\sqrt{n}}}{4n}=\frac {e^{\pi\sqrt{\frac{2n}{3}}}} {4\sqrt{3}n}\)

 

 

재미있는 사실

 

 

 

역사

 

 

메모

 

 

관련된 항목들

 

 

수학용어번역

 

 

사전 형태의 자료

 

 

관련논문

 

관련도서 및 추천도서

 

 

관련기사

 

 

블로그