Mittag-Leffler polynomials

Definition and examples

Generating functions

The Mittag-Leffler polynomials are defined respectively by the generating functions : \displaystyle \sum_{n=0}^{\infty} g_n(x)t^n :=\frac{1}{2}\Bigl(\frac{1+t}{1-t} \Bigr)^x and : \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). They also have the bivariate generating function : \displaystyle \sum_{n=1}^{\infty}\sum_{m=1}^{\infty} g_n(m)x^my^n =\frac{xy}{(1-x)(1-x-y-xy)}.

Examples

The first few polynomials are given in the following table. The coefficients of the numerators of the g_n(x) can be found in the OEIS, though without any references, and the coefficients of the M_n(x) are in the OEIS as well. :{| class="wikitable" !n !! gn(x) !! Mn(x) |- | 0 || \frac{1}{2} || 1 |- | 1 || x || 2x |- | 2 || x^2 || 4x^2 |- | 3 || {\frac{1}{3}} (x+2x^3) || 8x^3+4x |- | 4 || {\frac{1}{3}} (2x^2+x^4) || 16x^4+32x^2 |- | 5 || {\frac{1}{15}} (3x+10x^3+2x^5) || 32 x^5 + 160 x^3 + 48 x |- | 6 || {\frac{1}{45}} (23x^2+20x^4+2x^6) || 64 x^6 + 640 x^4 + 736 x^2 |- | 7 || {\frac{1}{315}} (45 x + 196 x^3 + 70 x^5 + 4 x^7) || 128 x^7 + 2240 x^5 + 6272 x^3 + 1440 x |- | 8 || {\frac{1}{315}} (132 x^2 + 154 x^4 + 28 x^6 + x^8) || 256 x^8 + 7168 x^6 + 39424 x^4 + 33792 x^2 |- | 9 || {\frac{1}{2835}} (315 x + 1636 x^3 + 798 x^5 + 84 x^7 + 2 x^9) || 512 x^9 + 21504 x^7 + 204288 x^5 + 418816 x^3 + 80640 x |- | 10 || {\frac{1}{14175}} (5067 x^2 + 7180 x^4 + 1806 x^6 + 120 x^8 + 2 x^{10}) || 1024 x^{10} + 61440 x^8 + 924672 x^6 + 3676160 x^4 + 2594304 x^2 |}

Properties

The polynomials are related by M_n(x)=2\cdot{n!} , g_n(x) and we have g_n(1)=1 for n\geqslant 1 . Also 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)} .

Explicit formulas

Explicit formulas are : 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} : g_n(x) = \sum_{k = 0}^{n-1} \binom{n-1}k\binom{k+x}n : 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} (the last one immediately shows ng_n(m)=mg_m(n) , a kind of reflection formula), and : M_n(x)=(n-1)!\sum_ {k = 1}^{n}k2^k\binom nk \binom xk , which can be also written as : M_n(x)=\sum_ {k = 1}^{n}2^k\binom nk(n-1)_{n-k}(x)_k, where (x)_n = n!\binom xn = x(x-1)\cdots(x-n+1) denotes the falling factorial. In terms of the Gaussian hypergeometric function, we have : g_n(x) = x!\cdot {}_2!F_1 (1-n,1-x; 2; 2).

Reflection formula

As stated above, for m,n\in\mathbb N , we have the reflection formula ng_n(m)=mg_m(n) .

Recursion formulas

The polynomials M_n(x) can be defined recursively by : M_n(x)=2xM_{n-1}(x)+(n-1)(n-2)M_{n-2}(x), starting with M_{-1}(x)=0 and M_{0}(x)=1. Another recursion formula, which produces an odd one from the preceding even ones and vice versa, is : M_{n+1}(x) = 2x \sum_{k=0}^{\lfloor n/2 \rfloor} \frac{n!}{(n-2k)!} M_{n-2k}(x), again starting with M_0(x) = 1.

As for the g_n(x), we have several different recursion formulas:

: \displaystyle (1)\quad g_n(x + 1) - g_{n-1}(x + 1)= g_n(x) + g_{n-1}(x)

: \displaystyle (2)\quad (n + 1)g_{n+1}(x) - (n - 1)g_{n-1}(x) = 2xg_n(x)

: (3)\quad x\Bigl(g_n(x+1)-g_n(x-1)\Bigr) = 2ng_n(x)

: (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)

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

Initial values

The table of the initial values of g_n(m) (these values are also called the "figurate numbers for the n-dimensional cross polytopes" in the OEIS) 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. g_5(3)=51=33+8+10. It also illustrates the reflection formula ng_n(m)=mg_m(n) with respect to the main diagonal, e.g. 3\cdot44=4\cdot33 . :{| class="wikitable" ! !! 1!! 2 !! 3 !! 4 !! 5 !! 6 !! 7 !! 8 !! 9 !! 10 |- ! 1 |- ! 2 |- ! 3 |- ! 4 |- ! 5 |- ! 6 |- ! 7 |- ! 8 |- ! 9 |- ! 10 |- |}

Orthogonality relations

For m,n\in\mathbb N the following orthogonality relation holds: : \int_{-\infty}^{\infty}\frac{g_n(-iy)g_m(iy)}{y \sinh \pi y} dy=\frac 1{2n}\delta_{mn}. (Note that this is not a complex integral. As each g_n is an even or an odd polynomial, the imaginary arguments just produce alternating signs for their coefficients. Moreover, if m and n have different parity, the integral vanishes trivially.)

Binomial identity

Being a Sheffer sequence of binomial type, the Mittag-Leffler polynomials M_n(x) also satisfy the binomial identity : M_n(x+y)=\sum_{k=0}^n\binom nk M_k(x)M_{n-k}(y).

Integral representations

Based on the representation as a hypergeometric function, there are several ways of representing g_n(z) for |z| directly as integrals, some of them being even valid for complex z, e.g.

:(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

:(27)\qquad g_n(z) = \frac{\sin(\pi z)}{2\pi} \int_{-\infty}^{\infty} e^{uz}\frac{(\tanh \frac u2)^n}{\sinh u} du

:(32)\qquad g_n(z) = \frac1\pi\int _0^\pi \cot^z (\frac u2) \cos (\frac{\pi z}2) \cos (nu)du

:(33)\qquad g_n(z) = \frac1\pi\int _0^\pi \cot^z (\frac u2) \sin (\frac{\pi z}2) \sin (nu)du

:(34)\qquad g_n(z) = \frac1{2\pi}\int _0^{2\pi} (1+e^{it})^z (2+e^{it})^{n-1} e^{-int}dt.

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 \tan^{\pm n} or \tanh^{\pm n}, and the degree of the Mittag-Leffler polynomial varies with n. 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, define for n\geqslant m \geqslant 2 :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. These integrals have the closed form :(1)\quad I(n,m)=\frac{n!}{2^{n-1}}\zeta^{n+1}~g {m-1}(\frac1{\zeta} ) in umbral notation, meaning that after expanding the polynomial in \zeta, each power \zeta^k has to be replaced by the zeta value \zeta(k). E.g. from g_6(x)={\frac{1}{45}} (23x^2+20x^4+2x^6)\ we get \ I(n,7)=\frac{n!}{2^{n-1}}\frac{23~\zeta(n-1)+20~\zeta(n-3)+2~\zeta(n-5)}{45}\ for n\geqslant 7.

2. Likewise take for n\geqslant m \geqslant 2 : 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.

In umbral notation, where after expanding, \eta^k has to be replaced by the Dirichlet eta function \eta(k):=\left(1-2^{1-k}\right)\zeta(k), those have the closed form : (2)\quad J(n,m)=\frac{n!}{2^{n-1}} \eta^{n+1}~g_ {m-1}(\frac1{\eta} ).

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

Note that for n\geqslant m \geqslant 2, this also yields a closed form for the integrals

: \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.

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

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

Note that only odd zeta values (odd k) occur here (unless the denominators are cast as even zeta values), e.g. :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}, : 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).

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

Note that by virtue of the logarithmic derivative \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) of Riemann's functional equation, taken after applying Euler's reflection formula, these expressions in terms of the s_k can be written in terms of \frac{\zeta'(2j) }{\zeta(2j) }, e.g. :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}.

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

References

References

1. "see the formula section of OEIS A142978".
2. "see OEIS A064984".
3. "see OEIS A137513".
4. Özmen, Nejla. (2019). "On The Mittag-Leffler Polynomials and Deformed Mittag-Leffler Polynomials".
5. "see the comment section of OEIS A142983".
6. "see OEIS A142978".
7. Stankovic, Miomir S.. (2010). "Deformed Mittag–Leffler Polynomials".
8. "Mathworld entry "Mittag-Leffler Polynomial"".
9. (1940). "The polynomial of Mittag-Leffler". [[Proceedings of the National Academy of Sciences.
10. "see at the end of this question on Mathoverflow".
11. "answer on math.stackexchange".
12. "similar to this question on Mathoverflow".
13. "method used in this answer on Mathoverflow".
14. or see formula (14) in https://mathworld.wolfram.com/RiemannZetaFunction.html