A175999 -id:A175999 - cp's OEIS frontend

A351885 Decimal expansion of lim_{n -> infinity} (Sum_{x=1..n} x^(1/x) - Integral_{k=0..n} x^(1/x) dx).

5, 6, 8, 1, 8, 0, 0, 1, 2, 3, 5, 9, 0, 6, 6, 4, 5, 2, 5, 1, 2, 3, 1, 4, 7, 2, 6, 5, 2, 1, 8, 8, 3, 0, 7, 4, 4, 4, 0, 4, 4, 9, 1, 3, 0, 5, 1, 4, 4, 0, 1, 4, 8, 6, 5, 9, 0, 0, 7, 6, 6, 3, 3, 2, 5, 1, 5, 8, 3, 4, 2, 7, 6, 8, 0, 7, 3, 5, 1, 0, 0, 4, 2, 2, 1, 7, 5

Offset: 0

Daniel Hoyt, Feb 23 2022

The limiting difference between the integral and sum of x^(1/x). The limit converges slowly.

			0.5681800123590664525123147265218830744...

Daniel Hoyt, Graphical representation of the limit.
Daniel Hoyt, Computing the limiting difference between the sum and integral of x^(1/x).
Wikipedia, Stieltjes Constants

Cf. A350862, A329117, A175999, A001620.

# Gives 15 correct digits
from mpmath import stieltjes,fac,quad
def limgen(n):
    terms = []
    for y in range(3, n):
        for x in range(y, n):
            terms.append((((-1)**y)*stieltjes(x)*(x-(y-1))**(y-2))/(fac(x-(y-2))*fac(y-2)))
    return terms
f = lambda x: x**(1/x)
int01 = quad(f, [0,1])
limit = sum(limgen(60)) + 1.5 - stieltjes(0) - int01
print(limit)

Equals 3/2 - A001620 - A175999 + Sum_{k>=3} Sum_{n>=k} (((-1)^k)*Stieltjes(n)*(n-k+1)^(k-2))/((n-k+2)!*(k-2)!).

A350358 Value of -F(0), where F(x) is the indefinite integral of x^(1/x).

Original entry on oeis.org

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

Robert B Fowler, Dec 26 2021

The indefinite integral of x^(1/x) can be derived by expanding
    x^(1/x) = exp(log(x)/x) = Sum_{n>=0} (log(x)/x)^n/n!,
then integrating term-by-term to get the double summation
    F(x) = x + (1/2)*(log(x))^2
             - (Sum_{n>=2} (n-1)^(-n-1) * x^(-n+1)/n!
             * (Sum_{k=0..n} A350297(n,k)*(log(x))^k)).
We assume the constant of integration in F(x) is zero.
To compute definite integrals of x^(1/x) for ranges of x starting at x=0 or x=1, we would need the values
    F(0) = lim {x->0} F(x) = -0.4203695887832022981324...
    F(1) = 1 - Sum_{n>=2} (n-1)^(-n-1) = -0.06687278808178024266...
Note that the definite Integral_{t=0..1} x^(1/x) = F(1)-F(0) = A175999.
The calculation of F(0) requires some care. See the FORMULA below.
Since x^(1/x) is the inverse of the infinite exponentiation function E(y) = y^(y^(y^(...))) = x, the definite integrals of these two functions are related by
    (Integral_{t=0..y} E(t) dt) + (Integral_{t=0..x} t^(1/t) dt) = x*y.
Note that the repeated exponentiation in E(y) converges for 0 <= y < e^(1/e), but diverges for y > e^(1/e).

			0.4203695887832022981324...

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.

Cf. A350297, A175999, A176000, A073229.

# 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

# 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

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 *)

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.

More terms from Hugo Pfoertner, Dec 26 2021

cp's OEIS Frontend

A176000 Decimal expansion of 1 - A175999.

Views

Author

Keywords

Comments

Crossrefs

A351885 Decimal expansion of lim_{n -> infinity} (Sum_{x=1..n} x^(1/x) - Integral_{k=0..n} x^(1/x) dx).

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Python

Formula

A350358 Value of -F(0), where F(x) is the indefinite integral of x^(1/x).

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Julia

Maple

Mathematica

Formula

Extensions