Mittag-Leffler polynomials

Template:Short description Template:Distinguish In mathematics, the Mittag-Leffler polynomials are the polynomials g_n(x) or M_n(x) studied by Template:Harvs.

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

Definition and examples

Generating functions

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

$\sum _{n=0}^{\mathrm{\infty }}{g}_n(x)t^n:=\frac{1}{2}(\frac{1+t}{1-t}{)}^x$ and

$\sum _{n=0}^{\mathrm{\infty }}{M}_n(x)\frac{t^n}{n!}:=(\frac{1+t}{1-t}{)}^x=(1+t{)}^x(1-t{)}^{-x}=\exp (2x\text{ artanh }t).$

They also have the bivariate generating function

$\sum _{n=1}^{\mathrm{\infty }}\sum _{m=1}^{\mathrm{\infty }}{g}_n(m)x^m{y}^n=\frac{xy}{(1-x)(1-x-y-xy)}.$

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.

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

Properties

The polynomials are related by $M_n(x)=2\cdot n!g_n(x)$ and we have $g_n(1)=1$ for $n⩾1$. Also $g_{2k}(\frac{1}{2})=g_{2k+1}(\frac{1}{2})=\frac{1}{2}\frac{(2k-1)!!}{(2k)!!}=\frac{1}{2}\cdot \frac{1\cdot 3\cdots (2k-1)}{2\cdot 4\cdots (2k)}$.

Explicit formulas

Explicit formulas are

$g_n(x)=\sum _{k=1}^n{2}^{k-1}(\genfrac{}{}{0ex}{}{n-1}{n-k})(\genfrac{}{}{0ex}{}{x}{k})=\sum _{k=0}^{n-1}{2}^k(\genfrac{}{}{0ex}{}{n-1}{k})(\genfrac{}{}{0ex}{}{x}{k+1})$

$g_n(x)=\sum _{k=0}^{n-1}(\genfrac{}{}{0ex}{}{n-1}{k})(\genfrac{}{}{0ex}{}{k+x}{n})$

$g_n(m)=\frac{1}{2}\sum _{k=0}^m(\genfrac{}{}{0ex}{}{m}{k})(\genfrac{}{}{0ex}{}{n-1+m-k}{m-1})=\frac{1}{2}\sum _{k=0}^{\min (n,m)}\frac{m}{n+m-k}(\genfrac{}{}{0ex}{}{n+m-k}{k,n-k,m-k})$

(the last one immediately shows $n{g}_n(m)=m{g}_m(n)$, a kind of reflection formula), and

$M_n(x)=(n-1)!\sum _{k=1}^{n}k{2}^k(\genfrac{}{}{0ex}{}{n}{k})(\genfrac{}{}{0ex}{}{x}{k})$, which can be also written as

$M_n(x)=\sum _{k=1}^n{2}^k(\genfrac{}{}{0ex}{}{n}{k})(n-1{)}_{n-k}(x{)}_k$, where $(x{)}_n=n!(\genfrac{}{}{0ex}{}{x}{n})=x(x-1)\cdots (x-n+1)$ 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 $n{g}_n(m)=m{g}_m(n)$.

Recursion formulas

The polynomials $M_n(x)$ can be defined recursively by

$M_n(x)=2x{M}_{n-1}(x)+(n-1)(n-2)M_{n-2}(x)$, 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:

$(1)g_n(x+1)-g_{n-1}(x+1)=g_n(x)+g_{n-1}(x)$

$(2)(n+1)g_{n+1}(x)-(n-1)g_{n-1}(x)=2x{g}_n(x)$

$(3)x(g_n(x+1)-g_n(x-1))=2n{g}_n(x)$

$(4)g_{n+1}(m)=g_n(m)+2\sum _{k=1}^{m-1}{g}_n(k)=g_n(1)+g_n(2)+\cdots +g_n(m)+g_n(m-1)+\cdots +g_n(1)$

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 $n{g}_n(m)=m{g}_m(n)$ with respect to the main diagonal, e.g. $3\cdot 44=4\cdot 33$.

Template:Diagonal split header	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 $m,n\in \mathbb{N}$ the following orthogonality relation holds:

$\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}\frac{g_n(-iy)g_m(iy)}{y\sinh \pi y}dy=\frac{1}{2n}{\delta }_{mn}.$

(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

$M_n(x+y)=\sum _{k=0}^n(\genfrac{}{}{0ex}{}{n}{k})M_k(x)M_{n-k}(y)$.

Integral representations

Based on the representation as a hypergeometric function, there are several ways of representing $g_n(z)$ for $|z|<1$ directly as integrals, some of them being even valid for complex $z$, e.g.

$(26)g_n(z)=\frac{\sin (\pi z)}{2\pi }\underset{-1}{\overset{1}{\int }}{t}^{n-1}(\frac{1+t}{1-t}{)}^{z}dt$

$(27)g_n(z)=\frac{\sin (\pi z)}{2\pi }\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}{e}^{uz}\frac{(\tanh \frac{u}{2}{)}^n}{\sinh u}du$

$(32)g_n(z)=\frac{1}{\pi }\underset{0}{\overset{\pi }{\int }}{\cot }^z(\frac{u}{2})\cos (\frac{\pi z}{2})\cos (nu)du$

$(33)g_n(z)=\frac{1}{\pi }\underset{0}{\overset{\pi }{\int }}{\cot }^z(\frac{u}{2})\sin (\frac{\pi z}{2})\sin (nu)du$

$(34)g_n(z)=\frac{1}{2\pi }\underset{0}{\overset{2\pi }{\int }}(1+e^{it}{)}^z(2+e^{it}{)}^{n-1}{e}^{-int}dt$.

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 $n⩾m⩾2$

$I(n,m):=\underset{0}{\overset{1}{\int }}\frac{{\text{artanh}}^{n}x}{x^m}dx=\underset{0}{\overset{1}{\int }}{\log }^{n/2}\left(\frac{1+x}{1-x}\right)\frac{dx}{x^m}=\underset{0}{\overset{\mathrm{\infty }}{\int }}{z}^n\frac{{\coth }^{m-2}z}{{\sinh }^{2}z}dz.$

These integrals have the closed form

$(1)I(n,m)=\frac{n!}{2^{n-1}}{\zeta }^{n+1}{g}_{m-1}(\frac{1}{\zeta })$

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 $n⩾7$.

2. Likewise take for $n⩾m⩾2$

$J(n,m):=\underset{1}{\overset{\mathrm{\infty }}{\int }}\frac{{\text{arcoth}}^{n}x}{x^m}dx=\underset{1}{\overset{\mathrm{\infty }}{\int }}{\log }^{n/2}\left(\frac{x+1}{x-1}\right)\frac{dx}{x^m}=\underset{0}{\overset{\mathrm{\infty }}{\int }}{z}^n\frac{{\tanh }^{m-2}z}{{\cosh }^{2}z}dz.$

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

$(2)J(n,m)=\frac{n!}{2^{n-1}}{\eta }^{n+1}{g}_{m-1}(\frac{1}{\eta })$.

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

$(3)\underset{0}{\overset{\pi /2}{\int }}\frac{x^n}{{\tan }^{m}x}dx=\cos \left(\frac{m}{2}\pi \right)\frac{(\pi /2{)}^{n+1}}{n+1}+\cos \left(\frac{m-n-1}{2}\pi \right)\frac{n!m}{2^n}{\zeta }^{n+2}{g}_m(\frac{1}{\zeta })+\sum _{v=0}^n\cos \left(\frac{m-v-1}{2}\pi \right)\frac{n!m{\pi }^{n-v}}{(n-v)!2^n}{\eta }^{n+2}{g}_m(\frac{1}{\eta }).$

Note that for $n⩾m⩾2$, this also yields a closed form for the integrals

$\underset{0}{\overset{\mathrm{\infty }}{\int }}\frac{{\arctan }^{n}x}{x^m}dx=\underset{0}{\overset{\pi /2}{\int }}\frac{x^n}{{\tan }^{m}x}dx+\underset{0}{\overset{\pi /2}{\int }}\frac{x^n}{{\tan }^{m-2}x}dx.$

4. For $n⩾m⩾2$, define $K(n,m):=\underset{0}{\overset{\mathrm{\infty }}{\int }}\frac{{\tanh }^n(x)}{x^m}dx$.

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 $h_k$,

$(4)K(n,m):=\underset{0}{\overset{\mathrm{\infty }}{\int }}\frac{{\tanh }^n(x)}{x^m}dx=\frac{n\cdot 2^{m-1}}{(m-1)!}(-h{)}^{m-1}{g}_n(h).$

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

$K(5,3)=-\frac{2}{3}(3h_3+10h_5+2h_7)=-7\frac{\zeta (3)}{{\pi }^2}+310\frac{\zeta (5)}{{\pi }^4}-1905\frac{\zeta (7)}{{\pi }^6},$

$K(6,2)=\frac{4}{15}(23h_3+20h_5+2h_7),K(6,4)=\frac{4}{45}(23h_5+20h_7+2h_9).$

5. If $n+m$ is odd, the same integral is much more involved to evaluate, including the initial one $\underset{0}{\overset{\mathrm{\infty }}{\int }}\frac{{\tanh }^3(x)}{x^2}dx$. Yet it turns out that the pattern subsists if we define $s_k:={\eta }^{\prime }(-k)=2^{k+1}\zeta (-k)\ln 2-(2^{k+1}-1){\zeta }^{\prime }(-k)$, equivalently $s_k=\frac{\zeta (-k)}{{\zeta }^{\prime }(-k)}\eta (-k)+\zeta (-k)\eta (1)-\eta (-k)\eta (1)$. Then $K(n,m)$ has the following closed form in umbral notation, replacing $s^k$ by $s_k$:

$(5)K(n,m)=\underset{0}{\overset{\mathrm{\infty }}{\int }}\frac{{\tanh }^n(x)}{x^m}dx=\frac{n\cdot 2^m}{(m-1)!}(-s{)}^{m-2}{g}_n(s)$, e.g.

$K(5,4)=\frac{8}{9}(3s_3+10s_5+2s_7),K(6,3)=-\frac{8}{15}(23s_3+20s_5+2s_7),K(6,5)=-\frac{8}{45}(23s_5+20s_7+2s_9).$

Note that by virtue of the logarithmic derivative $\frac{{\zeta }^{\prime }}{\zeta }(s)+\frac{{\zeta }^{\prime }}{\zeta }(1-s)=\log \pi -\frac{1}{2}\frac{{\mathrm{\Gamma }}^{\prime }}{\mathrm{\Gamma }}\left(\frac{s}{2}\right)-\frac{1}{2}\frac{{\mathrm{\Gamma }}^{\prime }}{\mathrm{\Gamma }}\left(\frac{1-s}{2}\right)$ 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 $\frac{{\zeta }^{\prime }(2j)}{\zeta (2j)}$, e.g.

$K(5,4)=\frac{8}{9}(3s_3+10s_5+2s_7)=\frac{1}{9}\{\frac{1643}{420}-\frac{16}{315}\ln 2+3\frac{{\zeta }^{\prime }(4)}{\zeta (4)}-20\frac{{\zeta }^{\prime }(6)}{\zeta (6)}+17\frac{{\zeta }^{\prime }(8)}{\zeta (8)}\}.$

6. For $n<m$, 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.

$(6)K(n-1,n)-K(n,n+1)=\underset{0}{\overset{\mathrm{\infty }}{\int }}(\frac{{\tanh }^{n-1}(x)}{x^n}-\frac{{\tanh }^n(x)}{x^{n+1}})dx=-\frac{1}{n}+\frac{(n+1)\cdot 2^n}{(n-1)!}{s}^{n-2}{g}_n(s)$.

See also

• Bernoulli polynomials of the second kind
• Stirling polynomials
• Poly-Bernoulli number

References

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