|
%I #348 Dec 06 2023 08:34:23
%S 1,6,4,4,9,3,4,0,6,6,8,4,8,2,2,6,4,3,6,4,7,2,4,1,5,1,6,6,6,4,6,0,2,5,
%T 1,8,9,2,1,8,9,4,9,9,0,1,2,0,6,7,9,8,4,3,7,7,3,5,5,5,8,2,2,9,3,7,0,0,
%U 0,7,4,7,0,4,0,3,2,0,0,8,7,3,8,3,3,6,2,8,9,0,0,6,1,9,7,5,8,7,0
%N Decimal expansion of Pi^2/6 = zeta(2) = Sum_{m>=1} 1/m^2.
%C "In 1736 he [Leonard Euler, 1707-1783] discovered the limit to the infinite series, Sum 1/n^2. He did it by doing some rather ingenious mathematics using trigonometric functions that proved the series summed to exactly Pi^2/6. How can this be? ... This demonstrates one of the most startling characteristics of mathematics - the interconnectedness of, seemingly, unrelated ideas." - Clawson [See Hardy and Wright, Theorems 332 and 333. - _N. J. A. Sloane_, Jan 20 2017]
%C Also dilogarithm(1). - _Rick L. Shepherd_, Jul 21 2004
%C Also Integral_{x>=0} x/(exp(x)-1) dx. [Abramowitz-Stegun, 23.2.7., for s=2, p. 807]
%C For the partial sums see the fractional sequence A007406/A007407.
%C Pi^2/6 is also the length of the circumference of a circle whose diameter equals the ratio of volume of an ellipsoid to the circumscribed cuboid. Pi^2/6 is also the length of the circumference of a circle whose diameter equals the ratio of surface area of a sphere to the circumscribed cube. - _Omar E. Pol_, Oct 07 2011
%C 1 < n^2/(eulerphi(n)*sigma(n)) < zeta(2) for n > 1. - _Arkadiusz Wesolowski_, Sep 04 2012
%C Volume of a sphere inscribed in a cube of volume Pi. More generally, Pi^x/6 is the volume of an ellipsoid inscribed in a cuboid of volume Pi^(x-1). - _Omar E. Pol_, Feb 17 2016
%C Surface area of a sphere inscribed in a cube of surface area Pi. More generally, Pi^x/6 is the surface area of a sphere inscribed in a cube of surface area Pi^(x-1). - _Omar E. Pol_, Feb 19 2016
%C zeta(2)+1 is a weighted average of the integers, n > 2, using zeta(n)-1 as the weights for each n. We have: Sum_{n >= 2} (zeta(n)-1) = 1 and Sum_{n >= 2} n*(zeta(n)-1) = zeta(2)+1. - _Richard R. Forberg_, Jul 14 2016
%C zeta(2) is the expected value of sigma(n)/n. - _Charlie Neder_, Oct 22 2018
%C Graham shows that a rational number x can be expressed as a finite sum of reciprocals of distinct squares if and only if x is in [0, Pi^2/6-1) U [1, Pi^2/6). See section 4 for other results and Theorem 5 for the underlying principle. - _Charles R Greathouse IV_, Aug 04 2020
%D M. Abramowitz and I. A. Stegun, eds., Handbook of Mathematical Functions, National Bureau of Standards Applied Math. Series 55, 1964 (and various reprintings), p. 811.
%D F. Aubonnet, D. Guinin and B. Joppin, Précis de Mathématiques, Analyse 2, Classes Préparatoires, Premier Cycle Universitaire, Bréal, 1990, Exercice 908, pages 82 and 91-92.
%D Calvin C. Clawson, Mathematical Mysteries, The Beauty and Magic of Numbers, Perseus Books, 1996, p. 97.
%D W. Dunham, Euler: The Master of Us All, The Mathematical Association of America, Washington, D.C., 1999, p. xxii.
%D Hardy and Wright, 'An Introduction to the Theory of Numbers'. See Theorems 332 and 333.
%D A. A. Markoff, Mémoire sur la transformation de séries peu convergentes en séries très convergentes, Mém. de l'Acad. Imp. Sci. de St. Pétersbourg, XXXVII, 1890.
%D G. F. Simmons, Calculus Gems, Section B.15, B.24, pp. 270-271, 323-325, McGraw Hill, 1992.
%D Arnold Walfisz, Weylsche Exponentialsummen in der neueren Zahlentheorie, Deutscher Verlag der Wissenschaften, Berlin, 1963, p. 99, Satz 1.
%D A. Weil, Number theory: an approach through history; from Hammurapi to Legendre, Birkhäuser, Boston, 1984; see p. 261.
%D David Wells, The Penguin Dictionary of Curious and Interesting Numbers, Revised Edition, Penguin Books, London, England, 1997, page 23.
%H Harry J. Smith, <a href="/A013661/b013661.txt">Table of n, a(n) for n = 1..20000</a>
%H M. Abramowitz and I. A. Stegun, eds., <a href="http://www.convertit.com/Go/ConvertIt/Reference/AMS55.ASP">Handbook of Mathematical Functions</a>, National Bureau of Standards, Applied Math. Series 55, Tenth Printing, 1972 [alternative scanned copy].
%H D. H. Bailey, J. M. Borwein and D. M. Bradley, <a href="https://arxiv.org/abs/math/0505270">Experimental determination of Apéry-like identities for zeta(4n+2)</a>, arXiv:math/0505270 [math.NT], 2005-2006.
%H Peter Bala, <a href="/A002117/a002117.pdf">New series for old functions</a>
%H Peter Bala, <a href="/A013661/a013661.txt">Formulas for A013661</a>
%H David Benko and John Molokach, <a href="http://www.jstor.org/stable/10.4169/college.math.j.44.3.171">The Basel Problem as a Rearrangement of Series</a>, The College Mathematics Journal, Vol. 44, No. 3 (May 2013), pp. 171-176.
%H R. Calinger, <a href="http://dx.doi.org/10.1006/hmat.1996.0015">Leonard Euler: The First St. Petersburg Years (1727-1741)</a>, Historia Mathematica, Vol. 23, 1996, pp. 121-166.
%H R. Chapman, <a href="http://www.maths.ex.ac.uk/~rjc/etc/zeta2.pdf">Evaluating Zeta(2):14 Proofs to Zeta(2)= (pi)^2/6</a>
%H R. W. Clickery, <a href="https://roywilliams.github.io/play/UPI/coprime.html">Probability of two numbers being coprime</a>
%H Alessio Del Vigna, <a href="https://arxiv.org/abs/2104.01710">On a solution to the Basel problem based on the fundamental theorem of calculus</a>, arXiv:2104.01710 [math.HO], 2021.
%H Leonhard Euler, <a href="https://arxiv.org/abs/math/0506415">On the sums of series of reciprocals</a>, arXiv:math/0506415 [math.HO], 2005-2008.
%H Leonhard Euler, <a href="http://eulerarchive.maa.org/pages/E041.html">De summis serierum reciprocarum</a>, E41.
%H R. L. Graham, <a href="http://www.math.ucsd.edu/~ronspubs/64_01_unit_fractions.pdf">On finite sums of unit fractions</a>, Proceedings of the London Mathematical Society, s3-14 (1964), pp. 193-207. doi:10.1112/plms/s3-14.2.193
%H Michael D. Hirschhorn, <a href="http://dx.doi.org/10.1007/s00283-011-9217-4">A simple proof that zeta(2) = Pi^2/6</a>, The Mathematical Intelligencer 33:3 (2011), pp 81-82.
%H Melissa Larson, <a href="https://www.d.umn.edu/~jgreene/masters_reports/BBP%20Paper%20final.pdf">Verifying and discovering BBP-type formulas</a>, 2008.
%H Alain Lasjaunias and Jean-Paul Tran, <a href="https://arxiv.org/abs/2312.02245">A note on the equality Pi^2/6 = Sum_{n>=1} 1/n^2</a>, arXiv:2312.02245 [math.HO], 2023.
%H Math. Reference Project, <a href="http://www.mathreference.com/lc-z,zeta2.html">The Zeta Function, Zeta(2)</a>
%H Math. Reference Project, <a href="http://www.mathreference.com/lc-z,cop.html">The Zeta Function, Odds That Two Numbers Are Coprime"</a>
%H R. Mestrovic, <a href="http://arxiv.org/abs/1202.3670">Euclid's theorem on the infinitude of primes: a historical survey of its proofs (300 BC--2012) and another new proof</a>, arXiv preprint arXiv:1202.3670 [math.HO], 2012.
%H Jon Perry, <a href="https://web.archive.org/web/20060515222108/http://www.users.globalnet.co.uk/~perry/maths/paradox/paradox.htm">Prime Product Paradox</a>
%H Simon Plouffe, Plouffe's Inverter, <a href="http://www.plouffe.fr/simon/constants/zeta2.txt">Zeta(2) or Pi**2/6 to 100000 digits</a>
%H Simon Plouffe, <a href="http://www.worldwideschool.org/library/books/sci/math/MiscellaneousMathematicalConstants/chap96.html">Zeta(2) or Pi**2/6 to 10000 places</a>
%H Simon Plouffe, <a href="/A293904/a293904_4096.gz">Zeta(2) to Zeta(4096) to 2048 digits each</a> (gzipped file)
%H A. L. Robledo, <a href="https://planetmath.org/ValueOfTheRiemannZetaFunctionAtS2">value of the Riemann zeta function at s=2</a>, PlanetMath.org.
%H E. Sandifer, How Euler Did It, <a href="http://www.maa.org/editorial/euler/How%20Euler%20Did%20It%2002%20Estimating%20the%20Basel%20Problem.pdf">Estimating the Basel Problem</a>
%H E. Sandifer, How Euler Did It, <a href="http://www.maa.org/editorial/euler/How%20Euler%20Did%20It%2005%20Basel%20with%20integrals.pdf">Basel Problem with Integrals</a>
%H C. Tooth, <a href="https://web.archive.org/web/20041029085212/http://www.pisquaredoversix.force9.co.uk:80/">Pi squared over six</a>
%H Eric Weisstein's World of Mathematics, <a href="http://mathworld.wolfram.com/Dilogarithm.html">Dilogarithm</a>.
%H Eric Weisstein's World of Mathematics, <a href="http://mathworld.wolfram.com/RiemannZetaFunctionZeta2.html">Riemann Zeta Function zeta(2)</a>.
%H Wikipedia, <a href="https://en.wikipedia.org/wiki/Basel_problem">Basel Problem</a>
%H Wikipedia, <a href="https://en.wikipedia.org/wiki/Bailey%E2%80%93Borwein%E2%80%93Plouffe_formula">Bailey-Borwein-Plouffe formula</a>.
%H Herbert S. Wilf, <a href="https://doi.org/10.46298/dmtcs.265">Accelerated series for universal constants, by the WZ method</a>, Discrete Mathematics & Theoretical Computer Science, Vol 3, No 4 (1999).
%H <a href="/index/Tra#transcendental">Index entries for transcendental numbers</a>
%H <a href="/index/Z#zeta_function">Index entries for zeta function</a>.
%F Limit_{n->oo} (1/n)*(Sum_{k=1..n} frac((n/k)^(1/2))) = zeta(2) and in general we have lim_{n->oo} (1/n)*(Sum_{k=1..n} frac((n/k)^(1/m))) = zeta(m), m >= 2. - _Yalcin Aktar_, Jul 14 2005
%F Equals Integral_{x=0..1} (log(x)/(x-1)) dx or Integral_{x>=1} (log(x/(x-1))/x) dx. - _Jean-François Alcover_, May 30 2013
%F For s >= 2 (including Complex), zeta(s) = Product_{n >= 1} prime(n)^s/(prime(n)^s - 1). - _Fred Daniel Kline_, Apr 10 2014
%F Also equals 1 + Sum_{n>=0} (-1)^n*StieltjesGamma(n)/n!. - _Jean-François Alcover_, May 07 2014
%F zeta(2) = Sum_{n>=1} ((floor(sqrt(n)) - floor(sqrt(n-1)))/n). - _Mikael Aaltonen_, Jan 10 2015
%F zeta(2) = Sum_{n>=1} (((sqrt(5)-1)/2/sqrt(5))^n/n^2) + Sum_{n>=1} (((sqrt(5)+1)/2/sqrt(5))^n/ n^2) + log((sqrt(5)-1)/2/sqrt(5))log((sqrt(5)+1)/2/sqrt(5)). - _Seiichi Kirikami_, Oct 14 2015
%F The above formula can also be written zeta(2) = dilog(x) + dilog(y) + log(x)*log(y) where x = (1-1/sqrt(5))/2 and y=(1+1/sqrt(5))/2. - _Peter Luschny_, Oct 16 2015
%F zeta(2) = Integral_{x>=0} 1/(1 + e^x^(1/2)) dx, because (1 - 1/2^(s-1))*Gamma[1 + s]*Zeta[s] = Integral_{x>=0} 1/(1 + e^x^(1/s)) dx. After _Jean-François Alcover_ in A002162. - _Mats Granvik_, Sep 12 2016
%F zeta(2) = Product_{n >= 1} (144*n^4)/(144*n^4 - 40*n^2 + 1). - _Fred Daniel Kline_, Oct 29 2016
%F zeta(2) = lim_{n->oo} (1/n) * Sum_{k=1..n} A017665(k)/A017666(k). - _Dimitri Papadopoulos_, May 10 2019 [See the Walfisz reference, and a comment in A284648, citing also the Sándor et al. Handbook. - _Wolfdieter Lang_, Aug 22 2019]
%F Equals Sum_{k>=1} H(k)/(k*(k+1)), where H(k) = A001008(k)/A002805(k) is the k-th harmonic number. - _Amiram Eldar_, Aug 16 2020
%F Equals (8/3)*(1/2)!^4 = (8/3)*Gamma(3/2)^4. - _Gary W. Adamson_, Aug 17 2021
%F Equals ((m+1)/m) * Integral_{x=0..1} log(Sum _{k=0..m} x^k )/x dx, m > 0 (Aubonnet reference). - _Bernard Schott_, Feb 11 2022
%F Equals 1 + Sum_{n>=2} Sum_{i>=n+1} (zeta(i)-1). - _Richard R. Forberg_, Jun 04 2023
%e 1.6449340668482264364724151666460251892189499012067984377355582293700074704032...
%p evalf(Pi^2/6,120); # _Muniru A Asiru_, Oct 25 2018
%p # Calculates an approximation with n exact decimal places (small deviation
%p # in the last digits are possible). Goes back to ideas of A. A. Markoff 1890.
%p zeta2 := proc(n) local q, s, w, v, k; q := 0; s := 0; w := 1; v := 4;
%p for k from 2 by 2 to 7*n/2 do
%p w := w*v/k;
%p q := q + v;
%p v := v + 8;
%p s := s + 1/(w*q);
%p od; 12*s; evalf[n](%) end:
%p zeta2(1000); # _Peter Luschny_, Jun 10 2020
%t RealDigits[N[Pi^2/6, 100]][[1]]
%t RealDigits[Zeta[2],10,120][[1]] (* _Harvey P. Dale_, Jan 08 2021 *)
%o (PARI) default(realprecision, 200); Pi^2/6
%o (PARI) default(realprecision, 200); dilog(1)
%o (PARI) default(realprecision, 200); zeta(2)
%o (PARI) A013661(n)={localprec(n+2); Pi^2/.6\10^n%10} \\ Corrected and improved by _M. F. Hasler_, Apr 20 2021
%o (PARI) default(realprecision, 20080); x=Pi^2/6; for (n=1, 20000, d=floor(x); x=(x-d)*10; write("b013661.txt", n, " ", d)); \\ _Harry J. Smith_, Apr 29 2009
%o (PARI) sumnumrat(1/x^2, 1) \\ _Charles R Greathouse IV_, Jan 20 2022
%o (Maxima) fpprec : 100$ ev(bfloat(zeta(2)))$ bfloat(%); /* _Martin Ettl_, Oct 21 2012 */
%o (Magma) pi:=Pi(RealField(110)); Reverse(Intseq(Floor(10^105*pi^2/6))); // _Vincenzo Librandi_, Oct 13 2015
%o (Python) # Use some guard digits when computing.
%o # BBP formula (3 / 16) P(2, 64, 6, (16, -24, -8, -6, 1, 0)).
%o from decimal import Decimal as dec, getcontext
%o def BBPzeta2(n: int) -> dec:
%o getcontext().prec = n
%o s = dec(0); f = dec(1); g = dec(64)
%o for k in range(int(n * 0.5536546824812272) + 1):
%o sixk = dec(6 * k)
%o s += f * ( dec(16) / (sixk + 1) ** 2 - dec(24) / (sixk + 2) ** 2
%o - dec(8) / (sixk + 3) ** 2 - dec(6) / (sixk + 4) ** 2
%o + dec(1) / (sixk + 5) ** 2 )
%o f /= g
%o return (s * dec(3)) / dec(16)
%o print(BBPzeta2(2000)) # _Peter Luschny_, Nov 01 2023
%Y Cf. A001008 (H(n): numerators), A002805 (denominators), A013679 (continued fraction), A002117 (zeta(3)), A013631 (cont.frac. for zeta(3)), A013680 (cont.frac. for zeta(4)), 1/A059956, A108625, A142995, A142999.
%K cons,nonn,nice
%O 1,2
%A _N. J. A. Sloane_
%E Edited by _N. J. A. Sloane_, Nov 22 2023
| 