"숫자 163"의 두 판 사이의 차이

2020년 11월 1일 (일) 04:33 판

개요

\(e^{\pi \sqrt{163}}\)는 정수에 매우 가깝다

\[e^{\pi \sqrt{163}}=262537412640768743.9999999999992500725\cdots\approx 262537412640768744 \label{ex163}\]

\ref{ex163}으로부터 다음을 얻는다

\[\sqrt[3]{e^{\sqrt{163} \pi }-744}=640319.999999999999999999999999390317352\cdots \label{e163} \]

\ref{e163}으로부터 다음을 얻는다

\[ \left(\frac{\log \left(640320^3+744\right)}{\pi }\right)^2=163.0000000000000000000000000000232\cdots \]

\(e^{\pi \sqrt{43}} = 884736743.9997774660349066619374620785\approx 884736744\)
\(e^{\pi \sqrt{67}} = 147197952743.9999986624542245068292613\approx 147197952744\)
이 숫자들은 정수에 매우 가까우며, 셋 모두 끝 세 자리가 744
\(\sqrt[3]{e^{\sqrt{67} \pi }-744}=5279.999999999999984007382352249\cdots\)
\(\sqrt[3]{e^{\sqrt{43} \pi }-744}=959.99999999991951173\)

complex multiplication

타원곡선

j-invariant

j-invariant 항목을 참조

재미있는 사실

라마누잔은 \(e^{\pi \sqrt{163}}=262537412640768743.99999999999925\cdots\) 와 같은 계산을 많이 남겼음
이와 유사한 공식들을 \(\pi\) 의 근사공식에 사용. 라마누잔과 파이 항목을 참조
In his Fields Medallists' lecture, Richard Borcherds said that every mathematician should see once in his/her life why this should be the case (citation needed)
\(x^2+x+41\)는 정수 \(-40\leq x\leq 39\) 에 대하여, 모두 소수가 된다
겔폰드-슈나이더 정리 를 사용하면, \(e^{\pi \sqrt{163}}=(e^{-i\pi})^{\sqrt{-163}}=(-1)^{\sqrt{-163}}\) 이므로 초월수임을 알 수 있다
- http://mathdl.maa.org/images/upload_library/22/Ford/Davis311-320.pdf
마틴 가드너의 만우절 칼럼

\[Ellipsis]when the transcendental number e is raised to the power of \[Pi] times \[Sqrt]163, the result is an integer. The Indian mathematician Srinivasa Ramanujan had conjectured that e to the power of \[Pi]\[Sqrt]163 is integral in a note in the Quarterly Journal of Pure and Applied Mathematics (vol. 45, 1913-1914, p. 350). Working by hand, he found the value to be 262,537,412,640,768,743.999,999,999,999,\[Ellipsis]. The calculations were tedious, and he was unable to verify the next decimal digit. Modern computers extended the 9's much farther; indeed, a French program of 1972 went as far as two million 9's. Unfortunately, no one was able to prove that the sequence of 9's continues forever (which, of course, would make the number integral) or whether the number is irrational or an integral fraction.

In May 1974 John Brillo of the University of Arizona found an ingenious way of applying Euler's constant to the calculation and managed to prove that the number exactly equals 262,537,412,640,768,744. How the prime number 163 manages to convert the expression to an integer is not yet fully understood.

http://math-frolic.blogspot.kr/2014/04/the-joy-of-number-theory.html

매스매티카 파일 및 계산 리소스

https://docs.google.com/file/d/0B8XXo8Tve1cxU2lXWVJGdy1ldTg/edit

사전형태의 참고자료

리뷰논문, 에세이, 강의노트

B.J.Green, The Ramanujan Constant. An Essay on Elliptic Curves, Complex. Multiplication and Modular Forms
I.J. Good, What Is the Most Amazing Approximate Integer in the Universe?, Pi Mu Epsilon Journal, Vol. 5, Fall 1972, No. 7, pgs. 314-15

블로그

피타고라스의 창

@@ 1번째 줄: / 1번째 줄: @@
 ==개요==
-* $e^{\pi \sqrt{163}}$는 정수에 매우 가깝다
+* <math>e^{\pi \sqrt{163}}</math>는 정수에 매우 가깝다
 :<math>e^{\pi \sqrt{163}}=262537412640768743.9999999999992500725\cdots\approx 262537412640768744 \label{ex163}</math>
 * \ref{ex163}으로부터 다음을 얻는다
-$$\sqrt[3]{e^{\sqrt{163} \pi }-744}=640319.999999999999999999999999390317352\cdots \label{e163} $$
+:<math>\sqrt[3]{e^{\sqrt{163} \pi }-744}=640319.999999999999999999999999390317352\cdots \label{e163} </math>
 * \ref{e163}으로부터 다음을 얻는다
-$$
+:<math>
 \left(\frac{\log \left(640320^3+744\right)}{\pi }\right)^2=163.0000000000000000000000000000232\cdots
-$$
+</math>
 * <math>e^{\pi \sqrt{43}} = 884736743.9997774660349066619374620785\approx 884736744</math>
 * <math>e^{\pi \sqrt{67}} = 147197952743.9999986624542245068292613\approx 147197952744</math>
 * 이 숫자들은 정수에 매우 가까우며, 셋 모두 끝 세 자리가 744
-* $\sqrt[3]{e^{\sqrt{67} \pi }-744}=5279.999999999999984007382352249\cdots$
+* <math>\sqrt[3]{e^{\sqrt{67} \pi }-744}=5279.999999999999984007382352249\cdots</math>
-* $\sqrt[3]{e^{\sqrt{43} \pi }-744}=959.99999999991951173$
+* <math>\sqrt[3]{e^{\sqrt{43} \pi }-744}=959.99999999991951173</math>
@@ 42번째 줄: / 42번째 줄: @@
 * 마틴 가드너의 만우절 칼럼
 <blockquote>
-…when the transcendental number e is raised to the power of π times √163, the result is an integer. The Indian mathematician Srinivasa Ramanujan had conjectured that e to the power of π√163 is integral in a note in the Quarterly Journal of Pure and Applied Mathematics (vol. 45, 1913-1914, p. 350). Working by hand, he found the value to be  262,537,412,640,768,743.999,999,999,999,…. The calculations were tedious, and he was unable to verify the next decimal digit. Modern computers extended the 9's much farther; indeed, a French program of 1972 went as far as two million 9's. Unfortunately, no one was able to prove that the sequence of 9's continues forever (which, of course, would make the number integral) or whether the number is irrational or an integral fraction.
+\[Ellipsis]when the transcendental number e is raised to the power of \[Pi] times \[Sqrt]163, the result is an integer. The Indian mathematician Srinivasa Ramanujan had conjectured that e to the power of \[Pi]\[Sqrt]163 is integral in a note in the Quarterly Journal of Pure and Applied Mathematics (vol. 45, 1913-1914, p. 350). Working by hand, he found the value to be  262,537,412,640,768,743.999,999,999,999,\[Ellipsis]. The calculations were tedious, and he was unable to verify the next decimal digit. Modern computers extended the 9's much farther; indeed, a French program of 1972 went as far as two million 9's. Unfortunately, no one was able to prove that the sequence of 9's continues forever (which, of course, would make the number integral) or whether the number is irrational or an integral fraction.
 </blockquote>
 <blockquote>
@@ 53번째 줄: / 53번째 줄: @@
 ==관련된 항목들==
-* [[오일러의 소수생성다항식 x²+x+41]]
+* [[오일러의 소수생성다항식 x\.b2+x+41]]
 * [[가우스의 class number one 문제]]
 * [[라마누잔과 파이]]