Mittag-Leffler polynomials

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In mathematics, the Mittag-Leffler polynomials are the polynomials gn(x) or Mn(x) studied by Mittag-Leffler (1891).

Mn(x) is a special case of the Meixner polynomial Mn(x;b,c) at b = 0, c = -1.

Definition and examples

Generating functions

The Mittag-Leffler polynomials are defined respectively by the generating functions

[math]\displaystyle{ \displaystyle \sum_{n=0}^{\infty} g_n(x)t^n :=\frac{1}{2}\Bigl(\frac{1+t}{1-t} \Bigr)^x }[/math] and
[math]\displaystyle{ \displaystyle \sum_{n=0}^{\infty} M_n(x)\frac{t^n}{n!}:=\Bigl(\frac{1+t}{1-t} \Bigr)^x=(1+t)^x(1-t)^{-x}=\exp(2x\text{ artanh } t). }[/math]

They also have the bivariate generating function[1]

[math]\displaystyle{ \displaystyle \sum_{n=1}^{\infty}\sum_{m=1}^{\infty} g_n(m)x^my^n =\frac{xy}{(1-x)(1-x-y-xy)}. }[/math]

Examples

The first few polynomials are given in the following table. The coefficients of the numerators of the [math]\displaystyle{ g_n(x) }[/math] can be found in the OEIS,[2] though without any references, and the coefficients of the [math]\displaystyle{ M_n(x) }[/math] are in the OEIS[3] as well.

n gn(x) Mn(x)
0 [math]\displaystyle{ \frac{1}{2} }[/math] [math]\displaystyle{ 1 }[/math]
1 [math]\displaystyle{ x }[/math] [math]\displaystyle{ 2x }[/math]
2 [math]\displaystyle{ x^2 }[/math] [math]\displaystyle{ 4x^2 }[/math]
3 [math]\displaystyle{ {\frac{1}{3}} (x+2x^3) }[/math] [math]\displaystyle{ 8x^3+4x }[/math]
4 [math]\displaystyle{ {\frac{1}{3}} (2x^2+x^4) }[/math] [math]\displaystyle{ 16x^4+32x^2 }[/math]
5 [math]\displaystyle{ {\frac{1}{15}} (3x+10x^3+2x^5) }[/math] [math]\displaystyle{ 32 x^5 + 160 x^3 + 48 x }[/math]
6 [math]\displaystyle{ {\frac{1}{45}} (23x^2+20x^4+2x^6) }[/math] [math]\displaystyle{ 64 x^6 + 640 x^4 + 736 x^2 }[/math]
7 [math]\displaystyle{ {\frac{1}{315}} (45 x + 196 x^3 + 70 x^5 + 4 x^7) }[/math] [math]\displaystyle{ 128 x^7 + 2240 x^5 + 6272 x^3 + 1440 x }[/math]
8 [math]\displaystyle{ {\frac{1}{315}} (132 x^2 + 154 x^4 + 28 x^6 + x^8) }[/math] [math]\displaystyle{ 256 x^8 + 7168 x^6 + 39424 x^4 + 33792 x^2 }[/math]
9 [math]\displaystyle{ {\frac{1}{2835}} (315 x + 1636 x^3 + 798 x^5 + 84 x^7 + 2 x^9) }[/math] [math]\displaystyle{ 512 x^9 + 21504 x^7 + 204288 x^5 + 418816 x^3 + 80640 x }[/math]
10 [math]\displaystyle{ {\frac{1}{14175}} (5067 x^2 + 7180 x^4 + 1806 x^6 + 120 x^8 + 2 x^{10}) }[/math] [math]\displaystyle{ 1024 x^{10} + 61440 x^8 + 924672 x^6 + 3676160 x^4 + 2594304 x^2 }[/math]

Properties

The polynomials are related by [math]\displaystyle{ M_n(x)=2\cdot{n!} \, g_n(x) }[/math] and we have [math]\displaystyle{ g_n(1)=1 }[/math] for [math]\displaystyle{ n\geqslant 1 }[/math]. Also [math]\displaystyle{ g_{2k}(\frac12)=g_{2k+1}(\frac12)=\frac12\frac{(2k-1)!!}{(2k)!!}=\frac12\cdot \frac{1\cdot 3\cdots (2k-1) }{2\cdot 4\cdots (2k)} }[/math].

Explicit formulas

Explicit formulas are

[math]\displaystyle{ g_n(x) = \sum_ {k = 1}^{n} 2^{k-1}\binom{n-1}{n-k}\binom xk = \sum_ {k = 0}^{n-1} 2^{k}\binom{n-1}{k}\binom x{k+1} }[/math]
[math]\displaystyle{ g_n(x) = \sum_{k = 0}^{n-1} \binom{n-1}k\binom{k+x}n }[/math]
[math]\displaystyle{ g_n(m) = \frac12\sum_{k = 0}^m \binom mk\binom{n-1+m-k}{m-1}=\frac12\sum_{k = 0}^{\min(n,m)} \frac m{n+m-k}\binom{n+m-k}{k,n-k,m-k} }[/math]

(the last one immediately shows [math]\displaystyle{ ng_n(m)=mg_m(n) }[/math], a kind of reflection formula), and

[math]\displaystyle{ M_n(x)=(n-1)!\sum_ {k = 1}^{n}k2^k\binom nk \binom xk }[/math], which can be also written as
[math]\displaystyle{ M_n(x)=\sum_ {k = 1}^{n}2^k\binom nk(n-1)_{n-k}(x)_k }[/math], where [math]\displaystyle{ (x)_n = n!\binom xn = x(x-1)\cdots(x-n+1) }[/math] denotes the falling factorial.

In terms of the Gaussian hypergeometric function, we have[4]

[math]\displaystyle{ g_n(x) = x\!\cdot {}_2\!F_1 (1-n,1-x; 2; 2). }[/math]

Reflection formula

As stated above, for [math]\displaystyle{ m,n\in\mathbb N }[/math], we have the reflection formula [math]\displaystyle{ ng_n(m)=mg_m(n) }[/math].

Recursion formulas

The polynomials [math]\displaystyle{ M_n(x) }[/math] can be defined recursively by

[math]\displaystyle{ M_n(x)=2xM_{n-1}(x)+(n-1)(n-2)M_{n-2}(x) }[/math], starting with [math]\displaystyle{ M_{-1}(x)=0 }[/math] and [math]\displaystyle{ M_{0}(x)=1 }[/math].

Another recursion formula, which produces an odd one from the preceding even ones and vice versa, is

[math]\displaystyle{ M_{n+1}(x) = 2x \sum_{k=0}^{\lfloor n/2 \rfloor} \frac{n!}{(n-2k)!} M_{n-2k}(x) }[/math], again starting with [math]\displaystyle{ M_0(x) = 1 }[/math].


As for the [math]\displaystyle{ g_n(x) }[/math], we have several different recursion formulas:

[math]\displaystyle{ \displaystyle (1)\quad g_n(x + 1) - g_{n-1}(x + 1)= g_n(x) + g_{n-1}(x) }[/math]
[math]\displaystyle{ \displaystyle (2)\quad (n + 1)g_{n+1}(x) - (n - 1)g_{n-1}(x) = 2xg_n(x) }[/math]
[math]\displaystyle{ (3)\quad x\Bigl(g_n(x+1)-g_n(x-1)\Bigr) = 2ng_n(x) }[/math]
[math]\displaystyle{ (4)\quad g_{n+1}(m)= g_{n}(m)+2\sum_{k=1}^{m-1}g_{n}(k)=g_{n}(1)+g_{n}(2)+\cdots+g_{n}(m)+g_{n}(m-1) +\cdots+g_{n}(1) }[/math]

Concerning recursion formula (3), the polynomial [math]\displaystyle{ g_n(x) }[/math] is the unique polynomial solution of the difference equation [math]\displaystyle{ x(f(x+1)-f(x-1)) = 2nf(x) }[/math], normalized so that [math]\displaystyle{ f(1) = 1 }[/math].[5] Further note that (2) and (3) are dual to each other in the sense that for [math]\displaystyle{ x\in\mathbb N }[/math], we can apply the reflection formula to one of the identities and then swap [math]\displaystyle{ x }[/math] and [math]\displaystyle{ n }[/math] to obtain the other one. (As the [math]\displaystyle{ g_n(x) }[/math] are polynomials, the validity extends from natural to all real values of [math]\displaystyle{ x }[/math].)

Initial values

The table of the initial values of [math]\displaystyle{ g_n(m) }[/math] (these values are also called the "figurate numbers for the n-dimensional cross polytopes" in the OEIS[6]) may illustrate the recursion formula (1), which can be taken to mean that each entry is the sum of the three neighboring entries: to its left, above and above left, e.g. [math]\displaystyle{ g_5(3)=51=33+8+10 }[/math]. It also illustrates the reflection formula [math]\displaystyle{ ng_n(m)=mg_m(n) }[/math] with respect to the main diagonal, e.g. [math]\displaystyle{ 3\cdot44=4\cdot33 }[/math].

n
m
1 2 3 4 5 6 7 8 9 10
1 1 1 1 1 1 1 1 1 1 1
2 2 4 6 8 10 12 14 16 18
3 3 9 19 33 51 73 99 129
4 4 16 44 96 180 304 476
5 5 25 85 225 501 985
6 6 36 146 456 1182
7 7 49 231 833
8 8 64 344
9 9 81
10 10

Orthogonality relations

For [math]\displaystyle{ m,n\in\mathbb N }[/math] the following orthogonality relation holds:[7]

[math]\displaystyle{ \int_{-\infty}^{\infty}\frac{g_n(-iy)g_m(iy)}{y \sinh \pi y} dy=\frac 1{2n}\delta_{mn}. }[/math]

(Note that this is not a complex integral. As each [math]\displaystyle{ g_n }[/math] is an even or an odd polynomial, the imaginary arguments just produce alternating signs for their coefficients. Moreover, if [math]\displaystyle{ m }[/math] and [math]\displaystyle{ n }[/math] have different parity, the integral vanishes trivially.)

Binomial identity

Being a Sheffer sequence of binomial type, the Mittag-Leffler polynomials [math]\displaystyle{ M_n(x) }[/math] also satisfy the binomial identity[8]

[math]\displaystyle{ M_n(x+y)=\sum_{k=0}^n\binom nk M_k(x)M_{n-k}(y) }[/math].

Integral representations

Based on the representation as a hypergeometric function, there are several ways of representing [math]\displaystyle{ g_n(z) }[/math] for [math]\displaystyle{ |z|\lt 1 }[/math] directly as integrals,[9] some of them being even valid for complex [math]\displaystyle{ z }[/math], e.g.

[math]\displaystyle{ (26)\qquad g_n(z) = \frac{\sin(\pi z)}{2\pi}\int _{-1}^1 t^{n-1} \Bigl(\frac{1+t}{1-t}\Bigr)^z dt }[/math]
[math]\displaystyle{ (27)\qquad g_n(z) = \frac{\sin(\pi z)}{2\pi} \int_{-\infty}^{\infty} e^{uz}\frac{(\tanh \frac u2)^n}{\sinh u} du }[/math]
[math]\displaystyle{ (32)\qquad g_n(z) = \frac1\pi\int _0^\pi \cot^z (\frac u2) \cos (\frac{\pi z}2) \cos (nu)du }[/math]
[math]\displaystyle{ (33)\qquad g_n(z) = \frac1\pi\int _0^\pi \cot^z (\frac u2) \sin (\frac{\pi z}2) \sin (nu)du }[/math]
[math]\displaystyle{ (34)\qquad g_n(z) = \frac1{2\pi}\int _0^{2\pi} (1+e^{it})^z (2+e^{it})^{n-1} e^{-int}dt }[/math].

Closed forms of integral families

There are several families of integrals with closed-form expressions in terms of zeta values where the coefficients of the Mittag-Leffler polynomials occur as coefficients. All those integrals can be written in a form containing either a factor [math]\displaystyle{ \tan^{\pm n} }[/math] or [math]\displaystyle{ \tanh^{\pm n} }[/math], and the degree of the Mittag-Leffler polynomial varies with [math]\displaystyle{ n }[/math]. One way to work out those integrals is to obtain for them the corresponding recursion formulas as for the Mittag-Leffler polynomials using integration by parts.

1. For instance,[10] define for [math]\displaystyle{ n\geqslant m \geqslant 2 }[/math]

[math]\displaystyle{ I(n,m):= \int _0^1\dfrac{\text{artanh}^nx}{x^m}dx = \int _0^1\log^{n/2}\Bigl(\dfrac{1+x }{1-x}\Bigr)\dfrac{dx}{x^m} = \int _0^\infty z^n\dfrac{ \coth^{m-2}z }{\sinh^2z} dz. }[/math]

These integrals have the closed form

[math]\displaystyle{ (1)\quad I(n,m)=\frac{n!}{2^{n-1}}\zeta^{n+1}~g_ {m-1}(\frac1{\zeta} ) }[/math]

in umbral notation, meaning that after expanding the polynomial in [math]\displaystyle{ \zeta }[/math], each power [math]\displaystyle{ \zeta^k }[/math] has to be replaced by the zeta value [math]\displaystyle{ \zeta(k) }[/math]. E.g. from [math]\displaystyle{ g_6(x)={\frac{1}{45}} (23x^2+20x^4+2x^6)\ }[/math] we get [math]\displaystyle{ \ I(n,7)=\frac{n!}{2^{n-1}}\frac{23~\zeta(n-1)+20~\zeta(n-3)+2~\zeta(n-5)}{45}\ }[/math] for [math]\displaystyle{ n\geqslant 7 }[/math].

2. Likewise take for [math]\displaystyle{ n\geqslant m \geqslant 2 }[/math]

[math]\displaystyle{ J(n,m):=\int _1^\infty\dfrac{\text{arcoth}^nx}{x^m}dx =\int _1^\infty\log^{n/2}\Bigl(\dfrac{x+1}{x-1}\Bigr)\dfrac{dx}{x^m} = \int _0^\infty z^n\dfrac{\tanh^{m-2}z }{\cosh^2z} dz. }[/math]

In umbral notation, where after expanding, [math]\displaystyle{ \eta^k }[/math] has to be replaced by the Dirichlet eta function [math]\displaystyle{ \eta(k):=\left(1-2^{1-k}\right)\zeta(k) }[/math], those have the closed form

[math]\displaystyle{ (2)\quad J(n,m)=\frac{n!}{2^{n-1}} \eta^{n+1}~g_ {m-1}(\frac1{\eta} ) }[/math].

3. The following[11] holds for [math]\displaystyle{ n\geqslant m }[/math] with the same umbral notation for [math]\displaystyle{ \zeta }[/math] and [math]\displaystyle{ \eta }[/math], and completing by continuity [math]\displaystyle{ \eta(1):=\ln 2 }[/math].

[math]\displaystyle{ (3)\quad \int\limits_0^{\pi/2} \frac{x^n}{\tan^m x}dx = \cos\Bigl(\frac{ m}{2}\pi\Bigr)\frac{(\pi/2)^{n+1}}{n+1} +\cos\Bigl(\frac{ m-n-1}{2}\pi\Bigr) \frac{n!~m}{2^{n}}\zeta^{n+2}g_m(\frac1{\zeta}) +\sum\limits_{v=0}^n \cos\Bigl(\frac{ m-v-1}{2}\pi\Bigr)\frac{n!~m~\pi^{n-v}}{(n-v)!~2^{n}} \eta^{n+2}g_m(\frac1{\eta}). }[/math]

Note that for [math]\displaystyle{ n\geqslant m \geqslant 2 }[/math], this also yields a closed form for the integrals

[math]\displaystyle{ \int\limits_0^{\infty} \frac{\arctan^n x}{x^m} dx = \int\limits_0^{\pi/2} \frac{x^n}{\tan^m x} dx + \int\limits_0^{\pi/2} \frac{x^n}{\tan^{m-2} x} dx. }[/math]

4. For [math]\displaystyle{ n\geqslant m\geqslant 2 }[/math], define[12] [math]\displaystyle{ \quad K(n,m):=\int\limits_0^\infty\dfrac{\tanh^n(x)}{x^m}dx }[/math].

If [math]\displaystyle{ n+m }[/math] is even and we define [math]\displaystyle{ h_k:= (-1)^{\frac{k-1}2} \frac{(k-1)!(2^k-1)\zeta(k)}{2^{k-1}\pi^{k-1}} }[/math], we have in umbral notation, i.e. replacing [math]\displaystyle{ h^k }[/math] by [math]\displaystyle{ h_k }[/math],

[math]\displaystyle{ (4)\quad K(n,m):=\int\limits_0^\infty\dfrac{\tanh^n(x)}{x^m}dx = \dfrac{n\cdot 2^{m-1}}{ (m-1)!}(-h)^{m-1} g_n(h). }[/math]

Note that only odd zeta values (odd [math]\displaystyle{ k }[/math]) occur here (unless the denominators are cast as even zeta values), e.g.

[math]\displaystyle{ K(5,3)=-\frac{2}{3}(3h_3+10h_5+2h_7)=-7\frac{\zeta(3)}{\pi^2}+ 310 \frac{\zeta(5)}{\pi^4} -1905\frac{\zeta(7)}{\pi^6}, }[/math]
[math]\displaystyle{ K(6,2)=\frac{4}{15}(23h_3+20h_5+2h_7),\quad K(6,4)=\frac{4}{45}(23h_5+20h_7+2h_9). }[/math]

5. If [math]\displaystyle{ n+m }[/math] is odd, the same integral is much more involved to evaluate, including the initial one [math]\displaystyle{ \int\limits_0^\infty\dfrac{\tanh^3(x)}{x^2}dx }[/math]. Yet it turns out that the pattern subsists if we define[13] [math]\displaystyle{ s_k:=\eta'(-k)=2^{k+1}\zeta(-k)\ln2-(2^{k+1}-1)\zeta'(-k) }[/math], equivalently [math]\displaystyle{ s_k = \frac{\zeta(-k)}{\zeta'(-k)}\eta(-k)+\zeta(-k)\eta(1)-\eta(-k)\eta(1) }[/math]. Then [math]\displaystyle{ K(n,m) }[/math] has the following closed form in umbral notation, replacing [math]\displaystyle{ s^k }[/math] by [math]\displaystyle{ s_k }[/math]:

[math]\displaystyle{ (5)\quad K(n,m)=\int\limits_0^\infty\dfrac{\tanh^n(x)}{x^m}dx=\frac{n\cdot2^{m}}{(m-1)!}(-s)^{m-2}g_n(s) }[/math], e.g.
[math]\displaystyle{ K(5,4)=\frac{8}{9}(3s_3+10s_5+2s_7), \quad K(6,3)=-\frac{8}{15}(23s_3+20s_5+2s_7),\quad K(6,5)=-\frac{8}{45}(23s_5+20s_7+2s_9). }[/math]

Note that by virtue of the logarithmic derivative [math]\displaystyle{ \frac{\zeta'}{\zeta}(s)+\frac{\zeta'}{\zeta}(1-s)=\log\pi-\frac{1}{2}\frac{\Gamma'}{\Gamma}\left(\frac{s}{2}\right)-\frac{1}{2}\frac{\Gamma'}{\Gamma}\left(\frac{1-s}{2}\right) }[/math] of Riemann's functional equation, taken after applying Euler's reflection formula,[14] these expressions in terms of the [math]\displaystyle{ s_k }[/math] can be written in terms of [math]\displaystyle{ \frac{\zeta'(2j) }{\zeta(2j) } }[/math], e.g.

[math]\displaystyle{ K(5,4)=\frac{8}{9}(3s_3+10s_5+2s_7)=\frac 19\left\{ \frac{1643}{420}-\frac{16 }{315}\ln2+3\frac{\zeta'(4) }{\zeta(4) }-20\frac{\zeta'(6) }{\zeta(6) }+17\frac{\zeta'(8) }{\zeta(8) }\right\}. }[/math]

6. For [math]\displaystyle{ n\lt m }[/math], the same integral [math]\displaystyle{ K(n,m) }[/math] diverges because the integrand behaves like [math]\displaystyle{ x^{n-m} }[/math] for [math]\displaystyle{ x\searrow 0 }[/math]. But the difference of two such integrals with corresponding degree differences is well-defined and exhibits very similar patterns, e.g.

[math]\displaystyle{ (6)\quad K(n-1,n)-K(n,n+1)=\int\limits_0^\infty\left(\dfrac{\tanh^{n-1}(x)}{x^{n}}-\dfrac{\tanh^{n}(x)}{x^{n+1}}\right)dx= -\frac 1n + \frac{ (n+1)\cdot2^{n}}{(n-1)!}s^{n-2}g_n(s) }[/math].

See also

References

  1. see the formula section of OEIS A142978, https://oeis.org/A142978 
  2. see OEIS A064984, https://oeis.org/A064984 
  3. see OEIS A137513, https://oeis.org/A137513 
  4. Özmen, Nejla; Nihal, Yılmaz (2019) (in English). On The Mittag-Leffler Polynomials and Deformed Mittag-Leffler Polynomials. https://dergipark.org.tr/en/download/article-file/843805. 
  5. see the comment section of OEIS A142983, https://oeis.org/A142983 
  6. see OEIS A142978, https://oeis.org/A142978/table 
  7. Stankovic, Miomir S.; Marinkovic, Sladjana D.; Rajkovic, Predrag M. (2010) (in English). Deformed Mittag–Leffler Polynomials. 
  8. Mathworld entry "Mittag-Leffler Polynomial", https://mathworld.wolfram.com/Mittag-LefflerPolynomial.html 
  9. Bateman, H. (1940). "The polynomial of Mittag-Leffler". Proceedings of the National Academy of Sciences of the United States of America 26 (8): 491–496. doi:10.1073/pnas.26.8.491. ISSN 0027-8424. PMID 16588390. PMC 1078216. Bibcode1940PNAS...26..491B. http://authors.library.caltech.edu/8694/1/BATpnas40.pdf. 
  10. see at the end of this question on Mathoverflow, https://mathoverflow.net/questions/231964/how-to-prove-that-int-0-infty-frac-textarcsinhnxxmdx-is-a-rational 
  11. answer on math.stackexchange, https://math.stackexchange.com/a/2939452 
  12. similar to this question on Mathoverflow, https://math.stackexchange.com/q/1582943 
  13. method used in this answer on Mathoverflow, https://mathoverflow.net/a/271569 
  14. or see formula (14) in https://mathworld.wolfram.com/RiemannZetaFunction.html