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\begin{document}

\vspace{1.3cm}

\title {Degenerate Bernstein polynomials}

\author{Taekyun Kim}
\address{Department of Mathematics, Kwangwoon University, Seoul 139-701, Republic of Korea}
\email{tkkim@kw.ac.kr}

\author{Dae San Kim}
\address{Department of Mathematics, Sogang University, Seoul 121-742, Republic of Korea}
\email{dskim@sogang.ac.kr}

\subjclass[2010]{11B83}
\keywords{degenerate Bernstein polnomials}

\begin{abstract}
Here we consider the degenerate Bernstein polynomials as a degenerate version of Bernstein polynomials, which are motivated by Simsek's recent work 'Generating functions for unification of the multidimensional Bernstein polynomials and their applications'([15,16]) and Carlitz's degenerate Bernoulli polynomials. We derived thier generating function, symmetric identities, recurrence relations, and some connections with generalized falling factorial polynomials, higher-order degenerate Bernoulli polynomials and degenerate Stirling numbers of the second kind.
\end{abstract}

\maketitle

\section{Introduction}

For $\lambda \in \mathbb{R}$, the degenerate Bernoulli polynomials of order $k$ are defined by L. Cartliz as
\begin{equation}\begin{split}\label{01}
\left( \frac{t}{(1+\lambda t)^{\frac{1}{\lambda }}-1} \right)^k (1+\lambda t)^{\frac{x}{\lambda }} = \sum_{n=0}^\infty  \beta_{n,\lambda }^{(k)}      \frac{t^n}{n!},\quad (\text{see}\,\, [4,5]).
\end{split}\end{equation}

Note that $\lim_{\lambda \rightarrow 0} \beta_{n,\lambda }^{(k)} (x) = B_n^{(k)}(x)$ are the ordinary Bernoulli polynomials of order $k$ given by
\begin{equation*}\begin{split}
\left( \frac{t}{e^t-1}\right)^k e^{xt} = \sum_{n=0}^\infty    B_n^{(k)}(x)    \frac{t^n}{n!},\quad (\text{see}\,\, [1,12,13]).
\end{split}\end{equation*}

It is known that the Stirling number of the second kind is defined as
\begin{equation}\begin{split}\label{02}
x^n = \sum_{l=0}^n S_2(n,l) (x)_l,\quad (\text{see}\,\, [2,4,8,10]),
\end{split}\end{equation}
where $(x)_l = x(x-1) \cdots (x-l+1)$, $(l \geq 1),$ $(x)_0=1$.

For $\lambda \in \mathbb{R}$, the $(x)_{n,\lambda}$ is defined as
\begin{equation}\begin{split}\label{03}
(x)_{0,\lambda }=1,\,\,(x)_{n,\lambda }=x(x-\lambda )(x-2\lambda )\cdots(x-(n-1)\lambda ),\,\,(n \geq 1)
\end{split}\end{equation}

In [8,9,10], ${x \choose n}_\lambda$ is defined as
\begin{equation}\begin{split}\label{04}
{x \choose n}_\lambda  = \frac{(x)_{n,\lambda }}{n!} = \frac{x(x-\lambda )\cdots(x-(n-1)\lambda )}{n!},\,\,(n \geq 1),\,\, {x \choose 0}_\lambda  =1.
\end{split}\end{equation}

Thus, by \eqref{04}, we get
\begin{equation}\begin{split}\label{05}
(1+\lambda t)^{\frac{x}{\lambda }} = \sum_{n=0}^\infty {x \choose n}_\lambda  t^n,\quad (\text{see}\,\, [7]).
\end{split}\end{equation}

From \eqref{05}, we note that
\begin{equation}\begin{split}\label{06}
\sum_{m=0}^n {y \choose m}_\lambda  {x \choose n-m}_\lambda  = {x+y \choose n}_\lambda ,\,\,(n \geq 0).
\end{split}\end{equation}

The degenerate Stirling numbers of the second kind are defined by
\begin{equation}\begin{split}\label{07}
\frac{1}{k!} \big( (1+\lambda t)^{\frac{1}{\lambda }}-1\big)^k = \sum_{n=k}^\infty S_{2,\lambda }(n,k) \frac{t^n}{n!},\,\,(k \geq 0),\quad (\text{see}\,\, [7,8]).
\end{split}\end{equation}

By \eqref{07}, we easily get
\begin{equation*}\begin{split}
\lim_{\lambda  \rightarrow 0} S_{2,\lambda }(n,k) = S_2(n,k),\,\,(n \geq k \geq 0),\quad (\text{see}\,\, [8,10]).
\end{split}\end{equation*}

In this paper, we use the following notation.
\begin{equation}\begin{split}\label{08}
(x \oplus_\lambda  y)^n = \sum_{k=0}^n {n \choose k} (x)_{k,\lambda } (y)_{n-k,\lambda },\,\,(n \geq 0).
\end{split}\end{equation}

The Bernstein polynomials of degree $n$ is defined by
\begin{equation}\begin{split}\label{09}
B_{k,n}(x) = {n \choose k} x^k (1-x)^{n-k},\,\,(n \geq k \geq 0),\quad (\text{see}\,\, [6,11,17]).
\end{split}\end{equation}

Let $C[0,1]$ be the space of continuous functions on $[0,1]$. The Bernstein operator of order $n$ for $f$ is given by
\begin{equation}\begin{split}\label{10}
\mathbb{B}_n (f|x) = \sum_{k=0}^n f (\tfrac{k}{n} ) {n \choose k} x^k (1-x)^{n-k} = \sum_{k=0}^n  f (\tfrac{k}{n} ) B_{k,n}(x),
\end{split}\end{equation}
where $n \in \mathbb{N} \cup \{0\}$ and $f \in C[0,1]$, (see [3,6,14]).

A Bernoulli trial involves performing a random experiment and noting whether a particular event A occurs.
The outcome of Bernoulli trial is said to be "success" if A occurs and a "failure" otherwise. The probability $P_n(k)$ of $k$ successes in $n$ independent Bernoulli trials is given by the binomial probability law:
\begin{equation*}\begin{split}
P_n (k) = {n \choose k} p^k (1-p)^{n-k},\,\,\text{for}\,\,k=0,1,2,\cdots,
\end{split}\end{equation*}
From the definition of Bernstein polynomials we note that Bernstein basis is probability mass of binomial distribution with parameter $(n, x=p)$.

Let us assume that the probability of success in an experiment is $p$. We wondered if we can say the probability of success in the nineth trial is still $p$ after failing eight times in a ten trial experiment.  Because there's a psychological burden to be successful.

It seems plausible that the probability is less than $p$. This speculation motivated the study of the degenerate Bernstein polynomials associated with the probability distribution.

In this paper, we consider the degenerate Bernstein polynomials as a degenerate version of Bernstein polynomials. We derive thier generating function, symmetric identities, recurrence relations, and some connections with generalized falling factorial polynomials, higher-order degenerate Bernoulli polynomials and degenerate Stirling numbers of the second kind.

\section{Degenerate Bernstein polynomials}

For $\lambda \in \mathbb{R}$ and $k,n \in \mathbb{N}\cup \{0\}$, with $k \leq n$, we define the degenerate Bernstein polynomials of degree $n$ which are given by
\begin{equation}\begin{split}\label{11}
B_{k,n}(x|\lambda ) = {n \choose k} (x)_{k,\lambda }(1-x)_{n-k,\lambda },\,\,(x \in [0,1]).
\end{split}\end{equation}

Note that $\lim_{\lambda \rightarrow 0} B_{k,n}(x|\lambda ) = B_{k,n}(x)$, $(0 \leq k \leq n)$.
From \eqref{11}, we derive the generating function of $B_{k,n}(x|\lambda )$, which are given by
\begin{equation}\begin{split}\label{12}
\sum_{n=k}^\infty B_{k,n}(x|\lambda ) \frac{t^n}{n!} &= \sum_{n=k}^\infty {n \choose k} (x)_{k,\lambda }(1-x)_{n-k,\lambda } \frac{t^n}{n!}\\
&= \frac{(x)_{k,\lambda}}{k!} \sum_{n=k}^\infty \frac{1}{(n-k)!} (1-x)_{n-k,\lambda }t^n\\
&= \frac{(x)_{k,\lambda }}{k!} \sum_{n=0}^\infty \frac{(1-x)_{n,\lambda }}{n!} t^{n+k} \\
&= \frac{(x)_{k,\lambda }}{k!} t^k \sum_{n=0}^\infty {1-x \choose n}_\lambda  t^n \\
&= \frac{(x)_{k,\lambda }}{k!}  t^k (1+\lambda t)^{\frac{1-x}{\lambda }}.
\end{split}\end{equation}

Therefore, by \eqref{12}, we obtain the following theorem.
\begin{thm}
For $x \in [0,1]$ and $k=0,1,2,\cdots,$ we have
\begin{equation*}\begin{split}
\frac{1}{k!} (x)_{k,\lambda } t^k (1+\lambda t)^{\frac{1-x}{\lambda }} = \sum_{n=k}^\infty B_{k,n}(x|\lambda ) \frac{t^n}{n!}.
\end{split}\end{equation*}
\end{thm}

From \eqref{11}, we note that
\begin{equation}\begin{split}\label{13}
B_{k,n}(x|\lambda ) = {n \choose k} (x)_{k,\lambda } (1-x)_{n-k,\lambda } = {n \choose n-k} (x)_{k,\lambda }(1-x)_{n-k,\lambda }.
\end{split}\end{equation}

By replacing $x$ by $1-x$, we get
\begin{equation}\begin{split}\label{14}
B_{k,n}(1-x|\lambda ) = {n \choose n-k} (1-x)_{k,\lambda }(x)_{n-k,\lambda } = B_{n-k,n}(x|\lambda ),
\end{split}\end{equation}
where $n,k \in \mathbb{N} \cup \{0\}$, with $0 \leq k \leq n$.

Therefore, by \eqref{14}, we obtain the following theorem.
\begin{thm} (\textnormal{Symmetric identities})
For $n,k \in \mathbb{N} \cup \{0\}$, with $k \leq n$, and $x \in [0,1]$, we have
\begin{equation*}\begin{split}
B_{n-k,n}(x|\lambda )=B_{k,n}(1-x|\lambda )  .
\end{split}\end{equation*}
\end{thm}

Now, we observe that
\begin{equation}\begin{split}\label{15}
&\frac{n-k}{n} B_{k,n}(x|\lambda ) + \frac{k+1}{n} B_{k+1,n}(x|\lambda )\\
&= \frac{n-k}{n} {n \choose k} (x)_{k,\lambda } (1-x)_{n-k,\lambda } + \frac{k+1}{n} {n \choose k+1} (x)_{k+1,\lambda }(1-x)_{n-k-1,\lambda }\\
&= \frac{(n-1)!}{k! (n-k-1)!} (x)_{k,\lambda }(1-x)_{n-k,\lambda } + \frac{(n-1)!}{k! (n-k-1)!} (x)_{k+1,\lambda } (1-x)_{n-k-1,\lambda }\\
&=(1-x-(n-k-1)\lambda ) B_{k,n-1}(x|\lambda ) + (x-k\lambda )B_{k,n-1}(x|\lambda )\\
&= (1+ \lambda (1-n))B_{k,n-1}(x|\lambda).
\end{split}\end{equation}

Therefore, by \eqref{15}, we obtain the following theorem.
\begin{thm}
For $k \in \mathbb{N} \cup \{0\}$, $n \in \mathbb{N}$, with $k \leq n-1$, and $x \in [0,1]$, we have
\begin{equation}\begin{split}\label{16}
(n-k)B_{k,n}(x|\lambda ) + (k+1)B_{k+1,n}(x|\lambda )=(1+ \lambda (1-n))B_{k,n-1}(x|\lambda).
\end{split}\end{equation}
\end{thm}

From \eqref{11}, we have
\begin{equation}\begin{split}\label{17}
&\left( \frac{n-k+1}{k} \right) \left( \frac{n-(k-1)\lambda }{1-x-(n-k)\lambda } \right) B_{k-1,n}(x|\lambda ) \\
&=\left( \frac{n-k+1}{k} \right) \left( \frac{n-(k-1)\lambda }{1-x-(n-k)\lambda } \right)
{n \choose k-1} (x)_{k-1,\lambda }(1-x)_{n-k+1,\lambda }\\
&= \frac{n!}{k! (n-k)!} (x)_{k-1,\lambda }(1-x)_{n-k,\lambda } = B_{k,n}(x|\lambda ).
\end{split}\end{equation}

Therefore, by \eqref{17}, we obtain the following theorem.
\begin{thm} For $n,k \in \mathbb{N}$, with $k \leq n$, we have
\begin{equation*}\begin{split}
\left( \frac{n-k+1}{k} \right) \left( \frac{n-(k-1)\lambda }{1-x-(n-k)\lambda } \right) B_{k-1,n}(x|\lambda ) = B_{k,n}(x|\lambda ).
\end{split}\end{equation*}
\end{thm}

For $0 \leq k \leq n$, we get
\begin{equation}\begin{split}\label{18}
&(1-x-(n-k-1)\lambda ) B_{k,n-1}(x|\lambda ) + (x-(k-1)\lambda )B_{k-1,n-1}(x|\lambda ) \\
&=(1-x-(n-k-1)\lambda ) {n-1 \choose k} (x)_{k,\lambda }(1-x)_{n-1-k,\lambda } \\
&\quad + (x-(k-1)\lambda ){n-1 \choose k-1}(x)_{k-1,\lambda }(1-x)_{n-k,\lambda }\\
&={n-1 \choose k} (x)_{k,\lambda }(1-x)_{n-k,\lambda }+ {n-1 \choose k-1} (x)_{k,\lambda } (1-x)_{n-k,\lambda }\\
&=\left( {n-1 \choose k}+{n-1 \choose k-1} \right) (x)_{k,\lambda } (1-x)_{n-k,\lambda } = {n \choose k} (x)_{k,\lambda } (1-x)_{n-k,\lambda }.
\end{split}\end{equation}

Therefore, by \eqref{18}, we obtain the following theorem.
\begin{thm}(\textnormal{Recurrence formula}).
For $k,n \in \mathbb{N}$, with $k \leq n-1$, $x \in [0,1]$, we have
\begin{equation*}\begin{split}
(1-x-(n-k-1)\lambda ) B_{k,n-1}(x|\lambda ) + (x-(k-1)\lambda )B_{k-1,n-1}(x|\lambda )=B_{k,n}(x|\lambda ).
\end{split}\end{equation*}
\end{thm}

\begin{rmk}For $n\in \mathbb{N}$, we have
\begin{equation*}\begin{split}
&\sum_{k=0}^n \frac{k}{n} B_{k,n}(x|\lambda ) = \sum_{k=0}^n \frac{k}{n} {n \choose k} (x)_{k,\lambda } (1-x)_{n-k,\lambda }\\
&= \sum_{k=1}^n {n-1 \choose k-1} (x)_{k,\lambda } (1-x)_{n-k,\lambda } = \sum_{k=0}^{n-1} {n-1 \choose k} (x)_{k+1,\lambda }(1-x)_{n-1-k,\lambda }\\
&=(x-k\lambda ) \sum_{k=0}^{n-1} {n-1 \choose k} (x)_{k,\lambda } (1-x)_{n-1-k,\lambda } = (x-k\lambda ) (x \oplus_\lambda  (1-x) )^{n-1}.
\end{split}\end{equation*}
\end{rmk}

Now, we observe that
\begin{equation}\begin{split}\label{19}
\sum_{k=2}^n \frac{{k \choose 2}}{{n \choose 2}} B_{k,n}(x|\lambda ) &=
\sum_{k=2}^n \frac{k(k-1)}{n(n-1)} {n \choose k} (x)_{k,\lambda }(1-x)_{n-k,\lambda }\\
&= \sum_{k=2}^n \frac{k(k-1)}{n(n-1)} {n \choose k} (x)_{k,\lambda }(1-x)_{n-k,\lambda }\\
&=\sum_{k=2}^n {n-2 \choose k-2} (x)_{k,\lambda } (1-x)_{n-k,\lambda }\\
&= \sum_{k=0}^{n-2} {n-2 \choose k} (x)_{k+2,\lambda }(1-x)_{n-2-k,\lambda }\\
&= (x-k\lambda )(x-(k+1)\lambda ) \sum_{k=0}^{n-2} {n-2 \choose k} (x)_{k,\lambda } (1-x)_{n-2-k,\lambda }.
\end{split}\end{equation}

Similarly, we have
\begin{equation}\begin{split}\label{20}
\sum_{k=i}^n \frac{{k \choose i}}{{n \choose i}} B_{k,n}(x|\lambda ) &= (x-k\lambda )_{i,\lambda } \sum_{k=0}^{n-i} {n-i \choose k} (x)_{k,\lambda } (1-x)_{n-i-k,\lambda }\\
&= (x-k\lambda )_{i,\lambda } (x \oplus_\lambda  (1-x))^{n-i}.
\end{split}\end{equation}

From \eqref{20}, we note that
\begin{equation}\begin{split}\label{21}
(x-k\lambda)_{i,\lambda } = \frac{1}{(x \oplus_\lambda  (1-x))^{n-i}} \sum_{k=i}^n \frac{{k \choose i}}{{n \choose i}} B_{k,n}(x|\lambda ),
\end{split}\end{equation}
where $n,i \in \mathbb{N}$, with $ i \leq n$, and $x \in [0,1]$.

Therefore, by \eqref{21}, we obtain the following theorem.
\begin{thm}
For $n,i \in \mathbb{N}$, with $ i \leq n$, and $x \in [0,1]$, we have
\begin{equation*}\begin{split}
(x-k\lambda)_{i,\lambda } = \frac{1}{(x \oplus_\lambda  (1-x))^{n-i}} \sum_{k=i}^n \frac{{k \choose i}}{{n \choose i}} B_{k,n}(x|\lambda ).
\end{split}\end{equation*}
\end{thm}

From Theorem 2.1, we note that
\begin{equation}\begin{split}\label{22}
&\frac{t^k}{k!} (x)_{k,\lambda }(1+\lambda t)^{\frac{1-x}{\lambda }} = \frac{(x)_{k,\lambda }}{k!} \Big( (1+\lambda t)^{\frac{1}{\lambda }}-1 \Big)^k \left(\frac{t}{(1+\lambda t)^{\frac{1}{\lambda }}-1} \right)^k (1+\lambda t)^{\frac{1-x}{\lambda }}\\
&= (x)_{k,\lambda } \left( \sum_{m=k}^\infty S_{2,\lambda }(m,k) \frac{t^m}{m!} \right) \left( \sum_{l=0}^\infty \beta_{l,\lambda }^{(k)} (1-x) \frac{t^l}{l!} \right)\\
&= (x)_{k,\lambda } \sum_{n=k}^\infty \left( \sum_{m=k}^n {n \choose m} S_{2,\lambda }(m,k) \beta_{n-m,\lambda }^{(k)} (1-x) \right) \frac{t^n}{n!}.
\end{split}\end{equation}

On the other hand,
\begin{equation}\begin{split}\label{23}
\frac{(x)_{k,\lambda }}{k!} t^k (1+\lambda t)^{\frac{1-x}{\lambda }} = \sum_{n=k}^\infty B_{k,n}(x|\lambda ) \frac{t^n}{n!}.
\end{split}\end{equation}

Therefore, by \eqref{22} and \eqref{23}, we obtain the following theorem.

\begin{thm}
For $n,k \in \mathbb{N}\cup \{0\}$ with $n \geq k$, we have
\begin{equation*}\begin{split}
 B_{k,n}(x|\lambda ) = (x)_{k,\lambda } \sum_{m=k}^n {n \choose m} S_{2,\lambda }(m,k) \beta_{n-m,\lambda }^{(k)} (1-x).
 \end{split}\end{equation*}
\end{thm}

Let $\Delta$ be the shift difference operator with $\Delta f(x) = f(x+1)-f(x)$. Then we easily get
\begin{equation}\begin{split}\label{24}
\Delta^n f(0) = \sum_{k=0}^n {n \choose k} (-1)^{n-k} f(k), \,\,(n\in \mathbb{N}\cup \{0\}).
\end{split}\end{equation}

Let us take $f(x)=(x)_{m,\lambda }$, $(m \geq 0)$. Then, by \eqref{24}, we get
\begin{equation}\begin{split}\label{25}
\Delta^n (0)_{m,\lambda } = \sum_{k=0}^n {n \choose k} (-1)^{n-k} (k)_{m,\lambda }.
\end{split}\end{equation}

From \eqref{07}, we note that
\begin{equation}\begin{split}\label{26}
\sum_{n=k}^\infty S_{2,\lambda }(n,k) \frac{t^n}{n!}&=
\frac{1}{k!} \Big( (1+\lambda t)^{\frac{1}{\lambda }}-1 \Big)^k = \frac{1}{k!} \sum_{l=0}^k {k \choose l} (-1)^{k-l} (1+\lambda t)^{\frac{l}{\lambda }}\\
&= \sum_{n=0}^\infty \left( \frac{1}{k!} \sum_{l=0}^k  {k \choose l} (-1)^{k-l} (l)_{n,\lambda } \right) \frac{t^n}{n!}.
\end{split}\end{equation}

Thus, by comparing the coefficients on both sides of \eqref{26}, we have
\begin{equation}\begin{split}\label{27}
\frac{1}{k!} \Delta^k (0)_{n,\lambda } = \frac{1}{k!} \sum_{l=0}^k {k \choose l} (-1)^{k-l} (l)_{n,\lambda } = \begin{cases}
S_{2,\lambda }(n,k)&\text{if}\,\,n \geq k,\\
0&\text{if}\,\,n<k.
\end{cases}
\end{split}\end{equation}

By \eqref{27}, we get
\begin{equation}\begin{split}\label{28}
\frac{1}{k!} \Delta^k (0)_{n,\lambda } = S_{2,\lambda }(n,k),\,\,\text{if} \,\,n \geq k.
\end{split}\end{equation}

From Theorem 7 and \eqref{28}, we obtain the following corollary.
\begin{cor}
For $n,k \in \mathbb{N}\cup \{0\}$ with $n \geq k$, we have
\begin{equation*}\begin{split}
B_{k,n}(x|\lambda ) = (x)_{k,\lambda }\sum_{m=k}^n {n \choose m} \beta_{n-m,\lambda }^{(k)}(1-x) \frac{1}{k!} \Delta^k (0)_{m,\lambda }.
\end{split}\end{equation*}
\end{cor}

Now, we observe that
\begin{equation}\begin{split}\label{29}
(1+\lambda t)^{\frac{x}{\lambda }}&= \Big( (1+\lambda t)^{\frac{1}{\lambda }}-1+1)^x = \sum_{k=0}^\infty {x \choose k} \Big( (1+\lambda t)^{\frac{1}{\lambda }}-1 \Big)^k \\
&= \sum_{k=0}^\infty (x)_k\frac{1}{k!} \Big( (1+\lambda t)^{\frac{1}{\lambda }}-1 \Big)^k \\
&= \sum_{k=0}^\infty (x)_k \sum_{n=k}^\infty S_{2,\lambda } (n,k) \frac{t^n}{n!}\\
&=\sum_{n=0}^\infty   \left( \sum_{k=0}^n  (x)_k S_{2,\lambda }(n,k) \right)
     \frac{t^n}{n!}.
\end{split}\end{equation}

On the other hand,
\begin{equation}\begin{split}\label{30}
(1+\lambda t)^{\frac{x}{\lambda }} = \sum_{n=0}^\infty { \frac{x}{\lambda } \choose n }
\lambda ^n t^n = \sum_{n=0}^\infty (x)_{n,\lambda }\frac{t^n}{n!}.
\end{split}\end{equation}

Therefore, by \eqref{29} and \eqref{30}, we obtain the following theorem.
\begin{thm}For $n \geq 0$, we have
\begin{equation*}\begin{split}
(x)_{n,\lambda } =  \sum_{k=0}^n (x)_k S_{2,\lambda }(n,k).
\end{split}\end{equation*}
\end{thm}

By Theorem 2.9, we easily get
\begin{equation}\begin{split}\label{31}
(x-k\lambda )_{i,\lambda } = \sum_{l=0}^i (x-k\lambda)_l S_{2,\lambda }(i,l).
\end{split}\end{equation}

From Theorem 2.6, we have the following theorem.
\begin{thm}
For $n,i \in \mathbb{N}$, with $i \leq n$, and $x \in [0,1]$, we have
\begin{equation*}\begin{split}
 \sum_{l=0}^i (x-k\lambda)_l S_{2,\lambda }(i,l) =\frac{1}{(x \oplus_\lambda  (1-x))^{n-i}} \sum_{k=i}^n \frac{{k \choose i}}{{n \choose i}} B_{k,n}(x|\lambda ).
\end{split}\end{equation*}
\end{thm}


TITLE: Specific birthday problem.
QUESTION [0 upvotes]: Consider the the place where one year equal to $2^{n}$ days.
Now, let's represent our birthday problem into this place. Consider the $k$ person among people from this planet.
Now, we have:
$$p(n,k) = 1-\frac{\prod_{i = 1}^{k}(2^{n} - i)}{2^{nk}}$$, is probability that $2$ people from this group have the same birthday.
I guess , that it's easy to get, so I would to know: does there some close form of this expression. Actually I want to represent this into irreducible fraction.

REPLY [1 votes]: You probably want this to solve this problem in codeforces:
Here are the key ideas to solve it.
If $k>MOD$ then $A$ is $0$, if $k>2^n$ then $A=1,B=1$.
Otherwise you can use modular arithmetic to calculate $\prod_{k=0}^{-1}n-i\bmod MOD$.
You can also calculate $2^{2^{k\times n}}$ fast using modular exponentiation.
finally you have to calculate the GCD of the numerator and the denominator. To do this you can use polignac's formula, notice that the $2^n$ in the numerator clearly cancels out. And after that you can notice that the maximum of power of $2$ dividing $2^n-j$ is the same as the maximum power of $2$ dividing $j$, for $1\leq j\leq k-1$.
So the term that needs to be canceled is $2^{n+P((k-1)!)}$.
Where $P(k!)$ is the maximum power of $2$ dividing $n!$, this can be calculated using polignac's formula.
Here is my c++ code:
#include <bits/stdc++.h>
using namespace std;
typedef long long lli;

lli MOD=1000003;

lli power(lli b, lli e){
    lli res=1;
    while(e){
        if(e%2) res=(res*b)%MOD;
        e/=2;
        b=(b*b)%MOD;
    }
    return(res);
}

lli poder(lli b, lli e){
    lli res=1;
    while(e){
        if(e%2) res=(res*b);
        e/=2;
        b=(b*b);
    }
    return(res);
}

lli inv(lli b){
    return(power(b,MOD-2));
}

lli polig(lli k){
    lli res=0;
    while(k){
        res+=k/2;
        k/=2;
    }
    return(res);
}

int main(){
    lli n,k;
    scanf("%lld %lld",&n,&k);
    if(n<=62 && k>poder(2,n) ){
        printf("1 1\n");
        return(0);
    }
    lli B=power(2,n);
    lli A=1;
    if(k>=MOD) A=0;
    else{
        for(int i=0;i<k;i++){
            A= (A*(MOD+B- (i%MOD) )%MOD )%MOD;
        }
    }
    lli cancel=n;
    cancel+=polig(k-1);
    B=power(B,k);
    A=(MOD+B-A)%MOD;
    A=( A*inv(power(2,cancel) ) ) %MOD;
    B=( B*inv(power(2,cancel) ) ) %MOD;
    printf("%lld %lld\n",A,B);
}


text:
\section{Localization of conormal spaces}
\label{sec:localization}
The goal of this section is to use \cref{thm:mather} to obtain formulae for the localization of conormal spaces.

Denote by $\iota:G/P \hookrightarrow T^*(G/P)$ the zero section.
By equation \eqref{E:cmath},
\begin{equation}
\label{E:pbi}
\iota^*[T^*_{X_w^P}(G/P)]_{T \times \C^*} = (-1)^{\ell(w)}\cMaT(X_w^P)^\hbar = (-1)^{\ell(w)}\pi_*((c^T(\mathcal{T}_w) \cap [X_w^B])^\hbar) \/.
\end{equation}
This is a class in $A_0^{T \times \C^*}(G/P)$ and it is obtained by homogenizing the equivariant (homology) class $\cMaT(X_w^P)$.
We can use \eqref{E:pbi} to calculate the localization at the fixed points of the conormal space.
Fix $u \in W^P$ such that $u \le w$.
If we write
\[
\cMaT(X_w^T) = \sum a_v [X_v^P] \/,
\]
where $a_v \in A^*_T(pt)$, then localizing at $u \in W$ gives
\[
\cMaT(X_w^T)|_u = \sum a_v [X_v^P]|_u \/.
\]
The coefficients $a_v$ are not homogeneous, thus homogenizing each term $(a_v [X_v^P]|_u)$ amounts to multiplying each homogeneous component by the appropriate power of
$\hbar \in A^1_{T \times \C^*}(pt)$ such that the total cohomological degree equals $\dim G/P$ (i.e. homological degree $0$.)

A more explicit formula is obtained by analyzing the homogenization $(c^T(\mathcal{T}_w) \cap [X_w^B])^\hbar$.
Let $a_1, \ldots , a_{\ell(w)}$ be the $T$-equivariant Chern roots of $\mathcal{T}_w$.
Then
\[
c(\mathcal{T}_w) \cap [X_w^B] = \Bigl(\prod_{i=1}^{\ell(w)}(1 + a_i)\Bigr)\cap [X_w^B] = \sum_{i \ge 0} e_i (a_1, \ldots, a_{\ell(w)}) \cap [X_w^B] \/,
\]
where $e_i$ denote the elementary symmetric functions.
The term $e_i (a_1, \ldots, a_{\ell(w)}) \cap[X_w^B]$ belongs to the equivariant Chow group $A_{\ell(w) -i}^T(G/B)$, therefore its homogenization by $\hbar$ is
\begin{equation}
\label{E:chomog}
(c(\mathcal{T}_w) \cap [X_w^B])^\hbar = \sum_{i \ge 0} \hbar^{\ell(w) - i}e_i (a_1, \ldots, a_{\ell(w)}) \cap [X_w^B] = \Bigl(\prod_{i=1}^{\ell(w)}(\hbar + a_i)\Bigr) \cap [X_w^B] \/.
\end{equation}
One key observation is that the quantity $(-1)^{\ell(w)}\prod_{i=1}^{\ell(w)}(\hbar + a_i)$ has a geometric interpretation.
If one considers the $\C^*$-action on the cotangent bundle $T^*(G/B)$ with character $\hbar^{-1}$, then the elements $-\hbar -a_i$ are the $T \times \C^*$-equivariant Chern roots of $\mathcal{T}_w^*$.
Thus
\begin{equation}
\label{E:topTC}
(-1)^{\ell(w)}\prod_{i=1}^{\ell(w)}(\hbar + a_i) = c^{T \times \C^*}_{\ell(w)}(\mathcal{T}^*_w) \/.
\end{equation}
The fibre of $\mathcal{T}^*_w$ over the fixed point $e_v$ is

\[
v.\bigoplus_{\alpha \in I(w)}\mathfrak{g}_{\alpha}\otimes \C_{-\hbar}= \bigoplus_{\alpha \in I(w)}\mathfrak{g}_{v(\alpha)}\otimes \C_{-\hbar}\/,
\]
and from this we deduce the formula for the localization:
\[
(c^{T\times \C^*}_{\ell(w)}(\mathcal{T}_w^*))|_{v}= \prod_{\alpha \in I(w)}(-\hbar + v(\alpha)) \/.
\]
Combining with the equation \eqref{E:chomog}, this proves the following Lemma:
\begin{lemma}
\label{lemma:ctwloc}
Let $v \le w$.
Then the following holds in $A^{\dim G/B}_{T \times \C^*}(pt)$:
\[
(-1)^{\ell(w)}((c(\mathcal{T}_w) \cap [X_w^B])^\hbar)|_v =  \prod_{\alpha \in I(w)}(-\hbar + v(\alpha)) \cdot [X_w^B]|_v \/.
\]
\end{lemma}
We note that since $\C^*$ acts trivially on $G/B$, the $T \times \C^*$-localization $[X_w^B]|_v$ is the same as
the $T$-equivariant localization.
A formula for the latter can be found in \cite[Thm. 11.1.7]{kumar:book};
see also \cite[App. D]{AJS:representations} or \cite{billey:kostant}.

The last step to calculate the localization of the class of the conormal space $T^*_{X_w^P}(G/P)$ is to relate the localization of the class from \cref{lemma:ctwloc} to the localization of its push forward.
For that, we need a generalization of the localization formula.
For $v \in W$, let $[e_v]$ denote the $T$-equivariant class of the fixed point $e_v$.
\begin{lemma}
\label{lemma:pf}
Let $u \in W^P$ and let $\kappa \in H^*_{T}(G/B)$ be any equivariant cohomology class.
Then
\[
\pi_*(\kappa)|_{uW_P}= \sum \frac{[e_{uW_P}]|_{uW_P}}{[e_v]|_v}\kappa|_v \/
\]
in an appropriate localization of $A^*_T(G/B)$, where the sum is over $v \in W$ such that $uW_P = v W_P$.
\end{lemma}
\begin{proof}
By injectivity of localization map,
the classes $[e_v]$ form a basis for the $T$-equivariant cohomology of $G/B$,
{\em localized} at the prime ideal generated by the equivariant parameters $A^*_{T}(pt)$.
Thus we can expand $\kappa = \sum c_v [e_v]$ where the sum is over $v \in W$.
Localizing both sides at $v \in W$, we obtain that $\kappa|_v = c_v [e_v]|_v$.
Pushing forward and localizing at $uW_P$ one obtains
\[
\pi_*(\kappa)|_{uW_P}= \sum_{v \in W}c_v \pi_*[e_v]|_{uW_P}= \sum_{v \in W}c_v [e_{vW_P}]|_{uW_P}= \sum_{v \in W}\frac{[e_{vW_P}]|_{uW_P}}{[e_v]|_v}\kappa|_v\/.
\]
Since $[e_{vW_P}]|_{uW_P}$ is nonzero only when $uW_P = v W_P$, the last sum is as in the statement of the lemma, and this finishes the proof.
\end{proof}

\begin{theorem}
\label{thm:conormalloc}
Let $u,w \in W^P$ such that $u \le w$, and let $\C^*$ act on $T^*(G/P)$ by character $\hbar^{-1}$.
Then the $T \times \C^*$-localization of the conormal space $T^*_{X_w^P}(G/P)$ is given by:
\[
[T^*_{X_w^P}(G/P)]|_{uW_P}=
\sum\limits_{\substack{v \le w \\ v W_P = u W_P}}\frac{\prod_{\alpha \in I(w)}(-\hbar + v(\alpha)) \cdot \prod_{\alpha \in R^+ \setminus R_P^+}u(-\alpha)}{\prod_{\alpha \in R^+}v(-\alpha)}[X_w^B]|_v \/.
\]
\end{theorem}
\begin{proof}
Let $\kappa = (-1)^{\ell(w)}(c^T(\mathcal{T}_w) \cap [X_w^B])^\hbar$, regarded as a cohomology class in $A^{*}_{T}(G/B)[\hbar]$.
By equation \eqref{E:pbi}, the left hand side equals to $\pi_*(\kappa)|_{uW_P}$.
Since $\C^*$ acts trivially on $G/B$, we have $A^{*}_{T \times \C^*}(G/B)= A^{*}_{T}(G/B)[\hbar]$,
and further, the projection $\pi_*$ is $A^*_{T \times \C^*}(pt)$-linear.
To finish the proof, apply \cref{lemma:ctwloc,lemma:pf}, using the fact that the $T \times \C^*$ equivariant Euler classes $[e_v]|_v$ and $[e_{uW_P}]|_{uW_P}$ coincide with the $T$-equivariant ones; further, the latter equal $[e_v]|_v = \prod_{\alpha \in R^+}v(-\alpha)$ and $[e_{uW_P}]|_{uW_P}= \prod_{\alpha \in R^+ \setminus R_P^+}u(-\alpha)$.
\end{proof}

\begin{example}
Let $u=w$.
The only $v$ satisfying the requirements is $v=w$.
Then
\[
\iota_w^*[T^*_{X_w^P}(G/P)]_{T \times \C^*}= \frac{\prod_{\alpha \in I(w)}(-\hbar +v(\alpha)) \cdot [X_w^B]|_w}{\prod_{\alpha \in R_P^+}v(-\alpha)}=
\prod_{\alpha \in I(w)}(-\hbar + w(\alpha)) \cdot [X_w^P]|_w \/,
\]
where the last equality follows from standard manipulations of (equivariant) Euler classes, for example by using \cref{lemma:pf}.
\end{example}

In \cite{conormal2017}, Lakshmibai and Singh identified certain conormal spaces as open subsets of affine Schubert varieties.
It would interesting to obtain localization formulae for the conormal spaces using localization for affine Schubert varieties.

config: arxiv


TITLE: For an equality of two fractions, each has a different domain; how?
QUESTION [2 upvotes]: Take the following equality:
$$\frac{1}{\left(\frac{1}{y_0}-t^2\right)} = \frac{y_0}{1-\left(t^2y_0\right)}$$
For the left hand side, the function is undefined for $y_0 = 0$. But for the right hand side, the function equals $0$ when $y_0 = 0$
How can this be?

REPLY [4 votes]: The best way of looking at it is this: Fix $t \in \mathbb R$. Then, the expression $\dfrac 1{\frac 1{y_0} - t^2}$ is not defined in two cases : either $y_0 = 0$ or $y_0 = \frac 1{t^2}$, which does not occur if $t = 0$. So, if we define the function $f_t(y_0) = \frac{1}{\frac 1 {y_0} - t^2}$ then for $t=0$ the domain of $f_t$ is $\mathbb R - \{0\}$, and for $t \neq 0$ it is $\mathbb R \setminus \left\{0, \frac 1{t^2}\right\}$.
Now, the other expression $\frac {y_0}{1-t^2y_0}$ is well defined whenever $y_0 \neq \frac 1{t^2}$. That is, if we let $g_t(y_0) = \frac {y_0}{1 - t^2y_0}$, then $g_t$ is defined on all of $\mathbb R$ if $t=0$, and the domain misses the point $\frac 1{t^2}$ if $t \neq 0$.
Therefore, for all $t$, $g_t$ and $f_t$ have different domain! Equality of functions must imply the same domain, but this is not the case here.
Thankfully, the idea of restriction comes in: simply define $h_t(y_0)$, which has the same domain as $f_t$ but $h_t(y_0) = g_t(y_0)$. Then, $h=f$ is a correct statement, since $h$ and $f$ have the same domain.
Therefore, if you find two expressions which are equal except at a set of points where one side is undefined and the other is not, then equality between the functions is actually the equality between the restrictions of these functions to the intersection of their domains i.e. where both are well defined. It is in this sense that the above inequality should be interpreted, and not the naive way.


TITLE: Bizarre definition of convergent sequences
QUESTION [2 upvotes]: I've recently begun studying Analysis I, specifically the sequence and series part. I've just come across the definition of convergence:

A sequence $(a_n)$ converges
  to a real number $a$ if, for every real number $\epsilon > 0$, there exists a natural number $N$ such that, for every natural number $n > N$, we have $|a_n - a| < \epsilon$.

This has the specific order quantifiers $\forall \epsilon \,  \exists n \in \mathbb{N} \,\ldots$
(I've never studied formal logic before, so please do point out if the above is wrong)
I was wondering what would happen if the order of the quantifiers were reversed, that is, if we have the order $\exists \epsilon > 0 \, \forall n \in \mathbb{N} \ldots$
Then the definition would be

A sequence $(a_n)$ converges to a real number $a$ if there exists a real number $\epsilon > 0$ such that for all $N \in \mathbb{N}$ we have that $n > N$ implies $|a_n − a| < \epsilon$.

I've been thinking about this problem for a bit, and I came up with the sequence $$a_n = \frac{(-1)^n}{3} = -\frac{1}{3}, \frac{1}{3}, -\frac{1}{3}, \frac{1}{3}, \ldots$$
With our new reversed definition we could argue that the sequence $a_n$ converges to a limit $0$ by taking $\epsilon = 1$ and having $|a_n - 0| < 1$ for all natural $n$. Obviously, I know that this is a typically divergent sequence with no limit.
My question is in two parts:

What exactly is this bizarre definition describing?
Under what conditions would this definition give a correct proof for a convergent sequence?

REPLY [7 votes]: Bounded sequences
I suppose never - because 1.

To elaborate: Let $\let\epsilon\varepsilon(a_n)$ be bounded, say $|a_n|<M$. Then for $a=0$ we can let $\epsilon=M$ and see that this sequence "converges" to $0$. Or for arbitrary $a$, take $\epsilon = |a|+M$ and see that the sequence also "converges" to $a$.
For the other direction assume $a_n$ "converges" to $a$ according to your definition. Pick $\epsilon >0$ accordingly. Then $|a_n-a|<\epsilon$ for all $n$, hence $(a_n)$ is bounded by $\epsilon+|a|$.


TITLE: Lang's Algebra, pg 61, fibered products and coproducts of abelian groups
QUESTION [1 upvotes]: I have two distinct questions about this page:
First,

The fibered product of two homomorphisms $f:X\to Z$ and $g:Y\to Z$ is the subgroup of $X\times Y$ consisting of all pairs $(x,y)$ such that $f(x)=g(y)$.

My attempt: First, it is clearly a subgroup of $X\times Y$. So consider $(x,y)\in X\times_Z Y$. We require $(f\circ p_1)(x,y)=f(x)=(g\circ p_2)(x,y)=g(y)$ for the diagram to commute so clearly we must have $f(x)=g(y)$. Is this enough to prove that this fibered product exists?
Second,

The coproduct of two homomorphisms $f:Z\to X$ and $g:Z\to Y$ is the factor group $(X\oplus Y)/W$ where $W$ is the subgroup of $X\oplus Y$ consisting of all elements $(f(z),-g(z))$ with $z\in Z$.

Surely he means to say 'fibered coproduct' or 'pushout' instead of 'coproduct'? I thought the coproduct of abelian groups was a direct sum.
If it does mean pushout, I continue by choosing $z\in Z$ and dutifully following the diagram. To commute, we must have $(q_1\circ f)(z)=(f(z),e_Y)=(q_2\circ g)(z)=(e_X,g(z))$. But how does this translate into needing to quotient out $(f(z),-g(z))$?
I'm very much new to category theory so any help is appreciated.

REPLY [1 votes]: Second question first: the coproduct of morphisms is the same thing as the pushout. You're right that the coproduct of objects is the direct sum.
Your discussion of the fiber product is quite confused: it's not actually clear to me what you're trying to show. In a literal sense, it does not require proof that the fiber product exists, at least, if you consider that the fiber product is defined as the subgroup of the cartesian product on those pairs $x,y$ mapped to the same point by $f$, respectively, $g$.
So presumably by "exists" you would prefer to mean "satisfies the universal property," so that it's an actual fiber product categorically. If that's the right interpretation, then you need to show that $X\times_Z Y$ is the universal group mapping to $X$ and $Y$ compatibly with $f$ and $g$, i.e. you need to consider another group $W$ with maps $h,k:W\to X,Y$ and $fh=gk$ and show that $h$ and $k$ factor uniquely through the fiber product.
Your discussion of the pushout is similarly unclear, since I'm not sure what the $q_i$ are-probably addressing the fiber product will be enough at first.


TITLE: Relationship between BIBO, Marginal and Asymptotic stability: statements
QUESTION [3 upvotes]: I know that asymptotic stability implies BIBO stability.
But then, which of the following can be held true:
  1. If a system is BIBO stable would it definitely be asymptotic stable?

  2. If a system is marginally stable, it will definitely be
asymptotically stable?
  3. If a system is asymptotic stable, it will
definitely be marginally stable?

Kepeing in view the poles real part values in mind, what could be held true?

REPLY [4 votes]: A linear system is said to be asymptotically stable if $\lim_{t \to \infty} x(t) = 0$, where $x$ is the states of the system. A linear system is asymptotically stable if and only if real parts of all poles (or eigenvalues of the system matrix) are negative.
A linear system is said to be marginally stable if $\lim_{t \to \infty} x(t) \neq 0$ but $x$ is bounded. A linear system is marginally stable if and only if it has at least one simple pole (not repeated) with real part zero, and all other poles have negative real parts. Therefore, a system cannot be both asymptotically stable and marginally stable.
A linear system is said to be BIBO stable if the output is bounded for an arbitrary bounded input. If a linear system is asymptotically stable, then it is BIBO stable. If a linear system is BIBO stable and the state space representation is minimal, i.e. both controllable and observable, then the system is asymptotically stable. This also implies that a marginally stable system with minimal realization is not BIBO stable.


TITLE: Determine whether the series converge
QUESTION [0 upvotes]: $\sum\limits_{i=3}^\infty \frac{1}{n(log(n)(log(log(n))}$
and
$\sum\limits_{i=3}^\infty \frac{\sqrt{n+1}}{(2n^2-3n+1)(ln((n)+ln(n)^2}$
I suppose the first one would diverge via the comparison test, but I'm not sure which series to compare it to.
As for the second, I don't know how to approach them.

REPLY [2 votes]: Any form the second term (in the denominator of the second series) be; assuming the following limit: $$\lim_{n\to\infty}n^{3/2}u_n=0<\infty$$ the comparison test tells us it converges.


text:
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\def\FFRAME#1#2#3#4#5#6#7{

  \@ifundefined{floatstyle}
    {
     \begin{figure}[#1]
    }
    {
	 \ifx#1h
      \begin{figure}[H]
	 \else
      \begin{figure}[#1]
	 \fi
	}
  \let\QCTOptA\empty
  \let\QCTOptB\empty
  \let\QCBOptA\empty
  \let\QCBOptB\empty
  \ifOverFrame
    #4
    \ifx\QCTOptA\empty
    \else
      \ifx\QCTOptB\empty
        \caption{\QCTOptA}
      \else
        \caption[\QCTOptB]{\QCTOptA}
      \fi
    \fi
    \ifUnderFrame\else
      \label{#5}
    \fi
  \else
    \UnderFrametrue
  \fi
  \begin{center}\GRAPHIC{#7}{#6}{#2}{#3}{\z@}\end{center}
  \ifUnderFrame
    #4
    \ifx\QCBOptA\empty
      \caption{}
    \else
      \ifx\QCBOptB\empty
        \caption{\QCBOptA}
      \else
        \caption[\QCBOptB]{\QCBOptA}
      \fi
    \fi
    \label{#5}
  \fi
  \end{figure}
 }

\newcount\dispkind

\def\makeactives{
  \catcode`\"=\active
  \catcode`\;=\active
  \catcode`\:=\active
  \catcode`\'=\active
  \catcode`\~=\active
}
\bgroup
   \makeactives
   \gdef\activesoff{
      \def"{\string"}
      \def;{\string;}
      \def:{\string:}
      \def'{\string'}
      \def~{\string~}

    }
\egroup

\def\FRAME#1#2#3#4#5#6#7#8{
 \bgroup
 \ifnum\@msidraft=\@ne
   \wasdrafttrue
 \else
   \wasdraftfalse
 \fi
 \def\LaTeXparams{}
 \dispkind=\z@
 \def\LaTeXparams{}
 \doFRAMEparams{#1}
 \ifnum\dispkind=\z@\IFRAME{#2}{#3}{#4}{#7}{#8}{#5}\else
  \ifnum\dispkind=\@ne\DFRAME{#2}{#3}{#7}{#8}{#5}\else
   \ifnum\dispkind=\tw@
    \edef\@tempa{\noexpand\FFRAME{\LaTeXparams}}
    \@tempa{#2}{#3}{#5}{#6}{#7}{#8}
    \fi
   \fi
  \fi
  \ifwasdraft\@msidraft=1\else\@msidraft=0\fi{}
  \egroup
 }

\def\TEXUX#1{"texux"}

\def\BF#1{{\bf {#1}}}
\def\NEG#1{\leavevmode\hbox{\rlap{\thinspace/}{$#1$}}}

\def\limfunc#1{\mathop{\rm #1}}
\def\func#1{\mathop{\rm #1}\nolimits}

\def\unit#1{\mathord{\thinspace\rm #1}}

\long\def\QQQ#1#2{
     \long\expandafter\def\csname#1\endcsname{#2}}
\@ifundefined{QTP}{\def\QTP#1{}}{}
\@ifundefined{QEXCLUDE}{\def\QEXCLUDE#1{}}{}
\@ifundefined{Qlb}{\def\Qlb#1{#1}}{}
\@ifundefined{Qlt}{\def\Qlt#1{#1}}{}
\def\QWE{}
\long\def\QQA#1#2{}
\def\QTR#1#2{{\csname#1\endcsname {#2}}}
\long\def\TeXButton#1#2{#2}
\long\def\QSubDoc#1#2{#2}
\def\EXPAND#1[#2]#3{}
\def\NOEXPAND#1[#2]#3{}
\def\PROTECTED{}
\def\LaTeXparent#1{}
\def\ChildStyles#1{}
\def\ChildDefaults#1{}
\def\QTagDef#1#2#3{}

\@ifundefined{correctchoice}{\def\correctchoice{\relax}}{}
\@ifundefined{HTML}{\def\HTML#1{\relax}}{}
\@ifundefined{TCIIcon}{\def\TCIIcon#1#2#3#4{\relax}}{}
\if@compatibility
  \typeout{Not defining UNICODE  U or CustomNote commands for LaTeX 2.09.}
\else
  \providecommand{\UNICODE}[2][]{\protect\rule{.1in}{.1in}}
  \providecommand{\U}[1]{\protect\rule{.1in}{.1in}}
  \providecommand{\CustomNote}[3][]{\marginpar{#3}}
\fi

\@ifundefined{lambdabar}{
      \def\lambdabar{\errmessage{You have used the lambdabar symbol.
                      This is available for typesetting only in RevTeX styles.}}
   }{}

\@ifundefined{StyleEditBeginDoc}{\def\StyleEditBeginDoc{\relax}}{}

\def\QQfnmark#1{\footnotemark}
\def\QQfntext#1#2{\addtocounter{footnote}{#1}\footnotetext{#2}}

\@ifundefined{TCIMAKEINDEX}{}{\makeindex}

\@ifundefined{abstract}{
 \def\abstract{
  \if@twocolumn
   \section*{Abstract (Not appropriate in this style!)}
   \else \small
   \begin{center}{\bf Abstract\vspace{-.5em}\vspace{\z@}}\end{center}
   \quotation
   \fi
  }
 }{
 }
\@ifundefined{endabstract}{\def\endabstract
  {\if@twocolumn\else\endquotation\fi}}{}
\@ifundefined{maketitle}{\def\maketitle#1{}}{}
\@ifundefined{affiliation}{\def\affiliation#1{}}{}
\@ifundefined{proof}{\def\proof{\noindent{\bfseries Proof. }}}{}
\@ifundefined{endproof}{\def\endproof{\mbox{\ \rule{.1in}{.1in}}}}{}
\@ifundefined{newfield}{\def\newfield#1#2{}}{}
\@ifundefined{chapter}{\def\chapter#1{\par(Chapter head:)#1\par }
 \newcount\c@chapter}{}
\@ifundefined{part}{\def\part#1{\par(Part head:)#1\par }}{}
\@ifundefined{section}{\def\section#1{\par(Section head:)#1\par }}{}
\@ifundefined{subsection}{\def\subsection#1
 {\par(Subsection head:)#1\par }}{}
\@ifundefined{subsubsection}{\def\subsubsection#1
 {\par(Subsubsection head:)#1\par }}{}
\@ifundefined{paragraph}{\def\paragraph#1
 {\par(Subsubsubsection head:)#1\par }}{}
\@ifundefined{subparagraph}{\def\subparagraph#1
 {\par(Subsubsubsubsection head:)#1\par }}{}

\@ifundefined{therefore}{\def\therefore{}}{}
\@ifundefined{backepsilon}{\def\backepsilon{}}{}
\@ifundefined{yen}{\def\yen{\hbox{\rm\rlap=Y}}}{}
\@ifundefined{registered}{
   \def\registered{\relax\ifmmode{}\r@gistered
                    \else$\m@th\r@gistered$\fi}
 \def\r@gistered{^{\ooalign
  {\hfil\raise.07ex\hbox{$\scriptstyle\rm\text{R}$}\hfil\crcr
  \mathhexbox20D}}}}{}
\@ifundefined{Eth}{\def\Eth{}}{}
\@ifundefined{eth}{\def\eth{}}{}
\@ifundefined{Thorn}{\def\Thorn{}}{}
\@ifundefined{thorn}{\def\thorn{}}{}

\def\TEXTsymbol#1{\mbox{$#1$}}
\@ifundefined{degree}{\def\degree{{}^{\circ}}}{}

\newdimen\theight
\@ifundefined{Column}{\def\Column{
 \vadjust{\setbox\z@=\hbox{\scriptsize\quad\quad tcol}
  \theight=\ht\z@\advance\theight by \dp\z@\advance\theight by \lineskip
  \kern -\theight \vbox to \theight{
   \rightline{\rlap{\box\z@}}
   \vss
   }
  }
 }}{}

\@ifundefined{qed}{\def\qed{
 \ifhmode\unskip\nobreak\fi\ifmmode\ifinner\else\hskip5\p@\fi\fi
 \hbox{\hskip5\p@\vrule width4\p@ height6\p@ depth1.5\p@\hskip\p@}
 }}{}

\@ifundefined{cents}{\def\cents{\hbox{\rm\rlap c/}}}{}
\@ifundefined{tciLaplace}{\def\tciLaplace{\ensuremath{\mathcal{L}}}}{}
\@ifundefined{tciFourier}{\def\tciFourier{\ensuremath{\mathcal{F}}}}{}
\@ifundefined{textcurrency}{\def\textcurrency{\hbox{\rm\rlap xo}}}{}
\@ifundefined{texteuro}{\def\texteuro{\hbox{\rm\rlap C=}}}{}
\@ifundefined{euro}{\def\euro{\hbox{\rm\rlap C=}}}{}
\@ifundefined{textfranc}{\def\textfranc{\hbox{\rm\rlap-F}}}{}
\@ifundefined{textlira}{\def\textlira{\hbox{\rm\rlap L=}}}{}
\@ifundefined{textpeseta}{\def\textpeseta{\hbox{\rm P\negthinspace s}}}{}

\@ifundefined{miss}{\def\miss{\hbox{\vrule height2\p@ width 2\p@ depth\z@}}}{}

\@ifundefined{vvert}{\def\vvert{\Vert}}{}

\@ifundefined{tcol}{\def\tcol#1{{\baselineskip=6\p@ \vcenter{#1}} \Column}}{}

\@ifundefined{dB}{\def\dB{\hbox{{}}}}{}
\@ifundefined{mB}{\def\mB#1{\hbox{$#1$}}}{}
\@ifundefined{nB}{\def\nB#1{\hbox{#1}}}{}

\@ifundefined{note}{\def\note{$^{\dag}}}{}

\def\newfmtname{LaTeX2e}

\ifx\fmtname\newfmtname
  \DeclareOldFontCommand{\rm}{\normalfont\rmfamily}{\mathrm}
  \DeclareOldFontCommand{\sf}{\normalfont\sffamily}{\mathsf}
  \DeclareOldFontCommand{\tt}{\normalfont\ttfamily}{\mathtt}
  \DeclareOldFontCommand{\bf}{\normalfont\bfseries}{\mathbf}
  \DeclareOldFontCommand{\it}{\normalfont\itshape}{\mathit}
  \DeclareOldFontCommand{\sl}{\normalfont\slshape}{\@nomath\sl}
  \DeclareOldFontCommand{\sc}{\normalfont\scshape}{\@nomath\sc}
\fi

\def\alpha{{\Greekmath 010B}}
\def\beta{{\Greekmath 010C}}
\def\gamma{{\Greekmath 010D}}
\def\delta{{\Greekmath 010E}}
\def\epsilon{{\Greekmath 010F}}
\def\zeta{{\Greekmath 0110}}
\def\eta{{\Greekmath 0111}}
\def\theta{{\Greekmath 0112}}
\def\iota{{\Greekmath 0113}}
\def\kappa{{\Greekmath 0114}}
\def\lambda{{\Greekmath 0115}}
\def\mu{{\Greekmath 0116}}
\def\nu{{\Greekmath 0117}}
\def\xi{{\Greekmath 0118}}
\def\pi{{\Greekmath 0119}}
\def\rho{{\Greekmath 011A}}
\def\sigma{{\Greekmath 011B}}
\def\tau{{\Greekmath 011C}}
\def\upsilon{{\Greekmath 011D}}
\def\phi{{\Greekmath 011E}}
\def\chi{{\Greekmath 011F}}
\def\psi{{\Greekmath 0120}}
\def\omega{{\Greekmath 0121}}
\def\varepsilon{{\Greekmath 0122}}
\def\vartheta{{\Greekmath 0123}}
\def\varpi{{\Greekmath 0124}}
\def\varrho{{\Greekmath 0125}}
\def\varsigma{{\Greekmath 0126}}
\def\varphi{{\Greekmath 0127}}

\def\nabla{{\Greekmath 0272}}
\def\FindBoldGroup{
   {\setbox0=\hbox{$\mathbf{x\global\edef\theboldgroup{\the\mathgroup}}$}}
}

\def\Greekmath#1#2#3#4{
    \if@compatibility
        \ifnum\mathgroup=\symbold
           \mathchoice{\mbox{\boldmath$\displaystyle\mathchar"#1#2#3#4$}}
                      {\mbox{\boldmath$\textstyle\mathchar"#1#2#3#4$}}
                      {\mbox{\boldmath$\scriptstyle\mathchar"#1#2#3#4$}}
                      {\mbox{\boldmath$\scriptscriptstyle\mathchar"#1#2#3#4$}}
        \else
           \mathchar"#1#2#3#4
        \fi
    \else
        \FindBoldGroup
        \ifnum\mathgroup=\theboldgroup
           \mathchoice{\mbox{\boldmath$\displaystyle\mathchar"#1#2#3#4$}}
                      {\mbox{\boldmath$\textstyle\mathchar"#1#2#3#4$}}
                      {\mbox{\boldmath$\scriptstyle\mathchar"#1#2#3#4$}}
                      {\mbox{\boldmath$\scriptscriptstyle\mathchar"#1#2#3#4$}}
        \else
           \mathchar"#1#2#3#4
        \fi
	  \fi}

\newif\ifGreekBold  \GreekBoldfalse
\let\SAVEPBF=\pbf
\def\pbf{\GreekBoldtrue\SAVEPBF}

\@ifundefined{theorem}{\newtheorem{theorem}{Theorem}}{}
\@ifundefined{lemma}{\newtheorem{lemma}[theorem]{Lemma}}{}
\@ifundefined{corollary}{\newtheorem{corollary}[theorem]{Corollary}}{}
\@ifundefined{conjecture}{\newtheorem{conjecture}[theorem]{Conjecture}}{}
\@ifundefined{proposition}{\newtheorem{proposition}[theorem]{Proposition}}{}
\@ifundefined{axiom}{\newtheorem{axiom}{Axiom}}{}
\@ifundefined{remark}{\newtheorem{remark}{Remark}}{}
\@ifundefined{example}{\newtheorem{example}{Example}}{}
\@ifundefined{exercise}{\newtheorem{exercise}{Exercise}}{}
\@ifundefined{definition}{\newtheorem{definition}{Definition}}{}

\@ifundefined{mathletters}{

  \newcounter{equationnumber}
  \def\mathletters{
     \addtocounter{equation}{1}
     \edef\@currentlabel{\theequation}
     \setcounter{equationnumber}{\c@equation}
     \setcounter{equation}{0}
     \edef\theequation{\@currentlabel\noexpand\alph{equation}}
  }
  \def\endmathletters{
     \setcounter{equation}{\value{equationnumber}}
  }
}{}

\@ifundefined{BibTeX}{
    \def\BibTeX{{\rm B\kern-.05em{\sc i\kern-.025em b}\kern-.08em
                 T\kern-.1667em\lower.7ex\hbox{E}\kern-.125emX}}}{}
\@ifundefined{AmS}
    {\def\AmS{{\protect\usefont{OMS}{cmsy}{m}{n}
                A\kern-.1667em\lower.5ex\hbox{M}\kern-.125emS}}}{}
\@ifundefined{AmSTeX}{\def\AmSTeX{\protect\AmS-\protect\TeX\@}}{}

\def\@@eqncr{\let\@tempa\relax
    \ifcase\@eqcnt \def\@tempa{& & &}\or \def\@tempa{& &}
      \else \def\@tempa{&}\fi
     \@tempa
     \if@eqnsw
        \iftag@
           \@taggnum
        \else
           \@eqnnum\stepcounter{equation}
        \fi
     \fi
     \global\tag@false
     \global\@eqnswtrue
     \global\@eqcnt\z@\cr}

\def\TCItag{\@ifnextchar*{\@TCItagstar}{\@TCItag}}
\def\@TCItag#1{
    \global\tag@true
    \global\def\@taggnum{(#1)}
    \global\def\@currentlabel{#1}}
\def\@TCItagstar*#1{
    \global\tag@true
    \global\def\@taggnum{#1}
    \global\def\@currentlabel{#1}}

\def\QATOP#1#2{{#1 \atop #2}}
\def\QTATOP#1#2{{\textstyle {#1 \atop #2}}}
\def\QDATOP#1#2{{\displaystyle {#1 \atop #2}}}
\def\QABOVE#1#2#3{{#2 \above#1 #3}}
\def\QTABOVE#1#2#3{{\textstyle {#2 \above#1 #3}}}
\def\QDABOVE#1#2#3{{\displaystyle {#2 \above#1 #3}}}
\def\QOVERD#1#2#3#4{{#3 \overwithdelims#1#2 #4}}
\def\QTOVERD#1#2#3#4{{\textstyle {#3 \overwithdelims#1#2 #4}}}
\def\QDOVERD#1#2#3#4{{\displaystyle {#3 \overwithdelims#1#2 #4}}}
\def\QATOPD#1#2#3#4{{#3 \atopwithdelims#1#2 #4}}
\def\QTATOPD#1#2#3#4{{\textstyle {#3 \atopwithdelims#1#2 #4}}}
\def\QDATOPD#1#2#3#4{{\displaystyle {#3 \atopwithdelims#1#2 #4}}}
\def\QABOVED#1#2#3#4#5{{#4 \abovewithdelims#1#2#3 #5}}
\def\QTABOVED#1#2#3#4#5{{\textstyle
   {#4 \abovewithdelims#1#2#3 #5}}}
\def\QDABOVED#1#2#3#4#5{{\displaystyle
   {#4 \abovewithdelims#1#2#3 #5}}}

\def\tint{\mathop{\textstyle \int}}
\def\tiint{\mathop{\textstyle \iint }}
\def\tiiint{\mathop{\textstyle \iiint }}
\def\tiiiint{\mathop{\textstyle \iiiint }}
\def\tidotsint{\mathop{\textstyle \idotsint }}
\def\toint{\mathop{\textstyle \oint}}
\def\tsum{\mathop{\textstyle \sum }}
\def\tprod{\mathop{\textstyle \prod }}
\def\tbigcap{\mathop{\textstyle \bigcap }}
\def\tbigwedge{\mathop{\textstyle \bigwedge }}
\def\tbigoplus{\mathop{\textstyle \bigoplus }}
\def\tbigodot{\mathop{\textstyle \bigodot }}
\def\tbigsqcup{\mathop{\textstyle \bigsqcup }}
\def\tcoprod{\mathop{\textstyle \coprod }}
\def\tbigcup{\mathop{\textstyle \bigcup }}
\def\tbigvee{\mathop{\textstyle \bigvee }}
\def\tbigotimes{\mathop{\textstyle \bigotimes }}
\def\tbiguplus{\mathop{\textstyle \biguplus }}

\def\dint{\mathop{\displaystyle \int}}
\def\diint{\mathop{\displaystyle \iint}}
\def\diiint{\mathop{\displaystyle \iiint}}
\def\diiiint{\mathop{\displaystyle \iiiint }}
\def\didotsint{\mathop{\displaystyle \idotsint }}
\def\doint{\mathop{\displaystyle \oint}}
\def\dsum{\mathop{\displaystyle \sum }}
\def\dprod{\mathop{\displaystyle \prod }}
\def\dbigcap{\mathop{\displaystyle \bigcap }}
\def\dbigwedge{\mathop{\displaystyle \bigwedge }}
\def\dbigoplus{\mathop{\displaystyle \bigoplus }}
\def\dbigodot{\mathop{\displaystyle \bigodot }}
\def\dbigsqcup{\mathop{\displaystyle \bigsqcup }}
\def\dcoprod{\mathop{\displaystyle \coprod }}
\def\dbigcup{\mathop{\displaystyle \bigcup }}
\def\dbigvee{\mathop{\displaystyle \bigvee }}
\def\dbigotimes{\mathop{\displaystyle \bigotimes }}
\def\dbiguplus{\mathop{\displaystyle \biguplus }}

\if@compatibility\else

  \RequirePackage{amsmath}
\fi

\def\ExitTCILatex{\makeatother\endinput}

\bgroup
\ifx\ds@amstex\relax
   \message{amstex already loaded}\aftergroup\ExitTCILatex
\else
   \@ifpackageloaded{amsmath}
      {\if@compatibility\message{amsmath already loaded}\fi\aftergroup\ExitTCILatex}
      {}
   \@ifpackageloaded{amstex}
      {\if@compatibility\message{amstex already loaded}\fi\aftergroup\ExitTCILatex}
      {}
   \@ifpackageloaded{amsgen}
      {\if@compatibility\message{amsgen already loaded}\fi\aftergroup\ExitTCILatex}
      {}
\fi
\egroup

\typeout{TCILATEX defining AMS-like constructs in LaTeX 2.09 COMPATIBILITY MODE}

\let\DOTSI\relax
\def\RIfM@{\relax\ifmmode}
\def\FN@{\futurelet\next}
\newcount\intno@
\def\iint{\DOTSI\intno@\tw@\FN@\ints@}
\def\iiint{\DOTSI\intno@\thr@@\FN@\ints@}
\def\iiiint{\DOTSI\intno@4 \FN@\ints@}
\def\idotsint{\DOTSI\intno@\z@\FN@\ints@}
\def\ints@{\findlimits@\ints@@}
\newif\iflimtoken@
\newif\iflimits@
\def\findlimits@{\limtoken@true\ifx\next\limits\limits@true
 \else\ifx\next\nolimits\limits@false\else
 \limtoken@false\ifx\ilimits@\nolimits\limits@false\else
 \ifinner\limits@false\else\limits@true\fi\fi\fi\fi}
\def\multint@{\int\ifnum\intno@=\z@\intdots@
 \else\intkern@\fi
 \ifnum\intno@>\tw@\int\intkern@\fi
 \ifnum\intno@>\thr@@\int\intkern@\fi
 \int}
\def\multintlimits@{\intop\ifnum\intno@=\z@\intdots@\else\intkern@\fi
 \ifnum\intno@>\tw@\intop\intkern@\fi
 \ifnum\intno@>\thr@@\intop\intkern@\fi\intop}
\def\intic@{
    \mathchoice{\hskip.5em}{\hskip.4em}{\hskip.4em}{\hskip.4em}}
\def\negintic@{\mathchoice
 {\hskip-.5em}{\hskip-.4em}{\hskip-.4em}{\hskip-.4em}}
\def\ints@@{\iflimtoken@
 \def\ints@@@{\iflimits@\negintic@
   \mathop{\intic@\multintlimits@}\limits
  \else\multint@\nolimits\fi
  \eat@}
 \else
 \def\ints@@@{\iflimits@\negintic@
  \mathop{\intic@\multintlimits@}\limits\else
  \multint@\nolimits\fi}\fi\ints@@@}
\def\intkern@{\mathchoice{\!\!\!}{\!\!}{\!\!}{\!\!}}
\def\plaincdots@{\mathinner{\cdotp\cdotp\cdotp}}
\def\intdots@{\mathchoice{\plaincdots@}
 {{\cdotp}\mkern1.5mu{\cdotp}\mkern1.5mu{\cdotp}}
 {{\cdotp}\mkern1mu{\cdotp}\mkern1mu{\cdotp}}
 {{\cdotp}\mkern1mu{\cdotp}\mkern1mu{\cdotp}}}

\def\RIfM@{\relax\protect\ifmmode}
\def\text{\RIfM@\expandafter\text@\else\expandafter\mbox\fi}
\let\nfss@text\text
\def\text@#1{\mathchoice
   {\textdef@\displaystyle\f@size{#1}}
   {\textdef@\textstyle\tf@size{\firstchoice@false #1}}
   {\textdef@\textstyle\sf@size{\firstchoice@false #1}}
   {\textdef@\textstyle \ssf@size{\firstchoice@false #1}}
   \glb@settings}

\def\textdef@#1#2#3{\hbox{{
                    \everymath{#1}
                    \let\f@size#2\selectfont
                    #3}}}
\newif\iffirstchoice@
\firstchoice@true

\def\Let@{\relax\iffalse{\fi\let\\=\cr\iffalse}\fi}
\def\vspace@{\def\vspace##1{\crcr\noalign{\vskip##1\relax}}}
\def\multilimits@{\bgroup\vspace@\Let@
 \baselineskip\fontdimen10 \scriptfont\tw@
 \advance\baselineskip\fontdimen12 \scriptfont\tw@
 \lineskip\thr@@\fontdimen8 \scriptfont\thr@@
 \lineskiplimit\lineskip
 \vbox\bgroup\ialign\bgroup\hfil$\m@th\scriptstyle{##}$\hfil\crcr}
\def\Sb{_\multilimits@}
\def\endSb{\crcr\egroup\egroup\egroup}
\def\Sp{^\multilimits@}
\let\endSp\endSb

\newdimen\ex@
\ex@.2326ex
\def\rightarrowfill@#1{$#1\m@th\mathord-\mkern-6mu\cleaders
 \hbox{$#1\mkern-2mu\mathord-\mkern-2mu$}\hfill
 \mkern-6mu\mathord\rightarrow$}
\def\leftarrowfill@#1{$#1\m@th\mathord\leftarrow\mkern-6mu\cleaders
 \hbox{$#1\mkern-2mu\mathord-\mkern-2mu$}\hfill\mkern-6mu\mathord-$}
\def\leftrightarrowfill@#1{$#1\m@th\mathord\leftarrow
\mkern-6mu\cleaders
 \hbox{$#1\mkern-2mu\mathord-\mkern-2mu$}\hfill
 \mkern-6mu\mathord\rightarrow$}
\def\overrightarrow{\mathpalette\overrightarrow@}
\def\overrightarrow@#1#2{\vbox{\ialign{##\crcr\rightarrowfill@#1\crcr
 \noalign{\kern-\ex@\nointerlineskip}$\m@th\hfil#1#2\hfil$\crcr}}}
\let\overarrow\overrightarrow
\def\overleftarrow{\mathpalette\overleftarrow@}
\def\overleftarrow@#1#2{\vbox{\ialign{##\crcr\leftarrowfill@#1\crcr
 \noalign{\kern-\ex@\nointerlineskip}$\m@th\hfil#1#2\hfil$\crcr}}}
\def\overleftrightarrow{\mathpalette\overleftrightarrow@}
\def\overleftrightarrow@#1#2{\vbox{\ialign{##\crcr
   \leftrightarrowfill@#1\crcr
 \noalign{\kern-\ex@\nointerlineskip}$\m@th\hfil#1#2\hfil$\crcr}}}
\def\underrightarrow{\mathpalette\underrightarrow@}
\def\underrightarrow@#1#2{\vtop{\ialign{##\crcr$\m@th\hfil#1#2\hfil
  $\crcr\noalign{\nointerlineskip}\rightarrowfill@#1\crcr}}}
\let\underarrow\underrightarrow
\def\underleftarrow{\mathpalette\underleftarrow@}
\def\underleftarrow@#1#2{\vtop{\ialign{##\crcr$\m@th\hfil#1#2\hfil
  $\crcr\noalign{\nointerlineskip}\leftarrowfill@#1\crcr}}}
\def\underleftrightarrow{\mathpalette\underleftrightarrow@}
\def\underleftrightarrow@#1#2{\vtop{\ialign{##\crcr$\m@th
  \hfil#1#2\hfil$\crcr
 \noalign{\nointerlineskip}\leftrightarrowfill@#1\crcr}}}

\def\qopnamewl@#1{\mathop{\operator@font#1}\nlimits@}
\let\nlimits@\displaylimits
\def\setboxz@h{\setbox\z@\hbox}

\def\varlim@#1#2{\mathop{\vtop{\ialign{##\crcr
 \hfil$#1\m@th\operator@font lim$\hfil\crcr
 \noalign{\nointerlineskip}#2#1\crcr
 \noalign{\nointerlineskip\kern-\ex@}\crcr}}}}

 \def\rightarrowfill@#1{\m@th\setboxz@h{$#1-$}\ht\z@\z@
  $#1\copy\z@\mkern-6mu\cleaders
  \hbox{$#1\mkern-2mu\box\z@\mkern-2mu$}\hfill
  \mkern-6mu\mathord\rightarrow$}
\def\leftarrowfill@#1{\m@th\setboxz@h{$#1-$}\ht\z@\z@
  $#1\mathord\leftarrow\mkern-6mu\cleaders
  \hbox{$#1\mkern-2mu\copy\z@\mkern-2mu$}\hfill
  \mkern-6mu\box\z@$}

\def\projlim{\qopnamewl@{proj\,lim}}
\def\injlim{\qopnamewl@{inj\,lim}}
\def\varinjlim{\mathpalette\varlim@\rightarrowfill@}
\def\varprojlim{\mathpalette\varlim@\leftarrowfill@}
\def\varliminf{\mathpalette\varliminf@{}}
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text:
\section{Neural network architectures}
\label{architectures}
The StyleGAN architecture were taken directly from \cite{styleGAN}.
The architectures for the MNIST dataset are detailed below in Tables \ref{genArch} and \ref{disArch}.
We used the architecture guidelines provided by \cite{Radford2015} and the GANs tutorial in \cite{tensorflow}.
We shorten some of the common terms used to describe the layers of the networks.
A fully connected layer in a neural network is denoted by FC, while we have used upconv. to denote a convolution layer that is
specifically padded to increase the dimensions of its input image.
The bias in each upconv. layer was not used in an effort to reduce the amount of parameters and the computational training time.
The parameters of the neural networks were initialized by sampling a Gaussian random variable with mean $0$ and standard deviation
$0.01$.

\begin{table}[H]
\caption{The generator's architecture for MNIST dataset.}
\label{genArch}
\vskip 0.15in
\begin{center}
\begin{small}
\begin{sc}
\begin{tabular}{l}
\hline \hline
Generator \\
\hline
	Input multivariate Gaussian noise vector of size $784$ with mean $\mathbf{0}$ \\
	\ \ \ and
	a covariance matrix that is the identity matrix of size $784 \times 784$. \\
	FC to $12,544$ neurons. \\
        Reshape into $7 \times 7 \times 256$ image.\\
        $5 \times 5$ upconv. $128$ LeakyRELU, batchnorm. \\
        $5 \times 5$ upconv. $64$ LeakyRELU, stride 2, batchnorm. \\
        $5 \times 5$ upconv. $1$ channel, $\tanh$ activation. \\
\hline
\end{tabular}
\end{sc}
\end{small}
\end{center}
\vskip -0.1in
\end{table}

\begin{table}[H]
\caption{The discriminator's architecture for MNIST dataset.}
\label{disArch}
\vskip 0.15in
\begin{center}
\begin{small}
\begin{sc}
\begin{tabular}{l}
\hline \hline
Discriminator \\
\hline
        Input $28 \times 28 \times 1$ grey image. \\
        $5 \times 5$ conv. $64$ LeakyRELU, stride 2, batchnorm, \\
        \ \ \ \ dropout $0.3$. \\
        $5 \times 5$ conv. $128$ LeakyRELU, stride 2, batchnorm, \\
        \ \ \ \ dropout $0.3$. \\
	FC to one output, $\frac{1}{1 + e^{-x}}$ activation. \\
\hline
\end{tabular}
\end{sc}
\end{small}
\end{center}
\vskip -0.1in
\end{table}

\section{Algorithms}
\label{algorithms}
We present the algorithms for L$k$GANs below.

For the CelebA dataset, the constants of the algorithms are $n = 125$ epochs or 25 million images and the batch size $m = 126$.

We also present the algorithms for \renyi GANs.
For the MNIST dataset, the constants of the algorithms are $n = 100$ epochs or 6 million images and batch size $m = 100$.
For the CelebA dataset, the constants of the algorithms are $n = 125$ epochs or 25 million images and the batch size $m = 126$.

\begin{algorithm}[htb]
	\caption{Overview of L$k$GAN-v$1$-$k$ and L$k$GAN-v$2$-$k$ algorithms with and without the simplified gradient penalty.}
	\label{algo:lkgan}
\begin{algorithmic}[tb]
	\STATE {\bfseries Initialize} neural networks.
        \STATE {\bfseries Fix} number of epochs $n$.
        \FOR{$i = 0$ {\bfseries to } $n-1$}
		\STATE {\bfseries Sample} batch size of $m$ noise samples $\{\bm{z}_1, \ldots, \bm{z}_m\}$ from noise prior $p_\mathbf{Z}$
		\STATE {\bfseries Sample} batch size of $m$ examples $\{\mathbf{x}_1, \ldots, \mathbf{x}_m \}$ from the true distribution $p_\mathbf{X}$
                \STATE {\bfseries Update} the discriminator by descending its gradient without the simplified gradient penalty:
                        $$\nabla_{\bm{\tilde{\theta}}} \left(\frac{1}{m}\sum_{i = 1}^{m} \left[
				\frac{1}{2} (D(\mathbf{x}_i) - b)^2 + \frac{1}{2} (D(g(\bm{z}_i)) - a)^2
				\right]\right),$$
			or with the simplified gradient penalty:
			\begin{eqnarray*}
				&& \nabla_{\bm{\tilde{\theta}}} \Bigg(\frac{1}{m}\sum_{i = 1}^{m} \left[
					\frac{1}{2} (D(\mathbf{x}_i) - b)^2 + \frac{1}{2} (D(g(\bm{z}_i)) - a)^2
                                \right] \\
			&& + 5 \left ( \frac{1}{m} \sum_{i=1}^{m} \left
              				\Vert \nabla_{\mathbf{x}}
				\log\left(\frac{D(\mathbf{x})}{1 - D(\mathbf{x})} \right) \bigg \vert_{\mathbf{x}= \mathbf{x}_i}  \right \Vert_2 ^2  \right) \Bigg).
			\end{eqnarray*}
		\STATE{Update} the generator by descending its gradient:
                $$ \nabla_{\bm{\theta}}   \left(
			\frac{1}{m} \sum_{i = 1}^{m} \vert D(g(\bm{z}_i)) - c \vert^{k} \right),  $$
        \ENDFOR
\end{algorithmic}
\end{algorithm}

\begin{algorithm}[htb]
	\caption{Overview of \renyi GAN-$\alpha$, \renyi GAN-$\alpha$-$L_1$, and \renyi GAN-GP-$\alpha$-$L_1$ algorithms}
	\label{algo:renyigan}
\begin{algorithmic}[tb]
	\STATE {\bfseries Initialize} neural networks.
        \STATE {\bfseries Fix} number of epochs $n$.
        \FOR{$i = 0$ {\bfseries to } $n-1$}
                \STATE {\bfseries Sample} batch size of $m$ noise samples $\{\bm{z}_1, \ldots, \bm{z}_m\}$ from noise prior $p_{\mathbf{Z}}$
                \STATE {\bfseries Sample} batch size of $m$ examples $\{\mathbf{x}_1, \ldots, \mathbf{x}_m \}$ from the true distribution $p_{\mathbf{X}}$
                \STATE {\bfseries Update} the discriminator by descending its stochastic gradient without the simplified gradient penalty:

                        $$\nabla_{\bm{\tilde{\theta}}} \bigg(-\frac{1}{m}\sum_{i = 1}^{m} \big[
                                        \log D(\mathbf{x}_i)
                                        + \log(1 - D(g(\bm{z}_i)))
                                \big]\bigg),$$
                or with the simplified gradient penalty:
		\begin{eqnarray*}
			&&\nabla_{\bm{\tilde{\theta}}} \Bigg(-\frac{1}{m}\sum_{i = 1}^{m} \big[
                                        \log D(\mathbf{x}_i)
                                        + \log(1 - D(g(\bm{z}_i)))
				\big]  \\ && +
			5 \left ( \frac{1}{m} \sum_{i=1}^{m} \left \Vert \nabla_{\mathbf{x}}
			\log\left(\frac{D(\mathbf{x})}{1 - D(\mathbf{x})} \right) \bigg \vert_{\mathbf{x}= \mathbf{x}_i}  \right \Vert_2 ^2  \right) \Bigg),
		\end{eqnarray*}

                and update the generator by descending its stochastic gradient without $L_1$ normalization:
                $$ \nabla_{\bm{\theta}}  \frac{1}{\alpha - 1}\log \Bigg[ \left(
	\frac{1}{m} \sum_{i = 1}^{m} [1 - D(g(\bm{z}_i))]^{\alpha - 1} \right) \Bigg], $$
                or with $L_1$ normalization:

$$ \nabla_{\bm{\theta}} \Bigg \vert \frac{1}{\alpha - 1}\log \Bigg[ \left(
	\frac{1}{m} \sum_{i = 1}^{m} [1 - D(g(\bm{z}_i))]^{\alpha - 1} \right) \Bigg] + \log(2) \Bigg \vert,$$

        \ENDFOR
\end{algorithmic}
\end{algorithm}