Nested radicals

개요

• 황금비:<math>\sqrt{1+\sqrt{1+\sqrt{1+\sqrt{1+\sqrt{1+\cdots}}}}}=\varphi=\frac{1+\sqrt5}{2}=1.61803398874989\cdots</math>
• 비에타의 공식:<math>\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</math>
• nested radical 상수:<math>\sqrt{1+\sqrt{2+\sqrt{3+\sqrt{4+\sqrt{5+\sqrt{6+\cdots}}}}}}=1.75793275661800453270881963821820816125\cdots</math>
• 삼각함수의 값:<math>\cos \frac{\pi}{32}=\cos\frac{\pi}{2^5}= \frac{\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2}}}}}{2}</math>:<math>\cos \frac{\pi}{64}=\cos\frac{\pi}{2^6}= \frac{\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2}}}}}}{2}</math>

라마누잔이 제시한 문제

• 다음 수열의 극한

<math>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</math>

정리

<math>\sqrt{1+2\sqrt{1+3\sqrt{1+4\sqrt{1+5\sqrt{1+6\cdots}}}}} = 3</math>

수열의 크기 변화

<math>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</math>

함수방정식

• <math>f(x)=\sqrt{1+x \sqrt{1+(x+1) \sqrt{1+(x+2) \sqrt{\cdots}}}}</math>
• <math>[f(x)]^2=1+xf(x+1), f(x)\ge 0</math>
• <math>f(x)=x+1</math>
• Functional Equations and and How to Solve Them, Section 3.8 Functional equations and nested radicals

증명

먼저 수렴성을 증명하자. 다음과 같이 정의된 수열

<math>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</math> 은 위로 유계이다.

<math>\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</math>

<math>n=\sqrt{1+(n-1)(n+1)}</math>을 이용

<math>\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*}</math>

메모

• http://math.stackexchange.com/questions/435778/finding-the-value-of-sqrt12-sqrt23-sqrt34-sqrt45-sqrt5-dots
• http://www.dgp.toronto.edu/~mjmcguff/math/nestedRadicals.pdf
• http://fluxionsdividebyzero.com/p1/math/calculus/number/cr/sr_nroots.pdf

관련된 항목들

• 연분수

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.

사전형태의 참고자료

• http://en.wikipedia.org/wiki/Nested_radical
• http://mathworld.wolfram.com/NestedRadical.html

블로그

• New Start, Ens!, 2009-1-16 Ramanujan's infinitely nested radicals problem

메타데이터

위키데이터

• ID : Q2670069

Spacy 패턴 목록

• [{'LOWER': 'nested'}, {'LEMMA': 'radical'}]