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\title[On some generalizations of the sum of powers of natural numbers]
{On some generalizations of the sum of powers of natural numbers}

\author[Andrei K. Svinin]{Andrei K. Svinin}
\address{Andrei K. Svinin,
Matrosov Institute for System Dynamics and Control Theory of
Siberian Branch of Russian Academy of Sciences,
P.O. Box 292, 664033 Irkutsk, Russia}
\email{svinin@icc.ru}

\date{\today}

\begin{abstract}
In this paper   some generalizations of the sum of powers of natural numbers is considered. In particular,  the class of sums whose generating function is the power of the generating function for the classical sums of powers is studying. The so-called binomial sums are also considered. The problem of constructing polynomials that allow  to calculate the values of the corresponding sums in certain cases  is solved.
\end{abstract}

\maketitle

\section{Introduction}

Power-sum
\begin{equation}
S_m(n):=\sum_{q=1}^{n}q^{m},\;\; m\geq 0
\label{power-sums}
\end{equation}
starting with the works of Blaise Pascal, Johann Faulhaber, Jacob Bernoulli for centuries, has been the object of active research. The expression (\ref{power-sums}) defines a function on the set of natural numbers. Pascal's theorem asserts that for any integer degree $m\geq 0$, the value of the sum (\ref{power-sums}) can be obtained as the value of the corresponding polynomial $\hat{S} _m(n)$.
The exponential generating  function (e.g.f.) for sums of the form (\ref{power-sums}) is given by
\begin{eqnarray}
G(n, t)&=&\sum_{j\geq 0}S_j(n)\frac{t^j}{j!}=\sum_{j\geq 0}\left(\sum_{q=1}^{n}q^j\frac{t^j}{j!}\right)=\sum_{q=1}^{n}e^{qt}\nonumber\\
&=&\frac{e^{(n+1)t}-e^t}{e^t-1}.
\label{genf}
\end{eqnarray}
In turn, the expansion (\ref{genf}) into an infinite series
\[
G(n, t)=\sum_{q\geq 0}\hat{S}_{q}(n)\frac{t^q}{q!}
\]
gives an infinite  set of polynomials $\{\hat{S}_m (n): m\geq 0\} $ whose values for natural  $n$ give the values of the corresponding sums. In 1713, Jacob Bernoulli published  an expression for the sum of the degree $m$ of the first $n$ natural numbers as a polynomial of $(m+1)$-th degree in  $n$, which is determined by some infinite set of rational numbers $\{B_m: m\geq 0\}$, which are now called the Bernoulli numbers.

We will write this expression in the following form:
\begin{equation}
\hat{S}_{m}(n)=\frac{1}{m+1}\sum_{q=0}^m(-1)^q{m+1\choose q}B_{q}n^{m+1-q}.
\label{22}
\end{equation}
Faulhaber's theorem, in turn, asserts that, for odd values of $m$, the polynomial (\ref{22}) is actually a polynomial in $w:=n(n+1)$. To a certain extent this is due to the fact that all odd Bernoulli numbers starting with the third one are zero \cite{Knut}. In turn, it is known that for any $m$, any polynomial (\ref{22}) can be written as some polynomial in $w$ multiplied by $2n+1$ \cite{Jacobi}.

In applications there can arise not only sums of the form (\ref{power-sums}), but also other sums involving the powers of natural numbers. For example, Johann Faulhaber considered the so-called $k$-fold sums $S^k_m(n)$ defined by the recurrence relation
\[
S^{k}_m(n)=\sum_{q=1}^{n}S^{k-1}_m(q),\;\; \forall k\geq 1
\]
starting from $S^{0}_m(n):=n^m$. When considering such generalizations, the question always arises about of polynomials that, like (\ref{22}), help to calculate the values of corresponding sums.

In this paper we consider some generalizations of power-sum (\ref{power-sums}) and  set the problem of finding the corresponding polynomials. In the next section, we define a some class of sums whose e.g.f. is the $k$-th power of the generating function (\ref{genf}), with $k$ being an arbitrary natural number. We denote these sums by the symbol $S^{(k)}_m(n) $ and call them power-sums  of higher order. One of our results is that we found an expression for the associated  polynomials similar to the formula (\ref{22}). These polynomials, in contrast to (\ref{22}), are determined by higher-order Bernoulli numbers. This is, although not obvious, is quite natural. In the next section we give some information about higher-order Bernoulli numbers and about the Stirling numbers of the second kind, which are also involved in the formula for these polynomials. In the second section, we also investigate some rational roots of polynomials corresponding to higher order power-sums.
In the third section, we discuss some generalizations of the sums (\ref{power-sums}) arising in applications, namely, a some class of multiple sums.

\section{ higher order power-sums}

\subsection{Definition of higher order power-sums}

Let us consider the power of the e.g.f.  (\ref{genf}):
\[
\left(G(n, t)\right)^k:=\sum_{q\geq 0}S_{q}^{(k)}(n)\frac{t^q}{q!}.
\]
We have
\begin{equation}
\left(G(n, t)\right)^k=\left(\sum_{q=1}^{n}e^{qt}\right)^k=\sum_{q=0}^{k(n-1)}{k\choose q}_ne^{(k+q)t},
\label{1}
\end{equation}
where the symbol ${k\choose q}_n$  stands for   polynomial coefficients  defined through the relation
\[
\left(\sum_{q=1}^{n}t^q\right)^k:=\sum_{q=0}^{k(n-1)}{k\choose q}_nt^{k+q}.
\]
It should be noted that the coefficients ${k\choose q}_n$, naturally generalizing the binomial coefficients ($n=2$), originated  from Abraham De Moivre and Leonhard Euler works \cite{Moivre}, \cite{Euler}.
Then they were rediscovered in the works \cite{Montel} and \cite{Tremblay} and later were studied in detail in the literature due to their good applicability. By direct computation we derive
\begin{proposition}
The expression  (\ref{1}) is the e.g.f. for sums of the form
\begin{equation}
S_{m}^{(k)}(n)=\sum_{q=0}^{k(n-1)}{k\choose q}_{n}\left(k+q\right)^m.
\label{sums}
\end{equation}
\end{proposition}
It is evident, that these sums, by definition, are the result of successive binomial convolutions, that is,
\[
S_m^{(k)}(n)=\sum_{q=0}^{m} {m\choose q}S_{q}^{(k-1)}(n)S_{m-q}(n),\;\;
k\geq 2.
\]
It is also obvious that this is true for the corresponding polynomials $\hat{S}_m^{(k)}(n)$. The following property holds:
\begin{proposition} \label{proposition2.2}
The sums  $S_m^{(k)}(n)$ satisfy the recurrence relation
\begin{eqnarray}
&&\sum_{q=0}^{m}(-1)^{q}{m+k\choose q}S(m+k-q, k)S_q^{(r)}(n) \nonumber  \\
      &&\;\;\;\;\;=\frac{1}{{k\choose r}}\sum_{j=0}^{m}(-1)^j{m+k\choose m+k-r-j}S(m+k-r-j, k-r)S(r+j, r)n^{r+j}.
			\label{id}
\end{eqnarray}
where $S(n, k)$ are the Stirling numbers of the second kind.
\end{proposition}
\begin{proof}
The proposition is proved with the help of standard arguments. Let us make the change of the argument of the e.g.f. (\ref{genf}):  $t\rightarrow -t$.  Obviously, the following relation holds
\begin{equation}
(-1)^r\left(\frac{e^{-nt}-1}{e^{t}-1}\right)^{r}=\sum_{q\geq 0}(-1)^qS^{(r)}_{q}(n)\frac{t^q}{q!}.
\label{id1}
\end{equation}
Multiplying both sides of (\ref{id1}) by $(e^t-1)^k$ and taking into account that
\[
(e^t-1)^k=k!\left(\sum_{q\geq 0} S(q, k)\frac{t^q}{q!}\right),\;\;
\forall k\geq 0,
\]
we obtain, as a result, the relation (\ref{id}).
\end{proof}
As a special case of (\ref{id}), for $r=k$, we get
\[
\sum_{q=0}^{m}(-1)^{q}{m+k\choose q}S(m+k-q, k)S_q^{(k)}(n) =(-1)^mS(k+m, k)n^{m+k}.
\]
In turn, if $k=1$, then the last relation turns into a well-known relation for the classical sums of powers \cite{Riordan}. The following theorem gives a some representation of the polynomials $\hat{S} _{m}^{(k)}(n) $.
\begin{theorem} \label{theorem1}
The polynomials $\hat{S}_{m}^{(k)}(n)$ can be written as
\begin{equation}
\hat{S}_{m}^{(k)}(n)=\frac{1}{{m+k\choose k}}\sum_{q=0}^{m}(-1)^q{m+k\choose q}B_q^{(k)}S(m+k-q, k)n^{m+k-q},
\label{higher-polynomials}
\end{equation}
where $B_q^{(k)}$ are the Bernoulli numbers of higher order that are defined by the e.g.f.
\begin{equation}
\frac{t^k}{(e^t-1)^k}=\sum_{q\geq 0}B^{(k)}_q\frac{t^q}{q!}.
\label{Bern-high}
\end{equation}
\end{theorem}
Comparing (\ref {higher-polynomials}) with the formula (\ref{22}), we see that the Bernoulli numbers here are replaced by their higher analogs. Moreover we see that this expression, in contrast with (\ref{22}), involves the Stirling numbers of the second kind.

\subsection{Bernoulli numbers of higher order}

The classical Bernoulli numbers are known to be uniquely determined by the e.g.f. \cite{Graham}
\begin{equation}
\frac{t}{e^t-1}=\sum_{q\geq 0}B_q\frac{t^q}{q!}.
\label{Bernoulli}
\end{equation}
It follows from (\ref{Bernoulli}) the recurrence relation
\begin{equation}
\sum_{q=0}^{m}{m+1\choose q}B_{q}=\delta_{0, m}.
\label{rec-rel}
\end{equation}
This relation, in fact, is one of the many recurrence relations for the Bernoulli numbers (see, for example, \cite{Agoh}). For example, one can derive a countable set of recurrence relations of the form
\begin{equation}
\sum_{q=0}^{m}{m+k\choose q}S(m+k-q, k)B_{q}=\frac{m+k}{k}S(m+k-1, k-1),\;\;
\forall k\geq 1.
\label{rec-rel1}
\end{equation}
The numbers $B_n^{(k)} $ first appeared in the work of N\"orlund \cite{Norlund} in connection with the theory of finite differences and were subsequently investigated from various points of view (see, for example, \cite{Carlitz}). It is known that they are related to each other by the recurrence relation \cite{Norlund}
\begin{equation}
B_n^{(k+1)}=\frac{k-n}{k}B_n^{(k)}-nB_{n-1}^{(k)}.
\label{Norlund}
\end{equation}
It should be noted that $B^{(k)}_n$ for some fixed $n\geq 0$ is calculated as the value of some polynomial in $k$. In what follows we shall use for these polynomials the same notation $B^{(k)}_n$ in the hope that this will not lead to confusion. In the literature, polynomials of this type are called the N\"orlund polynomials. The first six of them are as follows
\[
B_0^{(k)}=1,\;\;
B_1^{(k)}=-\frac{1}{2}k,\;\;
B_2^{(k)}=\frac{1}{12}k\left(3k-1\right),\;\;
B_3^{(k)}=-\frac{1}{8}k^2\left(k-1\right),
\]
\[
B_4^{(k)}=\frac{k}{240}\left(15k^3-30k^2+5k+2\right),\;\;
B_5^{(k)}=-\frac{1}{96}k^2\left(k-1\right)\left(3k^2-7k-2\right).
\]
\begin{remark}
By standard arguments, using the e.g.f. (\ref{Bern-high}), we can derive the following recurrence relation:
\begin{equation}
\sum_{q=0}^{m}{m+k\choose q}S(m+k-q, k)B_{q}^{(r)}=\frac{{m+k\choose k}}{{m+k-r\choose k-r}}S(m+k-r, k-r),\;\; \forall k\geq r,
\label{impl1}
\end{equation}
the particular case of which is (\ref{rec-rel1}). Making use (\ref{impl1}), we can prove that polynomials defined by the formula (\ref{higher-polynomials}) satisfy the recurrence relation
(\ref{id}), but this is not necessary here.
\end{remark}

\subsection{Stirling numbers of the second kind}

The numbers $S(n, k)$ in the formula (\ref{higher-polynomials}), as mentioned above, are Stirling numbers of the second kind. As is known, they satisfy the identity
\begin{equation}
S(n, k)=S(n-1, k-1)+kS(n-1, k)
\label{rr}
\end{equation}
and some boundary conditions \cite{Graham}. The Stirling numbers $S(m + k, k)$ for some fixed $m\geq 0$ can be calculated as the values of some polynomial $f_m (k)$ of degree $2m$. These polynomials satisfy the recurrence identity
\begin{equation}
f_m(k)-f_m(k-1)=kf_{m-1}(k),
\label{Stirling}
\end{equation}
which easily follows from the identity (\ref{rr}). In the literature they are known as the Stirling polynomials \cite{Gessel}, \cite{Gessel1}, \cite{Jordan}. The comparison (\ref{Norlund}) and (\ref{Stirling}) shows that the N\"orlund and Stirling polynomials are related by the relation (see, for example, \cite{Adelberg})
\begin{equation}
f_m(k)={m+k\choose m}B_m^{(-k)}.
\label{Stirling1}
\end{equation}
Using (\ref{Stirling1}), we can rewrite (\ref{impl1}) in the form
\[
\sum_{q=0}^{m}{m\choose q}B_{m-q}^{(-k)}B_{q}^{(r)}=B_{m}^{(r-k)}.
\]
The last relation is completely obvious, since it is simply a overwriting of the relation $B^{r-k} = B^rB^{-k} $, where $B = B (t)$ is the e.g.f. of the Bernoulli numbers.

\subsection{Proof of the theorem \ref{theorem1}}

In the formula (\ref{higher-polynomials}) we can replace $S(m + k-q, k)$ by $f_ {m-q}(k)$ and get the following representation:
\begin{equation}
\hat{S}_{m}^{(k)}(n)=\frac{1}{{m+k\choose k}}\sum_{q=0}^{m}(-1)^q{m+k\choose q}{m-q+k\choose m-q}B_q^{(k)}B_{m-q}^{(-k)}n^{m-q+k}.
\label{18}
\end{equation}
In turn, using the binomial identity
\[
{m+k\choose q}{m-q+k\choose m-q}={m+k\choose k}{m\choose q},
\]
we obtain a simpler expression
\begin{equation}
\hat{S}_{m}^{(k)}(n)=n^k\sum_{q=0}^{m}(-1)^q{m\choose q}B_q^{(k)}B_{m-q}^{(-k)}n^{m-q}.
\label{19}
\end{equation}
The formula (\ref{19}) represents the binomial convolution of two sequences: $\{a_m:=(-1)^{m} B_m^{(k)}\}$ and $\{b_m:=B_m^{(- k)}n^{m + k}\}$.
In turn, the e.g.f.'s of these sequences, as can be easily verified, are the $k$-degree of two e.g.f.'s
\begin{equation}
A(t)=\frac{-t}{e^{-t}-1}\;\;\mbox{and}\;\; C(n, t)=\frac{e^{nt}-1}{t}.
\label{AB}
\end{equation}
It is obvious that the representation of the e.g.f. (\ref{genf}) in the factorized form $G(n, t)=A(t)C(n, t)$ gives the simplest way to express the polynomials $\hat{S}_{m}(n)$ in terms of Bernoulli numbers. Since the expression (\ref{19}) is equivalent (\ref{18}), the theorem \ref{theorem1} should be considered proven.

\subsection{Some properties of the polynomials $\hat{S}_{m}^{(k)}(n)$}

In what follows we will consider these polynomials in the continuous variable $z\in\mathbb{C}$, so that we can talk about their derivatives. It is more convenient to investigate the polynomials\footnote{Obviously, $\hat{S}_{m}^{(k)}(z)=z^kQ_{m}^{(k)}(z)$.}
\[
Q_{m}^{(k)}(z)=\sum_{q=0}^{m}(-1)^q{m\choose q}B_q^{(k)}B_{m-q}^{(-k)}z^{m-q}.
\]
The following lemma will be useful for what follows.
\begin{lemma} \label{lemma1}
The following two identities are valid:
\begin{equation}
\sum_{q=0}^{m-1}{m\choose q}(m-q)B_q^{(k)}B_{m-q}^{(-k)}=\left\{
\begin{array}{l}
-kB_1,\;\;m= 1,\\[0.2cm]
kB_m,\;\;\forall m\geq 2
\end{array}
\right.
\label{191}
\end{equation}
and
\begin{equation}
\sum_{q=0}^{m-2}{m\choose q}(m-q)(m-q-1)B_q^{(k)}B_{m-q}^{(-k)}=\left\{
\begin{array}{l}
-k(3k-1)B_2
-2k^2B_{1},\;\;m= 2,\\[0.2cm]
-k((m+1)k-m+1)B_m
+mk^2B_{m-1},\;\;
\forall m\geq 3.
\end{array}
\right.
\label{1911}
\end{equation}
\end{lemma}
\begin{proof}
Let $B^{(k)}=B^{(k)}(t)$ be the e.g.f. of the Bernoulli numbers of higher order (\ref{Bern-high}). We have
\[
t\frac{d B^{(-k)}}{d t}B^{(k)}=kt\frac{e^t}{e^t-1}-k=k\left(A-1\right),
\]
where $A=A(t)$ is supposed to be the e.g.f. defined in (\ref{AB}). This relation implies (\ref{191}). To show (\ref{1911}), one needs to calculate
\begin{eqnarray}
t^2\frac{d^2 B^{(-k)}}{d t^2}B^{(k)}&=&\frac{t^2(k^2-k)e^{2t}}{(e^t-1)^2}+\frac{(t-2k)kte^t}{e^t-1}+k^2+k\nonumber\\
&=&k^2\left(A-1\right)^2-k\left(A^2-tA-1\right). \label{21}
\end{eqnarray}
Taking into account the recurrence relation (\ref{Norlund}), we derive from (\ref{21}) the relation (\ref{1911}).
\end{proof}
\begin{remark}
We could write (\ref{191}) as
\[
\sum_{q=0}^{m-1}{m\choose q}(m-q)B_q^{(k)}B_{m-q}^{(-k)}=(-1)^mkB_m,\;\;\forall m\geq 1.
\]
However, considering the fact that the Bernoulli numbers are nonzero only for even $m$ and $m=1$, we can get rid of the minus sign. A similar remark applies to the formula (\ref{1911}).
\end{remark}
Now we in a position to formulate some statements about the rational zeros of the polynomials $Q_{m}^{(k)}(z)$.
\begin{theorem} \label{2} The following two statements are true:
\begin{itemize}
\item[1)]
The polynomial $Q_{m}^{(k)}(z)$ has a simple root $z_0=-1$ for even $m$ and $m=1$ and any $k\geq 1$;
\item[2)]
The polynomial $Q_{m}^{(k)}(z)$ has a double root $z_0=-1$ for odd  $m\geq 3$ and any $ k\geq 1$.
\end{itemize}
\end{theorem}
\begin{proof}
It is obvious that
\[
Q_{m}^{(k)}(-1)=(-1)^{m}\sum_{q=0}^{m}{m\choose q}B_q^{(k)}B_{m-q}^{(-k)}=0,\;\;\forall m\geq 1,
\]
i.e, for any $m\geq 1$ and $k\geq 1$, the polynomial $Q_{m}^{(k)}(z)$ has the root $z_0=-1$. Further, we have
\[
\frac{d Q_{m}^{(k)}(z)}{dz}\left|_{z=-1} \right.=(-1)^{m-1}\sum_{q=0}^{m-1}{m\choose q}(m-q)B_q^{(k)}B_{m-q}^{(-k)}
\]
and by (\ref{191}), we obtain
\[
\frac{d Q_{m}^{(k)}(z)}{dz}\left|_{z=-1} \right.=
-kB_m,
\]
We know that the Bernoulli numbers $B_m$ are nonzero for even $m$ and $m=1$. Thus, the first part of the sentence (\ref{2}) is proved. For odd $m\geq 3$, the Bernoulli numbers are equal
zero and therefore the multiplicity of the root $z_0=-1 $ is at least equal to two. Next, we calculate
\[
\frac{d^2 Q_{m}^{(k)}(z)}{dz^2}\left|_{z=-1} \right.=(-1)^{m}\sum_{q=0}^{m-2}{m\choose q}(m-q)(m-q-1)B_q^{(k)}B_{m-q}^{(-k)}.
\]
Now we need to use the identity (\ref{1911}). By virtue of this identity, the second derivative $Q_{m}^{(k)}(z)$  at $z=-1$ for any $m\geq 2$ and $k\geq 1$ is not zero. Thus the theorem is proved.
\end{proof}
\begin{remark}
It was shown in the paper \cite{Kimura} that the rational zeros of the polynomials $\hat{S}_{m}^{(1)}(z)$ are only $0, -1$, and $-1/2$, and the last value is a simple root for these polynomials with even numbers.
\end{remark}
The polynomials $Q_{2}^{(k)}(z)$ and $Q_{3}^{(k)}(z)$, as is easily seen, have the following form:
\[
Q_{2}^{(k)}(z)=\frac{k}{12}(z+1)\left((3k+1)z+3k-1\right)\;\;\mbox{and}\;\;
Q_{3}^{(k)}(z)=\frac{k^2}{8}(z+1)^2\left((k+1)z+k-1\right).
\]
Hence it is obvious that
\[
Q_{2}^{(k)}\left(-\frac{3k-1}{3k+1}\right)=0,\;\;\forall k\geq 1\;\;\mbox{and}\;\;  Q_{3}^{(k)}\left(-\frac{k-1}{k+1}\right)=0,\;\;\forall k\geq 2.
\]

\section{Other generalizations  of power-sums}

\subsection{Multiple sums}

In the paper \cite{Svinin}, we considered multiple sums of the form
\begin{equation}
\tilde{\mathcal{S}}_{m}^{(k)}(n):=\sum_{\{q\}\in B_{k, kn}}\left(q_1^{m}+(q_2-n)^{m}+\cdots+(q_k-kn+n)^{m}\right),
\label{sums11}
\end{equation}
where  the exponent $m$ is supposed to be odd. Here $B_{k, s}:=\{q_j : 1\leq q_1\leq \cdots \leq q_k\leq s\}$. Obviously, if $m$ is odd, then the value of $q^m$ with negative $q$ in (\ref{sums11}) is  $q^m = -|q|^m$. According to this rule, a sum of the form (\ref{sums11}) can be rewritten as
\begin{equation}
\tilde{\mathcal{S}}_{m}^{(k)}(n)=\sum_{q=1}^{kn}c_q(k, n)q^m
\label{defin}
\end{equation}
with some integer coefficients $c_q(k, n)$.
\begin{example}
In the case $k=2$, we have
\begin{eqnarray}
\tilde{\mathcal{S}}_{m}^{(2)}(n)&=&\sum_{\{q\}\in B_{2, 2n}}\left(q_1^{m}+(q_2-n)^{m}\right)\nonumber\\
                                &=&\sum_{j=1}^{n-1}\left(\sum_{q=1}^{j}q^m-j(n-j)^m\right)+\sum_{j=1}^{n}\left(\sum_{q=1}^{n+j}q^m+(n+j)j^m\right).\nonumber
\end{eqnarray}
Hence we get
\[
c_q(2, n)=\left\{
\begin{array}{l}
2n+q+1,\;\; 1\leq q\leq n, \\[0.2cm]
2n-q+1,\;\; n+1\leq q\leq 2n.
\end{array}
\right.
\]
\end{example}

At first glance, multiple sums (\ref{sums11}) have nothing to do with  higher order power-sums  (\ref{sums}). However, in the paper \cite{Svinin}, it was conjectured that, in the case of odd $m$, the sum $\tilde{\mathcal{S}}_{m}^{(k)}(n)$ can be expressed  as\footnote{In this paper, the sums $S_{m}^{(k)}(n)$ were not identified as sums of higher orders.}
\begin{equation}
\tilde{\mathcal{S}}_{m}^{(k)}(n)=\sum_{q=0}^{k-1}{k(n+1)\choose q} S_{m}^{(k-q)}(n).
\label{relationsh}
\end{equation}
This hypothesis was based on numerous calculations using computer algebra method. Note that for $k=1$ the hypothesis is clear, and for $k=2$ it is easy to prove. By the way, if (\ref{relationsh}) is true, then it follows that the integer coefficients $c_q(k, n)$ are, in addition, non-zero and positive.

\subsection{Binomial sums}

On the right-hand side of (\ref{relationsh}) there is a sum, which we, somewhat conditionally, call binomial. Let us, throughout the  rest of the paper, study a sum of the form \cite{Svinin}
\begin{equation}
\mathcal{S}_{m}^{(k)}(n)=\sum_{q=0}^{k-1}{k(n+1)\choose q} S_{m}^{(k-q)}(n).
\label{sums1}
\end{equation}
In what follows, it is useful to consider its particular case
\[
\mathcal{S}_{m}^{(k)}(1)=\sum_{q=0}^{k-1}{2k\choose q} (k-q)^m.
\]
Observe that in the case of an odd exponent $m$, we can write
\begin{equation}
2\mathcal{S}_{m}^{(k)}(1)=\sum_{q=0}^{2k}{2k\choose q}|k-q|^m.
\label{23}
\end{equation}
On the right-hand side of this identity there is the binomial sum which was investigated in the papers \cite{Strazdins}, \cite{Tuenter}. Using the relation (\ref{23}), we can transfer some results of \cite{Tuenter} to the sum $\mathcal{S}_{m}^{(k)}(1)$.
\begin {proposition}
The sum $\mathcal{S}_{m}^{(k)}(1)$ satisfies the recurrence relation
\begin {equation}
\mathcal{S}_{m+2}^{(k)}(1)=k^2\mathcal{S}_{m}^{(k)}(1)-2k(2k-1)\mathcal{S}_{m}^{(k-1)}(1).
\label{241}
\end{equation}
\end{proposition}
\begin{proof}
We have
\begin{eqnarray}
k^2\mathcal{S}_{m}^{(k)}(1)-\mathcal{S}_{m+2}^{(k)}(1)&=&\sum_{q=1}^{k-1}{2k\choose q} (k-q)^m\left(k^2-(k-q)^2\right)\nonumber\\
&=&\sum_{q=1}^{k-1}{2k\choose q} (k-q)^mq\left(2k-q\right).\nonumber
\end{eqnarray}
Using the binomial identity
\[
{2k\choose q}={2k-2\choose q-1}\frac{2k(2k-1)}{q(2k-q)},
\]
we obtain, as a result, the relation (\ref{241}).
\end{proof}
\begin{proposition} There is an infinite set of polynomials $\{P_m(k): m\geq 0 \}$ such that,
\begin{equation}
\mathcal{S}_{2r+1}^{(k)}(1)=P_r(k)\frac{k}{2}{2k\choose k}.
\label{251}
\end{equation}
\end{proposition}
\begin{proof}
Let us first calculate $\mathcal{S}_{1}^{(k)}(1)$. We have
\begin{eqnarray}
\mathcal{S}_{1}^{(k)}(1)&=&\sum_{q=0}^{k}{2k\choose q} (k-q)=\frac{1}{2}\sum_{q=0}^{k}{2k\choose q} \left((2k-q)-q\right)\nonumber\\
&=&\frac{1}{2}\sum_{q=0}^{k}{2k\choose q} (2k-q)-\frac{1}{2}\sum_{q=1}^{k}{2k\choose q}q.\nonumber
\end{eqnarray}
Using binomial identities
\[
{2k\choose q}=\frac{2k}{2k-q}{2k-1\choose q}=\frac{2k}{q}{2k-1\choose q-1},
\]
we get
\[
\mathcal{S}_{1}^{(k)}(1)=k\sum_{q=0}^{k}{2k-1\choose q}-k\sum_{q=1}^{k}{2k-1\choose q-1}=k{2k-1\choose k}=\frac{k}{2}{2k\choose k}.
\]
Thus, $\mathcal{S}_{1}^{(k)}(1) $ is actually calculated by the formula (\ref{251}), with $P_0(k)=1$. Now, using the identity (\ref{241}) and the binomial identity
\[
2k(2k-1){2k-2\choose k-1}=k^2{2k\choose k},
\]
we see that if $\mathcal{S}_{2r+1}^{(k)}(1)$ is expressed by the formula (\ref{251}) with some polynomial $P_r(k)$, then $\mathcal{S}_{2r+3}^{(k)}(1)$ is also expressed by this expression with some polynomial $P_{r+1}(k)$, defined by
\begin{equation}
P_{r+1}(k)=k^2P_{r}(k)-k(k-1)P_{r}(k-1).
\label{pol}
\end{equation}
Therefore, by the method of mathematical induction, we prove the proposition.
\end{proof}
It is important to observe that the polynomials $P_{r}(k)$ are closely related to the well-known Gandhi polynomials \cite{Gandhi}, that are defined by the recurrence relation
\[
F_{r+1}(k)=(k+1)^2F_{r}(k+1)-k^2F_{r}(k),
\]
starting with $F_1(k)=1$. Indeed, it is not hard to see that $P_r(k)=(-1)^{r+1}kF_{r}(-k)$ for $r\geq 1$. The first six Gandhi polynomials are as follows:
\[
F_1(k)=1,\;\;
F_2(k)=2k+1,\;\;
F_3(k)=6k^2+8k+3,\;\;
F_4(k)=24k^3+60k^2+54k+17,
\]
\[
F_5(k)=120k^4+480k^3+762k^2+556k+155,
\]
\[
F_6(k)= 720k^5+4200k^4+10248k^3+12840k^2+8146k+2073.
\]
\begin{remark}
The Gandhi polynomials appeared in the paper \cite{Gandhi}, in which the hypothesis was formulated that
$F_r(0)=|G_{2r}|$, where $G_{2r}=2(1-4^r)B_{2r}$ are the alternating Genocci numbers determined by the e.g.f.
\[
\frac{2t}{e^t+1}=t+\sum_{q \geq 1}G_{2q}\frac{t^{2q}}{(2q)!}.
\]
This hypothesis was proved in the papers \cite{Carlitz2}, \cite{Riordan1}. It should be noted that the Genocci numbers, unlike the Bernoulli numbers, are integers.
\end{remark}
\begin{remark}
The Gandhi polynomials can be written as $F_{r}(k)=F_r(1, 1, k)$, where $F_r(x, y, z)$ are some polynomials of three variables that are defined by the recurrence relation
\[
F_{r+1}(x, y, z)=(z+x)(z+y)F_{r}(x, y, z+1)-z^2F_{r}(x, y, z),
\]
starting with $F_1(x, y, z)=1$. In the literature, these polynomials are known as the Dumont-Foath polynomials \cite{Carlitz1}, \cite{Dumont}. It turns out that these polynomials are symmetric with respect to the variables $x, y, z$. The main property of these polynomials is that $F_r(1, 1, 1)=|G_{2r+2}|$.
\end{remark}

We observe now that  the formula (\ref{251}) can be rewritten in the form
\[
\mathcal{S}_{2r+1}^{(k)}(1)=(-1)^{r+1}F_r(-k)k^2{2k-1\choose k-1}
\]
for $r\geq 1$.

\subsection{Polynomials corresponding to binomial sums}

Let us define the set of polynomials $\{\hat{\mathcal{S}}_m^{(k)}(z): m\geq 1\} $ by the relation
\[
\hat{\mathcal{S}}_m^{(k)}(z)=\sum_{q=0}^{k-1}{k(z+1)\choose q} \hat{S}_{m}^{(k-q)}(z).
\]
Our goal is to describe these polynomials. For the degree $m=1$ we can formulate the following result:
\begin{proposition} \label{pr:3.2}
The polynomial $\hat{\mathcal{S}}_1^{(k)}(z)$ can be  defined by
\[
\hat{\mathcal{S}}_1^{(k)}(z)=\frac{z(z+1)}{2}k{k(z+1)-1\choose k-1},\;\;
\forall k\geq 1.
\]
\end{proposition}
To prove this proposition, it suffices to use the following lemma:
\begin{lemma} \label{lemma2}
The relation
\begin{equation}
{k(z+1)-1\choose k-1}=\frac{1}{k}\sum_{q=0}^{k-1}(k-q){k(z+1)\choose q}z^{k-q-1}
\label{10}
\end{equation}
is an identity.
\end{lemma}
\begin{proof}
Let us denote
\[
g_k(z):=k{k(z+1)-1\choose k-1}\;\;\mbox{and}\;\;\tilde{g}_k(z):=\sum_{q=0}^{k-1}(k-q){k(z+1)\choose q}z^{k-q-1}.
\]
Since $g_k(z)$ and $\tilde{g}_k(z)$ are polynomials in $z$ of order $k-1$, it is obvious that to prove the identity (\ref{10}, it suffices to show that
\[
g_k^{(r)}(0)=\tilde{g}_k^{(r)}(0),\;\; \forall r=0,\ldots, k-1.
\]
For example, it is easy to see that $g_k(0)=\tilde{g}_k(0)=k$. Let $\xi:=k!$, while $\xi_{i_1,\ldots, i_m}$ denote the product $\prod_{i=1}^ki $, where there are no numbers $i_1<i_2<\ldots <i_m$ . With this notation, we can write
\begin{equation}
g_k^{(r)}(0)=\frac{r!k^r}{(k-1)!}\sum_{1\leq i_1<\cdots< i_r\leq k-1}\xi_{i_1, \ldots, i_r}.
\label{24}
\end{equation}
In turn, for $\tilde{g}_k(z)$, we get the following:
\begin{equation}
\tilde{g}_k^{(r)}(0)=r!\sum_{q=0}^{r}\frac{k^{r-q}}{q!(k-q-1)!}\sum_{q+1\leq i_1<\cdots< i_{r-q}\leq k-1}\xi_{i_1+1, \ldots, i_{r-q}+1}.
\label{25}
\end{equation}
Equating (\ref{24}) to (\ref{25}) and multiplying the resulting relation by $(k-1)!/r!$, we get the equality
\begin{equation}
k^r\sum_{1\leq i_1<\cdots< i_r\leq k-1}\xi_{i_1, \ldots, i_r}=\sum_{q=0}^{r}{k-1\choose q}k^{r-q}\sum_{q+1\leq i_1<\cdots< i_{r-q}\leq k-1}\xi_{i_1+1, \ldots, i_{r-q}+1},
\label{i0}
\end{equation}
that remains for us to prove.

For convenience, let us denote
\[
D_{q}:=\{i_j : q+1\leq i_1<\cdots< i_{r-q}\leq k-1\}.
\]
Obviously, the partition
\[
D_{0}=D^{(1)}_{0}\bigsqcup D^{(2)}_{0}
\]
with the parts
\[
D^{(1)}_{0}:=\left\{i_j : i_1=1;\; 2\leq i_2<\cdots< i_r\leq k-1\right\}
\]
and
\[
D^{(2)}_{0}:=\left\{i_j :  2\leq i_1<\cdots< i_r\leq k-1\right\}
\]
gives the relation
\begin{eqnarray}
\sum_{\{i_j\}\in D_{0}}\xi_{i_1, \ldots, i_r}&=&\sum_{2\leq i_2<\cdots< i_r\leq k-1}\xi_{1, i_2, \ldots, i_{r}}+\sum_{2\leq i_1<\cdots< i_r\leq k-1}\xi_{i_1, \ldots, i_{r}}\nonumber\\
&=&\sum_{\{i_j\}\in D_{1}}\xi_{i_1, \ldots, i_{r-1}}+\sum_{2\leq i_1<\cdots< i_r\leq k-1}\xi_{i_1, \ldots, i_{r}}.   \label{i1}
\end{eqnarray}
On the other hand, making  use  a partition
\[
D_{0}=\tilde{D}^{(1)}_{0}\bigsqcup \tilde{D}^{(2)}_{0}
\]
with
\[
\tilde{D}^{(1)}_{0}:=\left\{i_j : i_r=k-1;\; 1\leq i_1<\cdots< i_{r-1}\leq k-2\right\}
\]
and
\[
\tilde{D}^{(2)}_{0}:=\left\{i_j :  1\leq i_1<\cdots< i_r\leq k-2\right\},
\]
gives the relation
\begin{eqnarray}
\sum_{\{i_j\}\in D_{0}}\xi_{i_1+1, \ldots, i_r+1}&=&\sum_{1\leq i_1<\cdots< i_{r-1}\leq k-2}\xi_{i_1+1, \ldots, i_{r-1}+1, k}+\sum_{1\leq i_1<\cdots< i_r\leq k-2}\xi_{i_1+1, \ldots, i_{r}+1}\nonumber\\
&=&\frac{1}{k}\sum_{\{i_j\}\in D_{1}}\xi_{i_1, \ldots, i_{r-1}}+\sum_{2\leq i_1<\cdots< i_r\leq k-1}\xi_{i_1, \ldots, i_{r}}.
\label{i2}
\end{eqnarray}
Using (\ref{i1}) and (\ref{i2}), we can rewrite (\ref{i0}) as
\[
(k-1)k^{r-1}\sum_{\{i_j\}\in D_{1}}\xi_{i_1, \ldots, i_{r-1}}=\sum_{q=1}^{r}{k-1\choose q}k^{r-q}\sum_{\{i_j\}\in D_{q}}\xi_{i_1+1, \ldots, i_{r-q}+1}.
\]
Further, using similar partitionings of the sets $D_1, D_2,\ldots$, we continue step-by-step the transformations of the initial relation. At the $m$-th step we obtain this relation in the form
\[
{k-1\choose m}k^{r-m}\sum_{\{i_j\}\in D_{m}}\xi_{i_1, \ldots, i_{r-m}}=\sum_{q=m}^{r}{k-1\choose q}k^{r-q}\sum_{\{i_j\}\in D_{q}}\xi_{i_1+1, \ldots, i_{r-q}+1}.
\]
It is obvious that on the $r$-th step we obtain a trivial identity. Thus, the lemma is proved.
\end{proof}
Taking into account that $S^{(k)}_1(z)=k(z+1)z^k/2$, from the lemma \ref{lemma2} we obtain the proposition \ref{pr:3.2}.
This proposition  is consistent with the assumption made in the paper \cite{Svinin}, according to which the polynomials $\hat{\mathcal{S}}_{2r+1}^{(k)}(z)$, for $r\geq 0$ are determined by\footnote{Here this formula is written somewhat differently than in the work \cite{Svinin}.}
\begin{equation}
\hat{\mathcal{S}}_{2r+1}^{(k)}(z)=\left(-\frac{w}{2}\right)^{r+1}F_r(w, -k)k^2{k(z+1)-1\choose k-1},
\label{36}
\end{equation}
where $F_r(w, k)$ are certain polynomials in $k$ of degree $r-1$ with coefficients that are Laurent polynomials in the variable $w:=z(z+1) $. In particular, for $z=1$, we have
\[
\hat{\mathcal{S}}_{2r+1}^{(k)}(1)=(-1)^{r+1}F_r(2, -k)k^2{2k-1\choose k-1}=(-1)^{r+1}F_r(2, -k)\frac{k^2}{2}{2k\choose k}.
\]
The last expression agrees with (\ref{251}).

The first six polynomials $F_r(w, k)$ in $k $ are defined as follows: \cite{Svinin}:
\[
F_1(w, k)=1,\;\;
F_2(w, k)=2k+\frac{2}{3}\frac{w+1}{w},\;\;
\]
\[
F_3(w, k)=6k^2+\frac{16}{3}\frac{w+1}{w}k+\frac{4}{3}\frac{(w+1)^2}{w^2},\;\;
\]
\[
F_4(w, k)= 24k^3+40\frac{(w+1)}{w}k^2+24\frac{(w+1)^2}{w^2}k+\frac{24}{5}\frac{(w+1)^3}{w^3}+\frac{8}{5}\frac{1}{w},
\]
\begin{eqnarray}
F_5(w, k)&=& 120k^4+320\frac{(w+1)}{w}k^3+\frac{1016}{3}\frac{(w+1)^2}{w^2}k^2\nonumber\\
    &&+\left(160\frac{(w+1)^3}{w^3}+32\frac{1}{w}\right)k+\frac{80}{3}\left(\frac{(w+1)^4}{w^4}+\frac{w+1}{w^2}\right)\nonumber
\end{eqnarray}
\begin{eqnarray}
F_6(w, k)&=& 720k^5+2800\frac{(w+1)}{w}k^4+\frac{13664}{3}\frac{(w+1)^2}{w^2}k^3\nonumber\\
&&+\left(\frac{55936}{15}\frac{(w+1)^3}{w^3}+\frac{2544}{5}\frac{1}{w}\right)k^2+\left(\frac{22112}{15}\frac{(w+1)^4}{w^4}+\frac{13664}{15}\frac{(w+1)}{w^2}\right)k\nonumber\\
&&+\frac{22112}{105}\frac{(w+1)^5}{w^5}+\frac{44224}{105}\frac{(w+1)^2}{w^3}.\nonumber
\end{eqnarray}
A direct check shows that for these examples, $F_r(2, k)=F_r(k)$ is true, where the polynomials $F_r(k)$ are the Gandhi polynomials. The examples written out above suggest that polynomials
$F_r(w, k)$ should be written in the form
\[
F_r(w, k)=\sum_{q\geq 0}\frac{1}{w^q}\sum_{j=0}^{r-3q-1}F^{(q)}_{r,j}\frac{(w+1)^{r-j-3q-1}}{w^{r-j-3q-1}}k^j
\]
with some coefficients $ F^{(q)}_{r, j}$. For any fixed $r$, the sum over $q$ is in fact finite, since, by convention, if $r-3q-1\leq -1 $, then the sum over $j$ is assumed to be zero.


TITLE: Is there convenient notation for Viète's formula?
QUESTION [2 upvotes]: Is there a convenient way to write Viète's formula $\displaystyle \frac2\pi=
\frac{\sqrt2}2\cdot
\frac{\sqrt{2+\sqrt2}}2\cdot
\frac{\sqrt{2+\sqrt{2+\sqrt2}}}2\cdots$
using sigma and/or pi notation
a) without recursive notation;
b) even with recursive notation?
As many ways as possible would be appreciated. Thanks in advance!

REPLY [11 votes]: To reach the final result, you can follow these detailed steps:

Write the double-angle formula
$$
    \sin x = 2 \sin\frac x2 \cos\frac x2
$$
and apply it repeatedly to finally get
$$
    \sin x = 2^n \sin\frac{x}{2^n} \prod_{i = 1}^{n} \cos\frac{x}{2^i} \quad (n \in \mathbb{Z}^*). \tag{$*$}
$$

Divide both sides of $(*)$ by $x$ (given $x > 0$), then take the limit of both sides of the equation when $n \to \infty$. Note that the change of value of $n$ has no effects on LHS, so we get
$$
    \frac{\sin x}{x}
  = \lim_{n \to \infty} \Big(
        \frac{\sin(x / 2^n)}{x / 2^n} \prod_{i = 1}^{n} \cos\frac{x}{2^i}
    \Big)
  = \prod_{i = 1}^{\infty} \cos\frac{x}{2^i}.
$$
This is the result Tien Kha Pham provided directly. (Note that $\prod_{i = 1}^{\infty} \cos(x / 2^i)$ makes sense only if the corresponding limit $\lim_{n \to \infty} \prod_{i = 1}^{n} \cos(x / 2^i)$ exists.)

To relate this result to the form of Viète's formula you mentioned in the question, substituting $x = \pi / 2$ with formulas
$$
    \cos \Big(\frac{\pi}{4}\Big) = \frac{\sqrt{2}}{2}
    \quad\text{and}\quad
    \cos \Big(\frac{x}{2^k}\Big) = \frac 12 \sqrt{2 + 2\cos \Big(\frac{x}{2^{k - 1}}\Big)} \quad (k \ge 2),
$$
we finally conclude with
$$
    \frac{\pi}{2} = \prod_{k = 2}^{\infty} a_k,
$$
where $a_2 = \sqrt{2} / 2$, $a_3 = \sqrt{2 + 2a_2} / 2 = \sqrt{2 + \sqrt{2}} / 2$ and $a_{k + 1} = \sqrt{2 + 2a_k} / 2$ for $k \ge 3$.

Once there was a minor mistake in the answer made by Tien Kha Pham. The typo no longer existed. But another consequent mistake, in the comment to that answer, will remain. Therefore I keep the following words visible only as a reminder for those who may get confused by the inconsistency between formulas.

Just a correction to the answer of Tien Kha Pham (since I don't have
enough reputation to add a comment): it should be $$ \frac{2}{\pi} =
 \prod_{k = 2}^{\infty} \cos \Big( \frac{\pi}{2^k} \Big), $$ of which
the $\pi$ appears in numerator, not denominator.


TITLE: Cis Graphs (Finding Values)
QUESTION [4 upvotes]: For this question:

If you guys are not familiar with the $\operatorname{cis}$ notation
$$ \operatorname{cis}(x) = (\cos(x) + i\sin(x)) $$
I've tried doing this:
By comparing arguments
$$ x^2 - x = 0\Longrightarrow x(x-1) = 0\Longrightarrow x = 0 ,~ x = 1 $$
But clearly from the graph there are more intersections and searching the internet I learnt you can do this
$$ x^2  = x + 2k\pi $$
But why can we plus $2k\pi$ and does the $2k\pi$ have to be on the right side of the equation why can't it be on the left hand side?

REPLY [1 votes]: $$\text{cis}\left(x^2\right)=\text{cis}(x)\Longleftrightarrow$$
$$\cos\left(x^2\right)+\sin\left(x^2\right)i=\cos(x)+\sin(x)i\Longleftrightarrow$$
$$e^{ix^2}=e^{ix}\Longleftrightarrow$$
$$\ln\left(e^{ix^2}\right)=\ln\left(e^{ix}\right)\Longleftrightarrow$$
$$ix^2=ix\Longleftrightarrow$$
$$x^2=x\Longleftrightarrow$$

Notice, if $a,b\in\mathbb{R}$ and $k\in\mathbb{Z}$
$a+bi=\sqrt{a^2+b^2}e^{\arg(a+bi)i}=\sqrt{a^2+b^2}e^{\left(2\pi k+\arg(a+bi)\right)i}$

$$x^2-x+2\pi k=0\Longleftrightarrow$$

Use the abc-formula:

$$x=\frac{1}{2}\pm\frac{1}{2}\sqrt{1-8\pi k}\space\space\space\space\space\space\space\space\space\space\space\space\text{with}\space k\in\mathbb{Z}$$


TITLE: Show that $2x=\tan x$ has infinitely many solutions using the Intermediate Value Theorem.
QUESTION [0 upvotes]: I understand what the Intermediate Value Theorem is, but am not sure how to apply it to a $\tan x$ function being that there are many points of discontinuity.

REPLY [0 votes]: $\tan(x)$ is continuous on $[n\pi, n\pi + \pi/2)$ for all $n \in \mathbb{N}$. Then, by the Intermediate Value Theorem, it attains every value in $[0,\infty)$ in that interval for each $n$. In particular, it attains the value $2x$ once in each interval. Thus, the equation $2x = \tan(x)$ has infinitely many solutions.


TITLE: transformation of differential equations
QUESTION [1 upvotes]: If I have this D.E:
$$y^{''}+P(x)y^{'}+Q(x)y=0$$
which transformation $y \to u(x)$, shall I take to transform the D.E to: $u^{''}+q(x)u=0$?

REPLY [2 votes]: The answer is yes. Let $y=\rho(x)u(x)$ and then put it in the equation to get
$$ \rho''(x)u(x)+2\rho'(x)u'(x)+\rho(x)u''(x)+P(x)(\rho'(x)u(x)+\rho(x)u'(x))+Q(x)\rho(x)u(x)=0. \tag1 $$
Write (1) as an equation of $u$:
$$ \rho(x)u''(x)+(2\rho'(x)+P(x)\rho(x))u'(x)+(\rho''(x)u(x)+P(x)\rho'(x)+Q(x)\rho(x))u(x)=0. \tag2$$
Letting
$$ 2\rho'(x)+P(x)\rho(x)=0 $$
which implies
$$ \rho(x)=\exp\bigg(-\frac12\int P(x)dx\bigg) $$
and then dividing both sides of (2) by $\rho(x)$, one has
$$ u''(x)+q(x)u(x)=0 $$
for some $q(x)$ depending on $P(x),Q(x)$.


TITLE: Is there a way of detecting nearby type 1a supernovas?
QUESTION [4 upvotes]: The SuperNova Early Warning System (SNEWS) consists of several neutrino detectors across the world that can detect the neutrinos generated by a nearby supernova.  As I understand, a Type 1a supernova would not generate a large amount of neutrinos and thus would not be detectable by SNEWS.
Is there some other way we'd be able to detect a nearby Type 1a supernova before the light from it becomes visable?

REPLY [5 votes]: In short, it is extremely challenging if not impossible to detect SNe Ia before we see them.
As you said, the neutrino signal would be even weaker than for core-collapse events, and even that is pretty weak. We only caught one or two dozen neutrinos from SN 1987A, and that essentially went off in our own galaxy.
No other particles would escape faster than light could diffuse out of the explosion, even if they could be accelerated to ultra-relativistic speeds.
The only other long-range signal would be gravitational waves. The problem is that if SNe Ia are triggered by slow mass accretion onto a white dwarf (the single degenerate channel), the quadrupole moment will be no stronger than for any other binary star system in the universe. Even if white dwarf mergers turn out to be responsible (the double degenerate channel), some speculate that they could be third-body-induced head-on collisions rather than inspirals, so again no quadrupole moment to speak of.
There is a chance most SNe Ia are the result of inspiralling white dwarfs, which is the best case for emitting gravitational radiation. In this case, though, the signal is even weaker than neutron-star/neutron-star mergers (which are in turn weaker than the black-hole/black-hole merger so far detected), simply because the masses are smaller. On top of that, the inspiral (and gravitational wave emission) ends shortly after the surfaces collide, which is well before the objects' Schwarzschild radii merge, so the chirp in the signal is cutoff before its strongest point.
Also, unlike in the core-collapse case, there is hardly anything worth looking at just before the explosion. White dwarfs are hard enough to spot nearby in our own galaxy; they are next to impossible to see in other galaxies.
On the bright side, SNe Ia actually grow brighter for the first week or two after explosion, so there is still a sense in which we can catch them early. Indeed this and the next generation of high-cadence optical surveys are making this more and more common, as there is important information to be learned from the rise of the light curve.


TITLE: Find a line with maximum points from N points
QUESTION [5 upvotes]: You are given N points and you want to draw a line such that maximum points lie on the line. What is the efficient way to find the number of maximum points?

REPLY [0 votes]: Follows a script in MATHEMATICA which implements a procedure to obtain the line passing through more points.
NOTE
This script admits many improvements regarding computational efficiency. It was released in this basic form with the proposal to show the geometric ideas involved in the algorithm. Calling $p_i, i=1,\cdots n$ the set of points the first step is to form a matrix $V$ such that $V_{i,j} = p_i-p_j$. After, we construct a list $\mathcal{L}$ (list0) with the element indexes in $V$ such that for $i\ne j\ne k$ we have $V_{i,k}\times V_{j,k}=0$. The last operation consists in joining elements on $\mathcal{L}=\{\mathcal{l}_1,\cdots,\mathcal{l}_p,\cdots,\mathcal{l}_q,\cdots,\mathcal{l}_m\}$ such that $\mathcal{l}_p,\mathcal{l}_q$ have at least two indexes (points) in common. The biggest resulting elements in this union are the solutions.
n = 30;
SeedRandom[5]
list = {};
points = Union[RandomInteger[{0, 10}, {n, 2}]];
n = Length[points]
V = Table[Join[points[[i]] - points[[j]], {0}], {i, 1, n}, {j, 1, n}];
For[i = 1, i <= n, i++,
 For[j = 1, j <= n, j++,
  For[k = 1, k <= n, k++,
   If[i != k && j != k && i != j,
    cross = Cross[V[[i, k]], V[[j, k]]];
    If[cross == {0, 0, 0}, AppendTo[list, Sort[{i, j, k}]]]]
  ]
 ]
]
list0 = Union[list];

lines = {};
npts = 0;
For[i = 1, i <= Length[list0], i++,
 lini = list0[[i]];
 line = lini;
 For[j = 1, j < i, j++,
  linj = list0[[j]];
  If[Length[Intersection[lini, linj]] == 2, line = Union[line,linj]]
 ];
 nl = Length[line];
 If[nl >= npts,
  AppendTo[lines, line];
  npts = nl
 ]
]
nlin = Length[lines]
LINES = {};
For[k = 1, k <= nlin, k++,
 If[Length[lines[[k]]] == npts, AppendTo[LINES, lines[[k]]]]]
LINES = Union[LINES]
NL = Length[LINES];
gr1 = ListPlot[points];
gr2 = Flatten[Table[Graphics[{Red, PointSize[0.02], Point[points[[list0[[j, k]]]]]}], {k, 1, 3}, {j, 1, Length[list0]}]];
gr2a = Flatten[Table[Table[ParametricPlot[lambda points[[LINES[[j, k]]]] + (1 - lambda) points[[LINES[[j, k + 1]]]], {lambda, 0, 1}], {k, 1, Length[LINES[[j]]] - 1}], {j, 1, NL}]];
Show[gr1, gr2a, gr2, PlotRange -> All, Axes -> False, AspectRatio -> 1]

with SeedRandom[51]

and with SeedRandom[52]


TITLE: Can we invoke "almost supermartingale" Theorem for deterministic sequences?
QUESTION [1 upvotes]: Perhaps stupid question.

Question: Can "almost supermartingale" theorem be equally applicable to prove the convergence of some algorithms solving non-random optimization problems?

Attempt for a non-random version of "almost supermartingale" theorem (without any proof):

Let the non-negative sequences be $\{V^k\}$, $\{S^k\}$, and $\{ U^k \}$ for $k=0,1,2,\ldots$. Let $\beta_0, \beta_1$ be non-negative scalars satisfying $\sum_{k=0}^\infty \beta_k < \infty$. Assume
$$V^{k+1} \leq \left( 1 + \beta_k \right) V^k - S^k + U^k$$
and
$$\sum_{i={\color{red}{0 \text{ or } 1?}}}^\infty U^i < \infty.$$
Then,

$V^k \rightarrow V^\infty$
$\sum_{k=0}^\infty S^k < \infty$.

EDIT:
Additionally, let
$V^k := \| x^k - x^\star \|_P^2$ where $P$ is a positive semidefinite matrix. Then, using the above theorem, can one say for a subsequence $n_k$ $\lim_{k \rightarrow \infty} x^{n_k} =  x^\star$? or do we need more information to prove this?

For the completeness, I have also copied the "almost supermartingale" theorem from [Theorem 30, page 300].

Theorem: Let $V^k$, $S^k$, and $ U^k$ be $\mathcal{F}_k$-measurable random variables satisfying $V^k \geq 0$, $S^k \geq 0$, and $ U^k\geq 0$ for $k=0,1,2,\ldots$. Let $\beta_0, \beta_1$ be non-negative scalars satisfying $\sum_{k=0}^\infty \beta_k < \infty$.
Assume
$$\mathbb{E}\left[V^{k+1} \mid \mathcal{F}_k \right] \leq \left( 1 + \beta_k \right) V^k - S^k + U^k$$
and
$$\sum_{i={\color{red}{0 \text{ or } 1?}}}^\infty U^i < \infty$$
almost surely. Then,

$V^k \rightarrow V^\infty$
$\sum_{k=0}^\infty S^k < \infty$.

REPLY [1 votes]: For $k=0,1,\dots$, let $v_k:=V^k$, $s_k:=S^k$, $u_k:=U^k$, and $b_k:=\beta_k$, so that the $v_k$'s, $s_k$'s, $u_k$'s, and $b_k$'s are nonnegative real numbers such that $\sum_{k=0}^\infty b_k<\infty$,
\begin{equation*}
    \sum_{k=0}^\infty u_k<\infty, \tag{1}
\end{equation*}
and
\begin{equation*}
    v_{k+1}\le c_k v_k-s_k+u_k \tag{2}
\end{equation*}
for all $k$, where
\begin{equation*}
    c_k:=1+b_k\ge1,
\end{equation*}
so that
\begin{equation*}
    \prod_{k=0}^\infty c_k \text{ converges (to a number in $[1,\infty)$).} \tag{3}
\end{equation*}
It follows by (2) that $v_{k+1}\le c_k v_k+u_k$ for all $k$ and hence, by induction on $k$,
\begin{equation*}
    v_k\le\Big(v_0+\sum_{j=0}^{k-1}u_j\Big)\prod_{j=0}^{k-1}c_j \tag{4}
\end{equation*}
and hence
\begin{equation*}
    0\le v_k\le M:=\Big(v_0+\sum_{j=0}^\infty u_j\Big)\prod_{j=0}^\infty c_j<\infty. \tag{5}
\end{equation*}
Similarly to (4),
\begin{equation*}
    v_k\le\Big(v_n+\sum_{j=n}^{k-1}u_j\Big)\prod_{j=n}^{k-1}c_j \tag{6}
\end{equation*}
for any natural $k$ and $n$ such that $k>n$. Therefore,
\begin{equation*}
    \limsup_{k\to\infty}v_k\le\Big(v_n+\sum_{j=n}^\infty u_j\Big)\prod_{j=n}^\infty c_j
\end{equation*}
and hence, in view of (1) and (3),
\begin{equation*}
    \limsup_{k\to\infty}v_k\le\big(\liminf_{n\to\infty}v_n+0\big)\times1
    = \liminf_{n\to\infty}v_n.
\end{equation*}
So, in view of (5), there exists
\begin{equation*}
    v_\infty:=\lim_{k\to\infty}v_k\in[0,M|\subset[0,\infty). \tag{7}
\end{equation*}
Similarly to (4), for all $k$
\begin{equation*}
    v_k\le\Big(v_0+\sum_{j=0}^{k-1}u_j\Big)\prod_{j=0}^{k-1}c_j - \sum_{j=0}^{k-1}s_j,
\end{equation*}
whence
\begin{equation*}
    \sum_{j=0}^{k-1}s_j\le\Big(v_0+\sum_{j=0}^{k-1}u_j\Big)\prod_{j=0}^{k-1}c_j.
\end{equation*}
Letting now $k\to\infty$ and recalling (1) and (3), we see that
\begin{equation*}
    \sum_{j=0}^\infty s_j<\infty. \tag{8}
\end{equation*}
Relations (7) and (8) are what was desired.


TITLE: Prove that limit of integral is 0 (without dominated convergence theorem)
QUESTION [0 upvotes]: Let $r,x,\sigma \in \mathbb{R}^+$.
Let $$f(x) = x\left(\frac{x^2}{2\sigma^2}-1\right)\exp\left(-\frac{x^2}{2\sigma^2} \right)$$
It can be shown that
$$ \int_0^{+\infty} f(x)~dx = 0$$
I would like to prove that
$$ \lim_{r\rightarrow +\infty} \int_0^{+\infty} f(x)\sqrt{r^2+x^2}~dx = 0, $$
How to do that ? (I guess I cannot use the dominated convergence theorem here, to interchange limit and integral, since $\lim_{r\rightarrow +\infty} f(x)\sqrt{r^2+x^2}=\pm\infty$, right ?)

REPLY [3 votes]: Under your assumptions,
$$\int^\infty_0f(x)\sqrt{r^2+x^2}\,dx=\int^\infty_0f(x)\left(\sqrt{r^2+x^2}-r\right)\,dx.$$ Now $\sqrt{r^2+x^2}-r\rightarrow0$ as $r\rightarrow\infty$, and it's dominated by $x$. So if you can prove $\int^\infty_0|x\,f(x)|\,dx<\infty$, you're done (and that should be easy).


TITLE: Binomial Theorem for Fractional Powers
QUESTION [5 upvotes]: We know that the binomial theorem and expansion extends to powers which are non-integers.
For integer powers the expansion can be proven easily as the expansion is finite. However what is the proof that the expansion also holds for fractional powers?
A simple an intuitive approach would be appreciated.

REPLY [3 votes]: You could calculate, for example, $(1+x)^{1/2}=a_0+a_1x+a_2x^2+\cdots$ by squaring both sides and comparing coefficients. For example we can get the first three coefficients by ignoring all degree $3$ terms and higher:
$$1+x=a_0^2+2a_0a_1x+2a_0a_2x^2+a_1^2x^2+\cdots$$
From here we can conclude that $a_0=\pm1$ (we'll take $+1$ to match what happens when $x=0$). Then comparing coefficients of $x$ we have $2a_1=1$, so $a_1=1/2$. Finally, comparing coefficients of $x^2$, we have $2a_0a_2+a_1^2=0$, so $2a_2+1/4=0$ and $a_2=-1/8$.
You can definitely get as many coefficients as you want this way, and I trust that you can even derive the binomial coefficient formula. However, this is not any easier than the Taylor series, where you take $(1+x)^{1/2}=a_0+a_1x+a_2x^{2}+\cdots$ and find the coefficients by saying the $n$th derivatives on both sides have to be equal at $0$.
For example, plugging in $0$ on both sides we conclude $a_0=1$. Calculating the first derivative of both sides, we have
$$\frac{1}{2}(x+1)^{-1/2}=a_1+2a_2x+\cdots$$
Plugging in $0$, we get $a_1=1/2$. Taking the derivative one more time, we see
$$(-1/2)(1/2)(1+x)^{-3/2}=2a_2+\cdots$$
Plugging in $x=0$, we have $(-1/2)(1/2)=2a_2$, or $a_2=-1/8$. The advantage to this way, is that it is much easier to see the pattern of coefficients!
Unfortunately, there is a big hole in both arguments. They will give you what the coefficients have to be, but they won't prove that the series expansion converges in the first place. We started off by assuming you could write $1+x$ as an infinite power series, but there is no guarentee that this exists, and actually it doesn't converge unless $|x|<1$, which we never used. So you need to estimate the error in Taylor's formula to complete the proof rigorously.


TITLE: Electric Field of Bar
QUESTION [0 upvotes]: Problem: Positive charge $+Q$ is distributed uniformly along the positive $x$ axis from $x = 0$ to $x = a$. Negative charge $-Q$ is distributed along the $-x$ axis from $x = 0$ to $x = -a$. A point charge $+q$ is on $(0,y)$. Find the force that the positive and negative charge distributions exert on the point charge.
My (Wrong) Solution: We only consider the positive charges. Let $\sigma$ denote the charge density on the section. $\sigma = \dfrac{Q}{a}$. Then, the charge on a segment is $dQ = \sigma \hspace{1mm} dx = \dfrac{Q}{a} dx$. Plugging all this into the differential form of the formula for electric field, $$dE = \frac{k\dfrac{Q}{a}\,dx}{x^2+y^2}.$$ After drawing a diagram, we see that the $y-$component cancels out from both the positive and negative charges. Hence, we write $$dE_x = \frac{k\dfrac{Q}{a}\,dx}{x^2+y^2} \cdot \frac{a}{\sqrt{x^2 + y^2}} =  \frac{k{Q}\,dx}{(x^2+y^2)^\frac{3}{2}}.$$ Then, we integrate as such $$\int^{a}_{0}dE_x = \frac{kQa}{y^2\sqrt{a^2 + y^2}}.$$ As the x components are the same for both the negative and positive charges and they add up in the negative x direction, the total force is $$F = \frac{2kQq}{y^2\sqrt{a^2 + y^2}}$$ in the $-x$ direction.
However, the correct answer is $$\frac{2qQ}{a}\left(\frac{1}{y} - \frac{1}{\sqrt{a^2 + y^2}}\right).$$
What have I done wrong in the calculations I've done in my attempt to solve the problem?

REPLY [1 votes]: When you calculated $dE_x$, you multiplied by $\frac{a}{\sqrt{x^2+y^2}}$. Why? Hint: try $\frac{x}{\sqrt{x^2+y^2}}$