A350358 Value of -F(0), where F(x) is the indefinite integral of x^(1/x).
4, 2, 0, 3, 6, 9, 5, 8, 8, 7, 8, 3, 2, 0, 2, 2, 9, 8, 1, 3, 2, 4, 3, 9, 3, 8, 1, 8, 1, 8, 0, 8, 8, 1, 8, 9, 9, 8, 1, 5, 4, 9, 5, 5, 3, 2, 8, 7, 1, 2, 2, 3, 9, 5, 1, 4, 5, 3, 5, 4, 0, 7, 3, 5, 4, 3, 6, 4, 0, 1, 2, 0, 8, 1, 2, 1, 8, 7, 0, 6, 2, 2, 7, 3, 1, 5, 1, 4
Offset: 0
Examples
0.4203695887832022981324...
Links
- Peter Luschny, Table of n, a(n) for n = 0..999
- Jonathan Sondow and Diego Marques, Algebraic and transcendental solutions of some exponential equations, Annales Mathematicae et Informaticae 37 (2010) 151-164; see Figure 5.
- Eric Weisstein's World of Mathematics, Steiner's Problem.
Programs
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Julia
# The chosen parameters give about 100 exact decimal places. using Nemo RR = RealField(1100) function F(b::Int, x::arb) lnx = log(x) s = sum(gamma_regularized(RR(n+2), RR(n)*lnx) * RR(n)^(-n-2) for n in 1:b) x + (lnx * lnx) / RR(2) - s end println( F(1400, RR(0.015)) ) # Peter Luschny, Dec 26 2021 -
Maple
# The chosen parameters give about 100 exact decimal places. Digits := 400: F := proc(b, x) local s, lnx; lnx := log(x); s := add(add((n*lnx)^k / k!, k = 0..n+1) / (n*x)^(n+2), n = 1..b); x - x^2 * s + lnx^2 / 2 end: F(2000, 0.01); # Peter Luschny, Dec 27 2021
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Mathematica
RealDigits[N[Sum[1/n^(n+2), {n, 1, 100}] + Integrate[x^(1/x), {x, 0, 1}] - 1, 110]][[1]] (* Amiram Eldar, Dec 29 2021 *)
Formula
The calculation of F(0) requires some care, because terms in the formula for F(x) diverge for x=0, but converge for all x > 0, although convergence is progressively slower as x approaches zero. To calculate F(0), first choose a desired precision d (absolute error). Then choose any x such that 0 < x^(1/x) < d, and evaluate F(x) as defined above.
Since F(x)-F(0) < x^(1/x) < d, F(0)=F(x) to the desired precision.
For small x, the main summation terms initially increase in absolute value (but with alternating signs), reach a maximum of about 1/d at n = -log(x)/x, then decrease at an accelerating rate and reach a value of around d at n = -3.5911*log(x)/x, at which point the large terms have mostly cancelled out, and further terms are below precision level.
For example, for a final precision of d = 10^-50, the calculation must allow for intermediate terms and sums as large as 1/d = 10^50, so these terms must be evaluated to at least 100 digits. For 50 digits, F(.03) is a suitable choice, because .03^(1/.03) = 1.7273...*10^-51 < 10^-50. A few extra digits should also be allowed for round off error.
Extensions
More terms from Hugo Pfoertner, Dec 26 2021
Comments