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A073229
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Decimal expansion of e^(1/e).
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32
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1, 4, 4, 4, 6, 6, 7, 8, 6, 1, 0, 0, 9, 7, 6, 6, 1, 3, 3, 6, 5, 8, 3, 3, 9, 1, 0, 8, 5, 9, 6, 4, 3, 0, 2, 2, 3, 0, 5, 8, 5, 9, 5, 4, 5, 3, 2, 4, 2, 2, 5, 3, 1, 6, 5, 8, 2, 0, 5, 2, 2, 6, 6, 4, 3, 0, 3, 8, 5, 4, 9, 3, 7, 7, 1, 8, 6, 1, 4, 5, 0, 5, 5, 7, 3, 5, 8, 2, 9, 2, 3, 0, 4, 7, 0, 9, 8, 8, 5, 1, 1, 4, 2, 9, 5
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OFFSET
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1,2
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COMMENTS
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e^(1/e) = 1/((1/e)^(1/e)) (reciprocal of A072364).
Let w(n+1)=A^w(n); then w(n) converges if and only if (1/e)^e <= A <= e^(1/e) (see the comments in A073230) for initial value w(1)=A. If A=e^(1/e) then lim_{n->infinity} w(n) = e. - Benoit Cloitre, Aug 06 2002; corrected by Robert FERREOL, Jun 12 2015
x^(1/x) is maximum for x = e and the maximum value is e^(1/e). This gives an interesting and direct proof that 2 < e < 4 as 2^(1/2) < e^(1/e) > 4^(1/4) while 2^(1/2) = 4^(1/4). - Amarnath Murthy, Nov 26 2002
Value of the unique base b > 0 for which the exponential curve y=b^x and its inverse y=log_b(x) kiss each other; the kissing point is (e,e). - Stanislav Sykora, May 25 2015
Actually, there is another base with such property, b=(1/e)^e with kiss point (1/e,1/e). - Yuval Paz, Dec 29 2018
The problem of finding the maximum of f(x) = x^(1/x) was posed and solved by the Swiss mathematician Jakob Steiner (1796-1863) in 1850. - Amiram Eldar, Jun 17 2021
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LINKS
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Ovidiu Furdui, Problem 11982, The American Mathematical Monthly, Vol. 124, No. 5 (2017), p. 465; A Limit of a Power of a Sum, Solution to Problem 11982 by Roberto Tauraso, ibid., Vol. 126, No. 2 (2019), p. 187.
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FORMULA
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Equals 1 + Integral_{x = 1/e..1} (1 + log(x))/x^x dx = 1 - Integral_{x = 0..1/e} (1 + log(x))/x^x dx. - Peter Bala, Oct 30 2019
Equals lim_{x->oo} (Sum_{n>=1} (x/n)^n)^(1/x) (Furdui, 2017). - Amiram Eldar, Mar 26 2022
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EXAMPLE
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1.44466786100976613365833910859...
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MAPLE
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MATHEMATICA
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RealDigits[ E^(1/E), 10, 110] [[1]]
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PROG
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(PARI) exp(1)^exp(-1)
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CROSSREFS
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KEYWORD
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AUTHOR
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STATUS
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approved
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