Does the function $\Delta_{\alpha}(z)=\frac{1}{\alpha}\sum\limits_{\omega:\,\omega^\alpha=1}\omega^z$ have a name?

Question

Context

For $\alpha\in\mathbb{N}^{+}$, I define $$\bbox[15px,#FFECEC,border:2px groove #E74C3C]{ \Delta_{\alpha}(z)=\frac{1}{\alpha}\sum_{\omega:\,\omega^{\alpha}=1}\omega^z}$$ where $\omega$ are the roots of unity such that $\text{arg}(\omega)\in(-\pi,\pi]$.

In particular:

• $\displaystyle \Delta_1(z)=1$
• $\displaystyle \Delta_2(z)=\frac{1}{2}\left(1+e^{\pi i z}\right)$
• $\displaystyle \Delta_3(z)=\frac{1}{3}\left(1+2\cos\left(\frac{2\pi}{3}z\right)\right)$
• $\displaystyle \Delta_4(z)=\frac{1}{4}\left(1+2\cos\left(\frac{\pi}{2}z\right)+e^{\pi i z}\right)$
• $\displaystyle \Delta_5(z)=\frac{1}{5}\left(1+2\cos\left(\frac{2\pi}{5}z\right)+2\cos\left(\frac{4\pi}{5}z\right)\right)$
• etc...

Properties

This function has a lot of properties:

$1)$ Explicit formula:
$$\bbox[15px,#FFF4E0,border:2px groove #F39C12]{\displaystyle \Delta_{\alpha}(z)=\frac{1}{\alpha}\left(1+2\sum_{k=1}^{\lfloor\frac{\alpha-1}{2}\rfloor}\cos\left(\frac{2k\pi}{\alpha}z\right)+\delta_{\alpha|2}\cdot e^{\pi iz}\right)}$$

$2)$ Specific value in $j\in\mathbb{Z}$
$$\bbox[15px,#E0FFFF,border:2px groove #1ABC9C]{\displaystyle \Delta_{\alpha}(j)=\delta_{j|\alpha}=\begin{cases}1&\text{if }j\text{ is a multiple of }\alpha\\ 0&\text{otherwise}\end{cases}}$$

$3)$ Conjugate symmetric
$$\bbox[15px,#F5EEFB,border:2px groove #9B59B6]{\displaystyle \Delta_{\alpha}(-z^{*})^{*}=\Delta_{\alpha}(z)}$$

$4)$ Periodicity:
$$\bbox[15px,#E0EFFF,border:2px groove #0058B1]{\displaystyle \Delta_{\alpha}(z-\alpha)=\Delta_{\alpha}(z)}$$

$5)$ Link with the sinc function:
$$\bbox[15px,#FFF4E0,border:2px groove #F39C12]{\displaystyle \lim_{\alpha\to\infty}\Delta_{\alpha}(z)=\text{sinc}(z)=\begin{cases}\frac{\sin(\pi z)}{\pi z}&z\neq 0\\1&z=0\end{cases}}$$

$6)$ Closed under circular convolution (discrete and continue) $$\bbox[15px,#F5EEFB,border:2px groove #9B59B6]{\displaystyle (\Delta_{\alpha}\ast\delta_\beta)(z)=\begin{cases} \displaystyle\sum_{k=1}^{\text{lcm}(\alpha,\beta)}\Delta_{\alpha}(k)\Delta_{\beta}(z-k)\\ \displaystyle\int_0^{\text{lcm}(\alpha,\beta)}\Delta_{\alpha}(t)\Delta_{\beta}(z-t)\mathrm{d}t\end{cases}=\Delta_{\gcd(\alpha,\beta)}(z)}$$

In particular,
$6.1)$ If $\beta=1$, convex combination (discrete and continue):
$$\bbox[15px,#E9F7EF,border:2px groove #27AE60]{\displaystyle \sum_{k=1}^{\alpha}\Delta_{\alpha}(z-k)=1\qquad\qquad \int_0^{\alpha}\Delta_{\alpha}(t)\mathrm{d}t=1}$$ $6.2)$ If $\alpha=\beta$, idempotent function under discrete and continue circular convolution:
$$\bbox[15px,#E0EFFF,border:2px groove #0058B1]{\displaystyle \sum_{k=1}^\alpha \Delta_{\alpha}(k)\Delta_{\alpha}(z-k) = \Delta_{\alpha}(z)\qquad\qquad\int_0^{\alpha} \Delta_{\alpha}(t)\Delta_{\alpha}(z-t)\mathrm{d}t=\Delta_{\alpha}(z)}$$ $6.3)$ If $\alpha=\beta$ and $z\in \mathbb{Z}$, I get that has orthogonal translations in $\ell^2([0,\alpha])$ and $L^2([0,\alpha])$:
$$\bbox[15px,#E0EFFF,border:2px groove #0058B1]{\displaystyle\sum_{k=1}^{\alpha}\Delta_{\alpha}(k)\Delta_{\alpha}(k-j)^{*}=\Delta_{\alpha}(j)\qquad\qquad\int_{0}^{\alpha}\Delta_{\alpha}(t)\Delta_{\alpha}(t-j)^{*}\mathrm{d}t=\Delta_{\alpha}(j)}$$ $6.4)$ If $\alpha$ and $\beta$ are coprime ($\gcd(\alpha,\beta)=1$): $$\bbox[15px,#F5EEFB,border:2px groove #9B59B6]{\displaystyle \sum_{k=1}^{\alpha\beta}\Delta_{\alpha}(k)\Delta_{\beta}(z-k)=1\qquad\qquad\int_0^{\alpha\beta}\Delta_{\alpha}(t)\Delta_{\beta}(z-t)\mathrm{d}t=1}$$ $6.5)$ For properties 3 and 4 it is closed under circular cross-correlation (discrete and continue) $$\bbox[15px,#E0FFFF,border:2px groove #1ABC9C]{\displaystyle \begin{array}{c} ( \Delta_{\alpha}\star \Delta_{\beta})(z)=\begin{cases} \displaystyle\sum_{k=1}^{\text{lcm}(\alpha,\beta)} \Delta_{\alpha}(k)^{*}\Delta_{\beta}(z+k)\\ \displaystyle\int_0^{\text{lcm}(\alpha,\beta)} \Delta_{\alpha}(t)^{*}\Delta_{\beta}(z+t)\mathrm{d}t\end{cases}=\Delta_{\text{gcd}(\alpha,\beta)} (z)\end{array}} $$ In particular,
$6.6)$ If $\alpha=\beta$ idempotent function under discrete and continue cross-correlation (similar to $6.2$)
$6.7$) For $z=0$: inner product in $\ell^2([0,\text{lcm}(\alpha,\beta)])$ and $L^2([0,\text{lcm}(\alpha,\beta)])$: $$\bbox[15px,#E0FFFF,border:2px groove #1ABC9C]{\displaystyle \begin{array}{c} \displaystyle\left\langle \Delta_{\alpha}, \Delta_{\beta} \right\rangle_{\ell^2([0,\text{lcm}(\alpha,\beta)])}=\sum_{k=1}^{\text{lcm}(\alpha,\beta)} \Delta_{\alpha}(k)\Delta_{\beta}(k)^{*}=1\\ \displaystyle\left\langle \Delta_{\alpha},\Delta_{\beta}\right\rangle_{L^2([0,\text{lcm}(\alpha,\beta)])}=\int_0^{\text{lcm}(\alpha,\beta)} \Delta_{\alpha}(t)\Delta_{\beta}(t)^{*}\mathrm{d}t=1 \end{array}} $$

$7)$ [From the comments] it can be seen as shifted inverse DFT of $(1,1,1,...)$ $$\bbox[15px,#FFF4E0,border:2px groove #F39C12]{\Delta_{\alpha}(z)=\frac{1}{\alpha}\sum_{k=\left\lceil\frac{1-\alpha}{2}\right\rceil}^{\left\lfloor\frac{\alpha}{2}\right\rfloor}e^{-\frac{2\pi ik}{\alpha}x}=\frac{1}{\alpha}\sum_{k=1}^{\alpha}\exp\left(i\arg\left(e^{\frac{2k\pi i}{\alpha}}\right)x\right)}$$

$\color{red}{\text{[Just of notation]}}$ in analogy with the continuous Fourier transform and the 2th property, I think it might make sense to indicate $\Delta_{\alpha}(z)$ as a sort of analytical continuation of a periodic Kroneker $\delta$ with period $\alpha$

8. Multidimensional extension I define $$\bbox[15px,#FFECEC,border:2px groove #E74C3C]{\displaystyle\Delta_{\alpha}(z_1,...z_d)=\prod_{j=1}^{d}\Delta_{\alpha}(z_j)}$$ The properties on individual arguments continue to apply, in particular I define $\mathbf{z}=(z_1,...,z_d)$, $\mathbf{k}=(k_1,...,k_d)$ and $\mathbf{t}=(t_1,...,t_d)$ $$\bbox[15px,#E0FFFF,border:2px groove #1ABC9C]{(\delta_{|n}\ast\delta_m)(\mathbf{z})=\begin{cases}\displaystyle\sum_{k_1=1}^{\text{lcm}(\alpha,\beta)}\cdots\sum_{k_d=1}^{\text{lcm}(m,n)}\Delta_{\alpha}(\mathbf{k})\Delta_{\beta}(\mathbf{z}-\mathbf{k})\\ \displaystyle\int_{[0,\text{lcm}(\alpha,\beta)]^d}\Delta_{\alpha}(\mathbf{t})\Delta_{\beta}(\mathbf{z}-\mathbf{t})\mathrm{d}\mathbf{t}\end{cases}=\Delta_{\gcd(m,n)}(\mathbf{z})}$$ $9)$ The absolute value is a ratio of sinc

$$\left|\Delta_{\alpha}(z)\right|=\left|\frac{\text{sinc}(z)}{\text{sinc}\left(\frac{z}{\alpha}\right)}\right|$$

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Application 1 - General boundary condition

$\Delta_{\alpha}(z)$ is the analytic solution of $$\begin{cases} f(z-\alpha)=f(z)&\forall z\\ f(z)=1&\text{for }z=0\\ f(z)=0&\text{for }z=1,...,\alpha \end{cases} $$ and $$\begin{cases} f(z)+f(z-1)+...+f(z-(\alpha-1))=1&\forall z\\ f(z)=1&\text{for }z=0\\ f(z)=0&\text{for }z=1,...,\alpha \end{cases} $$ that minimize oscillations, so in general, if I have a system like this $$\bbox[15px,#FFF4E0,border:2px groove #F39C12]{ \begin{array}{c} \begin{cases} f(z-\alpha)=f(z)&\forall z\\ f(z)=c_j&z=j\in\{0,...,\alpha-1\} \end{cases}\\ \text{The analytic solution that minimize oscillation is}\\ \displaystyle f(z)=\sum_{j=0}^{\alpha-1}c_j \cdot \Delta_{\alpha}(z-j)\\ \text{and if }c_j\neq 0\text{ I have an other possible solution:}\\ \displaystyle f(z)=\prod_{j=0}^{\alpha-1}c_j^{\Delta_{\alpha}(z-j)}\\ \end{array}}$$

I specified "that minimizes oscillations" because the first and second systems have infinite harmonic solutions, in particular the generic solution of the first system is $$\Delta_{\alpha}(\mathbf{C}|z)=\frac{1}{\alpha}\sum_{k=0}^{\alpha-1}\sum_{j\in\mathbb{Z}}c_{j,k}\exp\left(2\left(j+\frac{k}{\alpha}\right)\pi iz\right)\qquad\text{where }\sum_{j\in\mathbb{Z}}c_{j,k}=1\quad\forall k$$ and the generic solution of the second system is $$\Delta_{\alpha}(\mathbf{C}|z)=\frac{1}{\alpha}\left[1+\sum_{k=1}^{\alpha-1}\sum_{j\in\mathbb{Z}\setminus\{0\}}c_{j,k}\exp\left(2\left(j+\frac{k}{\alpha}\right)\pi iz\right)\right]\qquad\text{where }\sum_{j\in\mathbb{Z}\setminus\{0\}}c_{j,k}=1\quad\forall k$$

Application 2 - multifactorial

It can be also used to find an analytic continuation of the multifactorial

$$\bbox[15px,#E0FFFF,border:2px groove #1ABC9C]{\begin{cases} x!_{(\alpha)}=x(x-\alpha)!_{(\alpha)}&\forall x\\ x!_{(\alpha)}=k&k=1,...,\alpha \end{cases}}$$

(calculations)... a solution is

$$\bbox[15px,#E0FFFF,border:2px groove #1ABC9C]{x!_{(\alpha)}=\frac{\alpha^{1+\frac{x}{\alpha}}\Gamma\left(1+\frac{x}{\alpha}\right)}{\displaystyle\prod_{j=1}^{\alpha}\left(\alpha^{\frac{j}{\alpha}}\Gamma\left(\frac{j}{\alpha}\right)\right)^{\Delta_{\alpha}(x-j)}}}$$

And also respect the property

$$\prod_{j=0}^{\alpha}(x-j)!_{(\alpha)}=\Gamma(x+1)\qquad\forall x$$

Or alternatively, if you want the real solution, let $C_{\alpha}(x)=\Re[\Delta_{\alpha}(x)]$, then

$$\bbox[15px,#E0FFFF,border:2px groove #1ABC9C]{x!_{(\alpha)}=\frac{\alpha^{1+\frac{x}{\alpha}}\Gamma\left(1+\frac{x}{\alpha}\right)}{\displaystyle\prod_{j=1}^{\alpha}\left(\alpha^{\frac{j}{\alpha}}\Gamma\left(\frac{j}{\alpha}\right)\right)^{C_{\alpha}(x-j)}}}$$

Application 3 - general equation

In general if

$$\begin{cases} f(x)=g(x) f(x-\alpha)&\forall x\\ f(x)=c_j&x=j=\{1,...,\alpha\}\\ \end{cases}$$

then

$$f(x)=G_{\alpha}(x)\prod_{j=1}^{\alpha}\left(\frac{c_j}{G_{\alpha}(j)}\right)^{\Delta_{\alpha}\left(x-j\right)}\qquad\text{or}\qquad f(x)=G_{\alpha}(x)\sum_{j=1}^{\alpha}\frac{c_j}{G_{\alpha}(j)}\Delta_{\alpha}\left(x-j\right)$$

• If $g(x)=x^n$ then $G_{\alpha}(x)=\left(\alpha^{\frac{x}{\alpha}}\Gamma\left(1+\frac{x}{\alpha}\right)\right)^n$
• If $g(x)=e^x$ then $G_{\alpha}(x)=e^{\frac{x(x-\alpha)}{2\alpha}}$

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Question

Has this function a name? Or is it connected to some known function?

• This looks like the Dirichlet kernel but it is not the same.

• I also found something similar for $z\in\mathbb{N}$: here it is referred to as $\text{SR}(n)$ but it does not specify the type of root from which it is calculated and $n$ is just a natural number

• It seems quite interesting to me as a function because it works like a continuous Kroneker delta and in general the fact that it minimizes oscillations makes it a "better" solution than the others

• The way it's defined by the system $$\begin{cases} f(x-\alpha)=f(x)&\forall x\\ f(j)=\delta_{j|\alpha}&j=0,...,\alpha-1 \end{cases}$$ reminds me a lot of some kind of Green function.

Answer 1

(This was answered before $F_n(z)$ got renamed to $\delta_{z\vert n}$ and before the argument of $-1$ got changed from $-\pi$ to $\pi$.)

First note that

$$F_n(z)=\frac{1}{n}D_{\frac{n-1}{2}}\left(\frac{2\pi z}{n}\right)=\frac1 n\frac{\sin \pi z}{\sin\left(\frac{\pi z}{n}\right)}$$ for $n$ odd and $$F_n(z)=\frac{1}{n}D_{\frac{n-2}{2}}\left(\frac{2\pi z}{n}\right)+\frac{e^{-\pi i z}}{n}=\frac1 n\frac{\sin\left(\frac{(n-1)\pi z}{n}\right)}{\sin\left(\frac{\pi z}{n}\right)}+\frac{1}{n}e^{-\pi i z}$$ for $n$ even where $D$ is the Dirichlet kernel.

Looking at this I'd confidently say that these functions are unnamed yet - at least in the sense that there is no consensus on the name - though there might be someone out there who has given them a name or designated a letter to them for some purpose. (Since they are reparametrized Dirichlet kernels.)

Also note that by a straightforward calculation starting from the integral suggested by @Jack in the comments with the contour I've mentioned one gets for $-n<\Re(z)<0$ that $$F_n(z)=-\frac{\sin\pi z}{\pi} I_n(z)$$ for $n$ odd and $$F_n(z)=-\frac{\sin\pi z}{\pi} I_n(z)-\frac{\sin\pi z}{n}i$$

for $n$ even (which is to be understood as Cauchy's principal value) where $$I_n(z)=\int\limits_0^\infty\frac{x^{z+n-1}}{x^n-(-1)^n}dx.$$

In the region considered we get this by evaluation the pieces of the contour separately while letting the inner circle with radius $r$ got to zero and letting the outer circle with radius $R$ go to infinity. In said region the outer circle part vanishes for $R\to\infty$ and the part with the inner circle vanishes for $r\to0^+$. In the case of $n$ even we have two parts along the indent, say a half circle with radius $\tilde r$, which gives the term $-\frac{\sin\pi z}{n}i$ for $\tilde r\to0^+$. The two parts going along the negative real axis give the term $-\frac{\sin\pi z}{\pi} I_n(z)$ for $R\to\infty$, $r\to0^+$ and $\tilde r\to0^+$. (The arguments along the contour are to be chosen such that they vary continuously and are $0$ on the positive real axis. In the final integral the argument of $x$ is $0$.)

Note that by periodicity we can use the integral to evaluate $F_n$ at all $z\in\mathbb{C}$, by translating by a multiple of $n$, except if $\Re(z)$ is a multiple of $n$.