"Nested radicals"의 두 판 사이의 차이
Pythagoras0 (토론 | 기여) |
Pythagoras0 (토론 | 기여) |
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| 71번째 줄: | 71번째 줄: | ||
* https://docs.google.com/file/d/0B8XXo8Tve1cxU1hvM09SaThwN0E/edit | * https://docs.google.com/file/d/0B8XXo8Tve1cxU1hvM09SaThwN0E/edit | ||
* http://mathematica.stackexchange.com/questions/100591/how-to-evaluate-the-limit-of-a-function-consists-of-range | * http://mathematica.stackexchange.com/questions/100591/how-to-evaluate-the-limit-of-a-function-consists-of-range | ||
| − | + | * http://oeis.org/A072449 | |
==관련도서== | ==관련도서== | ||
2015년 11월 27일 (금) 21:41 판
개요
- 황금비\[\sqrt{1+\sqrt{1+\sqrt{1+\sqrt{1+\sqrt{1+\cdots}}}}}=\varphi=\frac{1+\sqrt5}{2}=1.61803398874989\cdots\]
- 비에타의 공식\[\frac{2}{\pi}=\frac{\sqrt{2}}{2}\frac{\sqrt{2+\sqrt{2}}}{2} \frac{\sqrt{2+\sqrt{2+\sqrt{2}}}}{2} \frac{\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2}}}}}{2} \frac{\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2}}}}}}{2} \frac{\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2}}}}}}}{2}\cdots\]
- nested radical 상수\[\sqrt{1+\sqrt{2+\sqrt{3+\sqrt{4+\sqrt{5+\sqrt{6+\cdots}}}}}}=1.75793275661800453270881963821820816125\cdots\]
- 삼각함수의 값\[\cos \frac{\pi}{32}=\cos\frac{\pi}{2^5}= \frac{\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2}}}}}{2}\]\[\cos \frac{\pi}{64}=\cos\frac{\pi}{2^6}= \frac{\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2}}}}}}{2}\]
라마누잔이 제시한 문제
- 다음 수열의 극한
\[1,\sqrt{1+2 },\sqrt{1+2 \sqrt{1+3 }},\sqrt{1+2 \sqrt{1+3 \sqrt{1+4 }}},\sqrt{1+2 \sqrt{1+3 \sqrt{1+4 \sqrt{1+5 }}}},\sqrt{1+2 \sqrt{1+3 \sqrt{1+4 \sqrt{1+5 \sqrt{1+6 }}}}}, \cdots\]
- 정리
\(\sqrt{1+2\sqrt{1+3\sqrt{1+4\sqrt{1+5\sqrt{1+6\cdots}}}}} = 3\)
수열의 크기 변화
\(1,\sqrt{1+2 },\sqrt{1+2 \sqrt{1+3 }},\sqrt{1+2 \sqrt{1+3 \sqrt{1+4 }}},\sqrt{1+2 \sqrt{1+3 \sqrt{1+4 \sqrt{1+5 }}}},\sqrt{1+2 \sqrt{1+3 \sqrt{1+4 \sqrt{1+5 \sqrt{1+6 }}}}}, \cdots\)
함수방정식
- $f(x)=\sqrt{1+x \sqrt{1+(x+1) \sqrt{1+(x+2) \sqrt{\cdots}}}}$
- $[f(x)]^2=1+xf(x+1), f(x)\ge 0$
- $f(x)=x+1$
- Functional Equations and and How to Solve Them, Section 3.8 Functional equations and nested radicals
- 증명
먼저 수렴성을 증명하자. 다음과 같이 정의된 수열
\(1,\sqrt{1+2 },\sqrt{1+2 \sqrt{1+3 }},\sqrt{1+2 \sqrt{1+3 \sqrt{1+4 }}},\sqrt{1+2 \sqrt{1+3 \sqrt{1+4 \sqrt{1+5 }}}},\sqrt{1+2 \sqrt{1+3 \sqrt{1+4 \sqrt{1+5 \sqrt{1+6 }}}}}, \cdots\) 은 위로 유계이다.
\(\sqrt{1+2 \sqrt{1+3\sqrt{1+\cdots+ (n-1)\sqrt{1+n} }}} \leq \sqrt{1+2 \sqrt{1+3\sqrt{1+\cdots+ (n-1)\sqrt{1+n(n+2)} }}}=3\)
\(n=\sqrt{1+(n-1)(n+1)}\)을 이용
\(\begin{eqnarray*}3 &=& \sqrt{1+2\cdot4}\\ &=& \sqrt{1+2\sqrt{1+3\cdot5}}\\ &=& \sqrt{1+2\sqrt{1+3\sqrt{1+4\cdot6}}}\\ &=& \cdots\end{eqnarray*}\)
메모
- http://www.dgp.toronto.edu/~mjmcguff/math/nestedRadicals.pdf
- http://fluxionsdividebyzero.com/p1/math/calculus/number/cr/sr_nroots.pdf
- http://math.stackexchange.com/questions/435778/finding-the-value-of-sqrt12-sqrt23-sqrt34-sqrt45-sqrt5-dots
관련된 항목들
매스매티카 파일 및 계산 리소스
- https://docs.google.com/file/d/0B8XXo8Tve1cxU1hvM09SaThwN0E/edit
- http://mathematica.stackexchange.com/questions/100591/how-to-evaluate-the-limit-of-a-function-consists-of-range
- http://oeis.org/A072449
관련도서
- https://books.google.com.au/books?id=TT1T8A94xNcC&pg=PA221&redir_esc=y#v=onepage&q&f=false
- Ramanujan, S. Collected Papers of Srinivasa Ramanujan (Ed. G. H. Hardy, P. V. S. Aiyar, and B. M. Wilson). Providence, RI: Amer. Math. Soc., 2000.
- Functional Equations and and How to Solve Them
- section 3.8
관련논문
- Campbell, Geoffrey B., and Aleksander Zujev. “Variations on Ramanujan’s Nested Radicals.” arXiv:1511.06865 [math], November 21, 2015. http://arxiv.org/abs/1511.06865.
- Herschfeld, Aaron. 1935. “On Infinite Radicals.” The American Mathematical Monthly 42 (7) (August 1): 419–429. doi:http://dx.doi.org/10.2307/2301294.
- Ramanujan, S. Question No. 298. J. Indian Math. Soc. 1911.
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