Английская Википедия:Harmonic number
Шаблон:Short description Шаблон:Other uses Шаблон:Use American English
In mathematics, the Шаблон:Mvar-th harmonic number is the sum of the reciprocals of the first Шаблон:Mvar natural numbers: <math display="block">H_n= 1+\frac{1}{2}+\frac{1}{3}+\cdots+\frac{1}{n} =\sum_{k=1}^n \frac{1}{k}.</math>
Starting from Шаблон:Math, the sequence of harmonic numbers begins: <math display="block">1, \frac{3}{2}, \frac{11}{6}, \frac{25}{12}, \frac{137}{60}, \dots</math>
Harmonic numbers are related to the harmonic mean in that the Шаблон:Mvar-th harmonic number is also Шаблон:Mvar times the reciprocal of the harmonic mean of the first Шаблон:Mvar positive integers.
Harmonic numbers have been studied since antiquity and are important in various branches of number theory. They are sometimes loosely termed harmonic series, are closely related to the Riemann zeta function, and appear in the expressions of various special functions.
The harmonic numbers roughly approximate the natural logarithm function[1]Шаблон:Rp and thus the associated harmonic series grows without limit, albeit slowly. In 1737, Leonhard Euler used the divergence of the harmonic series to provide a new proof of the infinity of prime numbers. His work was extended into the complex plane by Bernhard Riemann in 1859, leading directly to the celebrated Riemann hypothesis about the distribution of prime numbers.
When the value of a large quantity of items has a Zipf's law distribution, the total value of the Шаблон:Mvar most-valuable items is proportional to the Шаблон:Mvar-th harmonic number. This leads to a variety of surprising conclusions regarding the long tail and the theory of network value.
The Bertrand-Chebyshev theorem implies that, except for the case Шаблон:Math, the harmonic numbers are never integers.[2]
n | Harmonic number, Hn | |||
---|---|---|---|---|
expressed as a fraction | decimal | relative size | ||
1 | 1 | Шаблон:Bartable | ||
2 | 3 | /2 | Шаблон:Bartable | |
3 | 11 | /6 | ~Шаблон:Bartable | |
4 | 25 | /12 | ~Шаблон:Bartable | |
5 | 137 | /60 | ~Шаблон:Bartable | |
6 | 49 | /20 | Шаблон:Bartable | |
7 | 363 | /140 | ~Шаблон:Bartable | |
8 | 761 | /280 | ~Шаблон:Bartable | |
9 | 7 129 | /2 520 | ~Шаблон:Bartable | |
10 | 7 381 | /2 520 | ~Шаблон:Bartable | |
11 | 83 711 | /27 720 | ~Шаблон:Bartable | |
12 | 86 021 | /27 720 | ~Шаблон:Bartable | |
13 | 1 145 993 | /360 360 | ~Шаблон:Bartable | |
14 | 1 171 733 | /360 360 | ~Шаблон:Bartable | |
15 | 1 195 757 | /360 360 | ~Шаблон:Bartable | |
16 | 2 436 559 | /720 720 | ~Шаблон:Bartable | |
17 | 42 142 223 | /12 252 240 | ~Шаблон:Bartable | |
18 | 14 274 301 | /4 084 080 | ~Шаблон:Bartable | |
19 | 275 295 799 | /77 597 520 | ~Шаблон:Bartable | |
20 | 55 835 135 | /15 519 504 | ~Шаблон:Bartable | |
21 | 18 858 053 | /5 173 168 | ~Шаблон:Bartable | |
22 | 19 093 197 | /5 173 168 | ~Шаблон:Bartable | |
23 | 444 316 699 | /118 982 864 | ~Шаблон:Bartable | |
24 | 1 347 822 955 | /356 948 592 | ~Шаблон:Bartable | |
25 | 34 052 522 467 | /8 923 714 800 | ~Шаблон:Bartable | |
26 | 34 395 742 267 | /8 923 714 800 | ~Шаблон:Bartable | |
27 | 312 536 252 003 | /80 313 433 200 | ~Шаблон:Bartable | |
28 | 315 404 588 903 | /80 313 433 200 | ~Шаблон:Bartable | |
29 | 9 227 046 511 387 | /2 329 089 562 800 | ~Шаблон:Bartable | |
30 | 9 304 682 830 147 | /2 329 089 562 800 | ~Шаблон:Bartable | |
31 | 290 774 257 297 357 | /72 201 776 446 800 | ~Шаблон:Bartable | |
32 | 586 061 125 622 639 | /144 403 552 893 600 | ~Шаблон:Bartable | |
33 | 53 676 090 078 349 | /13 127 595 717 600 | ~Шаблон:Bartable | |
34 | 54 062 195 834 749 | /13 127 595 717 600 | ~Шаблон:Bartable | |
35 | 54 437 269 998 109 | /13 127 595 717 600 | ~Шаблон:Bartable | |
36 | 54 801 925 434 709 | /13 127 595 717 600 | ~Шаблон:Bartable | |
37 | 2 040 798 836 801 833 | /485 721 041 551 200 | ~Шаблон:Bartable | |
38 | 2 053 580 969 474 233 | /485 721 041 551 200 | ~Шаблон:Bartable | |
39 | 2 066 035 355 155 033 | /485 721 041 551 200 | ~Шаблон:Bartable | |
40 | 2 078 178 381 193 813 | /485 721 041 551 200 | ~Шаблон:Bartable |
Identities involving harmonic numbers
By definition, the harmonic numbers satisfy the recurrence relation <math display="block"> H_{n + 1} = H_{n} + \frac{1}{n + 1}.</math>
The harmonic numbers are connected to the Stirling numbers of the first kind by the relation <math display="block"> H_n = \frac{1}{n!}\left[{n+1 \atop 2}\right]. </math>
The functions <math display="block">f_n(x)=\frac{x^n}{n!}(\log x-H_n)</math> satisfy the property <math display="block">f_n'(x)=f_{n-1}(x).</math> In particular <math display="block"> f_1(x)=x(\log x-1)</math> is an integral of the logarithmic function.
The harmonic numbers satisfy the series identities <math display="block"> \sum_{k=1}^n H_k = (n+1) H_{n} - n</math> and <math display="block">\sum_{k=1}^n H_k^2 = (n+1)H_{n}^2 - (2 n +1) H_n + 2 n.</math> These two results are closely analogous to the corresponding integral results <math display="block">\int_0^x \log y \ d y = x \log x - x</math> and <math display="block">\int_0^x (\log y)^2\ d y = x (\log x)^2 - 2 x \log x + 2 x.</math>
Identities involving Шаблон:Pi
There are several infinite summations involving harmonic numbers and powers of [[Pi|Шаблон:Pi]]:[3]Шаблон:Better source <math display="block">\sum_{n=1}^\infty \frac{H_n}{n\cdot 2^n}=\frac{1}{12}\pi ^2</math> <math display="block">\sum_{n=1}^\infty \frac{H_n^2}{(n+1)^2}=\frac{11}{360}\pi^4</math> <math display="block">\sum_{n=1}^\infty \frac{H_n^2}{n^2}=\frac{17}{360}\pi^4</math> <math display="block">\sum_{n=1}^\infty \frac{H_n}{n^3}=\frac{1}{72}\pi^4</math>
Calculation
An integral representation given by Euler[4] is <math display="block"> H_n = \int_0^1 \frac{1 - x^n}{1 - x}\,dx. </math>
The equality above is straightforward by the simple algebraic identity <math display="block"> \frac{1-x^n}{1-x}=1+x+\cdots +x^{n-1}.</math>
Using the substitution Шаблон:Math, another expression for Шаблон:Math is <math display="block">\begin{align} H_n &= \int_0^1 \frac{1 - x^n}{1 - x}\,dx = \int_0^1\frac{1-(1-u)^n}{u}\,du \\[6pt] &= \int_0^1\left[-\sum_{k=1}^n(-1)^k \binom nk u^{k-1}\right]\,du = -\sum_{k=1}^n (-1)^k\binom nk \int_0^1u^{k-1}\,du \\[6pt] &= -\sum_{k=1}^n(-1)^k\frac{1}{k}\binom nk . \end{align} </math>
The Шаблон:Mvarth harmonic number is about as large as the natural logarithm of Шаблон:Mvar. The reason is that the sum is approximated by the integral <math display="block">\int_1^n \frac{1}{x}\, dx,</math> whose value is Шаблон:Math.
The values of the sequence Шаблон:Math decrease monotonically towards the limit <math display="block"> \lim_{n \to \infty} \left(H_n - \ln n\right) = \gamma,</math> where Шаблон:Math is the Euler–Mascheroni constant. The corresponding asymptotic expansion is <math display="block">\begin{align}
H_n &\sim \ln{n}+\gamma+\frac{1}{2n}-\sum_{k=1}^\infty \frac{B_{2k}}{2k n^{2k}}\\ &=\ln{n}+\gamma+\frac{1}{2n}-\frac{1}{12n^2}+\frac{1}{120n^4}-\cdots, \end{align}</math>
where Шаблон:Math are the Bernoulli numbers.
Generating functions
A generating function for the harmonic numbers is <math display="block">\sum_{n=1}^\infty z^n H_n = \frac {-\ln(1-z)}{1-z},</math> where ln(z) is the natural logarithm. An exponential generating function is <math display="block">\sum_{n=1}^\infty \frac {z^n}{n!} H_n = -e^z \sum_{k=1}^\infty \frac{1}{k} \frac {(-z)^k}{k!} = e^z \operatorname{Ein}(z)</math> where Ein(z) is the entire exponential integral. The exponential integral may also be expressed as <math display="block">\operatorname{Ein}(z) = \mathrm{E}_1(z) + \gamma + \ln z = \Gamma (0,z) + \gamma + \ln z</math> where Γ(0, z) is the incomplete gamma function.
Arithmetic properties
The harmonic numbers have several interesting arithmetic properties. It is well-known that <math display="inline">H_n</math> is an integer if and only if <math display="inline">n=1</math>, a result often attributed to Taeisinger.[5] Indeed, using 2-adic valuation, it is not difficult to prove that for <math display="inline">n \ge 2</math> the numerator of <math display="inline">H_n</math> is an odd number while the denominator of <math display="inline">H_n</math> is an even number. More precisely, <math display="block">H_n=\frac{1}{2^{\lfloor\log_2(n)\rfloor}}\frac{a_n}{b_n}</math> with some odd integers <math display="inline">a_n</math> and <math display="inline">b_n</math>.
As a consequence of Wolstenholme's theorem, for any prime number <math>p \ge 5</math> the numerator of <math>H_{p-1}</math>is divisible by <math display="inline">p^2</math>. Furthermore, Eisenstein[6] proved that for all odd prime number <math display="inline">p</math> it holds <math display="block">H_{(p-1)/2} \equiv -2q_p(2) \pmod p</math> where <math display="inline">q_p(2) = (2^{p-1} -1)/p</math> is a Fermat quotient, with the consequence that <math display="inline">p</math> divides the numerator of <math>H_{(p-1)/2}</math> if and only if <math display="inline">p</math> is a Wieferich prime.
In 1991, Eswarathasan and Levine[7] defined <math>J_p</math> as the set of all positive integers <math>n</math> such that the numerator of <math>H_n</math> is divisible by a prime number <math>p.</math> They proved that <math display="block">\{p-1,p^2-p,p^2-1\}\subseteq J_p</math> for all prime numbers <math>p \ge 5,</math> and they defined harmonic primes to be the primes <math display="inline">p</math> such that <math>J_p</math> has exactly 3 elements.
Eswarathasan and Levine also conjectured that <math>J_p</math> is a finite set for all primes <math>p,</math> and that there are infinitely many harmonic primes. Boyd[8] verified that <math>J_p</math> is finite for all prime numbers up to <math>p = 547</math> except 83, 127, and 397; and he gave a heuristic suggesting that the density of the harmonic primes in the set of all primes should be <math>1/e</math>. Sanna[9] showed that <math>J_p</math> has zero asymptotic density, while Bing-Ling Wu and Yong-Gao Chen[10] proved that the number of elements of <math>J_p</math> not exceeding <math>x</math> is at most <math>3x^{\frac{2}{3}+\frac1{25 \log p}}</math>, for all <math>x \geq 1</math>.
Applications
The harmonic numbers appear in several calculation formulas, such as the digamma function <math display="block"> \psi(n) = H_{n-1} - \gamma.</math> This relation is also frequently used to define the extension of the harmonic numbers to non-integer n. The harmonic numbers are also frequently used to define Шаблон:Mvar using the limit introduced earlier: <math display="block"> \gamma = \lim_{n \rightarrow \infty}{\left(H_n - \ln(n)\right)}, </math> although <math display="block"> \gamma = \lim_{n \to \infty}{\left(H_n - \ln\left(n+\frac{1}{2}\right)\right)} </math> converges more quickly.
In 2002, Jeffrey Lagarias proved[11] that the Riemann hypothesis is equivalent to the statement that <math display="block"> \sigma(n) \le H_n + (\log H_n)e^{H_n},</math> is true for every integer Шаблон:Math with strict inequality if Шаблон:Math; here Шаблон:Math denotes the sum of the divisors of Шаблон:Mvar.
The eigenvalues of the nonlocal problem <math display="block"> \lambda \varphi(x) = \int_{-1}^{1} \frac{\varphi(x)-\varphi(y)}{|x-y|} \, dy </math> are given by <math>\lambda = 2H_n</math>, where by convention <math>H_0 = 0</math>, and the corresponding eigenfunctions are given by the Legendre polynomials <math>\varphi(x) = P_n(x)</math>.[12]
Generalizations
Generalized harmonic numbers
The nth generalized harmonic number of order m is given by <math display="block">H_{n,m}=\sum_{k=1}^n \frac{1}{k^m}.</math>
(In some sources, this may also be denoted by <math display="inline">H_n^{(m)}</math> or <math display="inline">H_m(n).</math>)
The special case m = 0 gives <math>H_{n,0}= n.</math> The special case m = 1 reduces to the usual harmonic number: <math display="block">H_{n, 1} = H_n = \sum_{k=1}^n \frac{1}{k}.</math>
The limit of <math display="inline">H_{n, m}</math> as Шаблон:Math is finite if Шаблон:Math, with the generalized harmonic number bounded by and converging to the Riemann zeta function <math display="block">\lim_{n\rightarrow \infty} H_{n,m} = \zeta(m).</math>
The smallest natural number k such that kn does not divide the denominator of generalized harmonic number H(k, n) nor the denominator of alternating generalized harmonic number H′(k, n) is, for n=1, 2, ... :
- 77, 20, 94556602, 42, 444, 20, 104, 42, 76, 20, 77, 110, 3504, 20, 903, 42, 1107, 20, 104, 42, 77, 20, 2948, 110, 136, 20, 76, 42, 903, 20, 77, 42, 268, 20, 7004, 110, 1752, 20, 19203, 42, 77, 20, 104, 42, 76, 20, 370, 110, 1107, 20, ... Шаблон:OEIS
The related sum <math>\sum_{k=1}^n k^m</math> occurs in the study of Bernoulli numbers; the harmonic numbers also appear in the study of Stirling numbers.
Some integrals of generalized harmonic numbers are <math display="block">\int_0^a H_{x,2} \, dx = a \frac {\pi^2}{6}-H_{a}</math> and <math display="block">\int_0^a H_{x,3} \, dx = a A - \frac {1}{2} H_{a,2},</math> where A is Apéry's constant ζ(3), and <math display="block">\sum_{k=1}^n H_{k,m}=(n+1)H_{n,m}- H_{n,m-1} \text{ for } m \geq 0 .</math>
Every generalized harmonic number of order m can be written as a function of harmonic numbers of order <math>m-1</math> using <math display="block">H_{n,m} = \sum_{k=1}^{n-1} \frac {H_{k,m-1}}{k(k+1)} + \frac {H_{n,m-1}}{n} </math> for example: <math>H_{4,3} = \frac {H_{1,2}}{1 \cdot 2} + \frac {H_{2,2}}{2 \cdot 3} + \frac {H_{3,2}}{3 \cdot 4} + \frac {H_{4,2}}{4} </math>
A generating function for the generalized harmonic numbers is <math display="block">\sum_{n=1}^\infty z^n H_{n,m} = \frac {\operatorname{Li}_m(z)}{1-z},</math> where <math>\operatorname{Li}_m(z)</math> is the polylogarithm, and Шаблон:Math. The generating function given above for Шаблон:Math is a special case of this formula.
A fractional argument for generalized harmonic numbers can be introduced as follows:
For every <math>p,q>0</math> integer, and <math>m>1</math> integer or not, we have from polygamma functions: <math display="block">H_{q/p,m}=\zeta(m)-p^m\sum_{k=1}^\infty \frac{1}{(q+pk)^m}</math> where <math>\zeta(m)</math> is the Riemann zeta function. The relevant recurrence relation is <math display="block">H_{a,m}=H_{a-1,m}+\frac{1}{a^m}.</math> Some special values are <math display="block">H_{\frac{1}{4},2}=16-8G-\tfrac{5}{6}\pi^2</math> where G is Catalan's constant, <math display="block">H_{\frac{1}{2},2}=4-\tfrac{\pi^2}{3}</math> <math display="block">H_{\frac{3}{4},2}=8G+\tfrac{16}{9}-\tfrac{5}{6}\pi^2</math> <math display="block">H_{\frac{1}{4},3}=64-27\zeta(3)-\pi^3</math> <math display="block">H_{\frac{1}{2},3}=8-6\zeta(3)</math> <math display="block">H_{\frac{3}{4},3}={(\tfrac{4}{3})}^3-27\zeta(3)+\pi^3</math>
In the special case that <math>p = 1</math>, we get <math display="block">H_{n,m}=\zeta(m, 1) - \zeta(m, n+1),</math> where <math>\zeta(m, n)</math> is the Hurwitz zeta function. This relationship is used to calculate harmonic numbers numerically.
Multiplication formulas
The multiplication theorem applies to harmonic numbers. Using polygamma functions, we obtain <math display="block">H_{2x}=\frac{1}{2}\left(H_x+H_{x-\frac{1}{2}}\right)+\ln 2,</math> <math display="block">H_{3x}=\frac{1}{3}\left(H_x+H_{x-\frac{1}{3}}+H_{x-\frac{2}{3}}\right)+\ln 3,</math> or, more generally, <math display="block">H_{nx}=\frac{1}{n}\left(H_x+H_{x-\frac{1}{n}}+H_{x-\frac{2}{n}}+\cdots +H_{x-\frac{n-1}{n}} \right) + \ln n.</math>
For generalized harmonic numbers, we have <math display="block">H_{2x,2}=\frac{1}{2}\left(\zeta(2)+\frac{1}{2}\left(H_{x,2}+H_{x-\frac{1}{2},2}\right)\right)</math> <math display="block">H_{3x,2}=\frac{1}{9}\left(6\zeta(2)+H_{x,2}+H_{x-\frac{1}{3},2}+H_{x-\frac{2}{3},2}\right),</math> where <math>\zeta(n)</math> is the Riemann zeta function.
Hyperharmonic numbers
The next generalization was discussed by J. H. Conway and R. K. Guy in their 1995 book The Book of Numbers.[1]Шаблон:Rp Let <math display="block"> H_n^{(0)} = \frac1n. </math> Then the nth hyperharmonic number of order r (r>0) is defined recursively as <math display="block"> H_n^{(r)} = \sum_{k=1}^n H_k^{(r-1)}. </math> In particular, <math>H_n^{(1)}</math> is the ordinary harmonic number <math>H_n</math>.
Harmonic numbers for real and complex values
Шаблон:Unreferenced section The formulae given above, <math display="block"> H_x = \int_0^1 \frac{1-t^x}{1-t} \, dt= -\sum_{k=1}^\infty {x \choose k} \frac{(-1)^k}{k}</math> are an integral and a series representation for a function that interpolates the harmonic numbers and, via analytic continuation, extends the definition to the complex plane other than the negative integers x. The interpolating function is in fact closely related to the digamma function <math display="block">H_x = \psi(x+1)+\gamma,</math> where Шаблон:Math is the digamma function, and Шаблон:Math is the Euler–Mascheroni constant. The integration process may be repeated to obtain <math display="block">H_{x,2}=-\sum_{k=1}^\infty \frac {(-1)^k}{k} {x \choose k} H_k.</math>
The Taylor series for the harmonic numbers is <math display="block">H_x=\sum_{k=2}^\infty (-1)^{k}\zeta (k)\;x^{k-1}\quad\text{ for } |x| < 1</math> which comes from the Taylor series for the digamma function (<math>\zeta </math> is the Riemann zeta function).
Alternative, asymptotic formulation
When seeking to approximate Шаблон:Math for a complex number Шаблон:Math, it is effective to first compute Шаблон:Math for some large integer Шаблон:Math. Use that as an approximation for the value of Шаблон:Math. Then use the recursion relation Шаблон:Math backwards Шаблон:Math times, to unwind it to an approximation for Шаблон:Math. Furthermore, this approximation is exact in the limit as Шаблон:Math goes to infinity.
Specifically, for a fixed integer Шаблон:Math, it is the case that <math display="block">\lim_{m \rightarrow \infty} \left[H_{m+n} - H_m\right] = 0.</math>
If Шаблон:Math is not an integer then it is not possible to say whether this equation is true because we have not yet (in this section) defined harmonic numbers for non-integers. However, we do get a unique extension of the harmonic numbers to the non-integers by insisting that this equation continue to hold when the arbitrary integer Шаблон:Math is replaced by an arbitrary complex number Шаблон:Math,
<math display="block">\lim_{m \rightarrow \infty} \left[H_{m+x} - H_m\right] = 0\,.</math> Swapping the order of the two sides of this equation and then subtracting them from Шаблон:Math gives <math display="block"> \begin{align}H_x &= \lim_{m \rightarrow \infty} \left[H_m - (H_{m+x}-H_x)\right] \\[6pt] &= \lim_{m \rightarrow \infty} \left[\left(\sum_{k=1}^m \frac{1}{k}\right) - \left(\sum_{k=1}^m \frac{1}{x+k}\right) \right] \\[6pt] &= \lim_{m \rightarrow \infty} \sum_{k=1}^m \left(\frac{1}{k} - \frac{1}{x+k}\right) = x \sum_{k=1}^{\infty} \frac{1}{k(x+k)}\, . \end{align} </math>
This infinite series converges for all complex numbers Шаблон:Math except the negative integers, which fail because trying to use the recursion relation Шаблон:Math backwards through the value Шаблон:Math involves a division by zero. By this construction, the function that defines the harmonic number for complex values is the unique function that simultaneously satisfies (1) Шаблон:Math, (2) Шаблон:Math for all complex numbers Шаблон:Math except the non-positive integers, and (3) Шаблон:Math for all complex values Шаблон:Math.
This last formula can be used to show that <math display="block"> \int_0^1 H_x \, dx = \gamma, </math> where Шаблон:Math is the Euler–Mascheroni constant or, more generally, for every Шаблон:Math we have: <math display="block"> \int_0^nH_{x}\,dx = n\gamma + \ln(n!) .</math>
Special values for fractional arguments
There are the following special analytic values for fractional arguments between 0 and 1, given by the integral <math display="block">H_\alpha = \int_0^1\frac{1-x^\alpha}{1-x}\,dx\, .</math>
More values may be generated from the recurrence relation <math display="block"> H_\alpha = H_{\alpha-1}+\frac{1}{\alpha}\,,</math> or from the reflection relation <math display="block"> H_{1-\alpha}-H_\alpha = \pi\cot{(\pi\alpha)}-\frac{1}{\alpha}+\frac{1}{1-\alpha}\, .</math>
For example: <math display="block"> H_{\frac{1}{2}} = 2 -2\ln{2}</math> <math display="block"> H_{\frac{1}{3}} = 3-\tfrac{\pi}{2\sqrt{3}} -\tfrac{3}{2}\ln{3}</math> <math display="block"> H_{\frac{2}{3}} = \tfrac{3}{2}(1-\ln{3})+\sqrt{3}\tfrac{\pi}{6}</math> <math display="block"> H_{\frac{1}{4}} = 4-\tfrac{\pi}{2} - 3\ln{2}</math> <math display="block"> H_{\frac{3}{4}} = \tfrac{4}{3}-3\ln{2}+\tfrac{\pi}{2}</math> <math display="block"> H_{\frac{1}{6}} = 6-\tfrac{\pi}{2} \sqrt{3} -2\ln{2} -\tfrac{3}{2} \ln{3}</math> <math display="block"> H_{\frac{1}{8}} = 8-\tfrac{\pi}{2} - 4\ln{2} - \tfrac{1}{\sqrt{2}} \left\{\pi + \ln\left(2 + \sqrt{2}\right) - \ln\left(2 - \sqrt{2}\right)\right\}</math> <math display="block"> H_{\frac{1}{12}} = 12-3\left(\ln{2}+\tfrac{\ln{3}}{2}\right)-\pi\left(1+\tfrac{\sqrt{3}}{2}\right)+2\sqrt{3}\ln \left (\sqrt{2-\sqrt{3}} \right )</math>
For positive integers p and q with p < q, we have: <math display="block"> H_{\frac{p}{q}} = \frac{q}{p} +2\sum_{k=1}^{\lfloor\frac{q-1}{2}\rfloor} \cos\left(\frac{2 \pi pk}{q}\right)\ln\left({\sin \left(\frac{\pi k}{q}\right)}\right)-\frac{\pi}{2}\cot\left(\frac{\pi p}{q}\right)-\ln\left(2q\right)</math>
Relation to the Riemann zeta function
Some derivatives of fractional harmonic numbers are given by <math display="block"> \begin{align} \frac{d^n H_x}{dx^n} & = (-1)^{n+1}n!\left[\zeta(n+1)-H_{x,n+1}\right] \\[6pt] \frac{d^n H_{x,2}}{dx^n} & = (-1)^{n+1}(n+1)!\left[\zeta(n+2)-H_{x,n+2}\right] \\[6pt] \frac{d^n H_{x,3}}{dx^n} & = (-1)^{n+1}\frac{1}{2}(n+2)!\left[\zeta(n+3)-H_{x,n+3}\right]. \end{align} </math>
And using Maclaurin series, we have for x < 1 that <math display="block"> \begin{align} H_x & = \sum_{n=1}^\infty (-1)^{n+1}x^n\zeta(n+1) \\[5pt] H_{x,2} & = \sum_{n=1}^\infty (-1)^{n+1}(n+1)x^n\zeta(n+2) \\[5pt] H_{x,3} & = \frac{1}{2}\sum_{n=1}^\infty (-1)^{n+1}(n+1)(n+2)x^n\zeta(n+3). \end{align} </math>
For fractional arguments between 0 and 1 and for a > 1, <math display="block"> \begin{align} H_{1/a} & = \frac{1}{a}\left(\zeta(2)-\frac{1}{a}\zeta(3)+\frac{1}{a^2}\zeta(4)-\frac{1}{a^3} \zeta(5) + \cdots\right) \\[6pt] H_{1/a, \, 2} & = \frac{1}{a}\left(2\zeta(3)-\frac{3}{a}\zeta(4)+\frac{4}{a^2}\zeta(5)-\frac{5}{a^3} \zeta(6) + \cdots\right) \\[6pt] H_{1/a, \, 3} & = \frac{1}{2a}\left(2\cdot3\zeta(4)-\frac{3\cdot4}{a}\zeta(5)+\frac{4\cdot5}{a^2}\zeta(6)-\frac{5\cdot6}{a^3}\zeta(7)+\cdots\right). \end{align} </math>
See also
- Watterson estimator
- Tajima's D
- Coupon collector's problem
- Jeep problem
- 100 prisoners problem
- Riemann zeta function
- List of sums of reciprocals
- False discovery rate#Benjamini–Yekutieli procedure
- Block-stacking problem
Notes
References
- Шаблон:Cite journal
- Шаблон:Cite book
- Ed Sandifer, How Euler Did It — Estimating the Basel problem Шаблон:Webarchive (2003)
- Шаблон:Cite journal
- Шаблон:Cite journal
- Шаблон:Cite journal
External links
- ↑ 1,0 1,1 Ошибка цитирования Неверный тег
<ref>
; для сносокConwayGuy
не указан текст - ↑ Шаблон:Cite book
- ↑ Sondow, Jonathan and Weisstein, Eric W. "Harmonic Number." From MathWorld--A Wolfram Web Resource. http://mathworld.wolfram.com/HarmonicNumber.html
- ↑ Шаблон:Citation.
- ↑ Шаблон:Cite book
- ↑ Шаблон:Cite journal
- ↑ Шаблон:Cite journal
- ↑ Шаблон:Cite journal
- ↑ Шаблон:Cite journal
- ↑ Шаблон:Cite journal
- ↑ Шаблон:Cite journal
- ↑ Шаблон:Cite journal