text:
\begin{document}
\title[] {Some identities on the $q$-Bernstein polynomials, $q$-Stirling numbers and $q$-Bernoulli numbers}
\author{Taekyun Kim}
\address{Taekyun Kim. Division of General Education-Mathematics \\
Kwangwoon University, Seoul 139-701, Republic of Korea  \\}
\email{tkkim@kw.ac.kr}
\author{Jongsung Choi}
\address{Jongsung Choi. Division of General Education-Mathematics \\
Kwangwoon University, Seoul 139-701, Republic of Korea  \\}
\email{jeschoi@kw.ac.kr}
\author{Young-Hee Kim}
\address{Young-Hee Kim. Division of General Education-Mathematics\\
Kwangwoon University, Seoul 139-701, Republic of Korea  \\}
\email{yhkim@kw.ac.kr}
\thanks{
{\it 2000 Mathematics Subject Classification}  : 11B68, 11B73,
41A30, 05A30, 65D15}
\thanks{\footnotesize{\it Key words and
phrases} :  Bernstein polynomial, Bernstein operator, Bernoulli
numbers and polynomials, Stirling number}

\maketitle

{\footnotesize {\bf Abstract} \hspace{1mm} {
In this paper, we consider the $q$-Bernstein polynomials on $\Bbb Z_p$ and
investigate some interesting properties of $q$-Bernstein polynomials related to
$q$-Stirling numbers and Carlitz's type $q$-Bernoulli numbers.
}

\section{Introduction}

Let $C[0, 1]$ denote the set of continuous functions on $[0, 1]$. Then Berstein operator for
$f \in C[0, 1]$ is defined as
$$ \mathbb{B}_n(f)(x)= \sum_{k=0}^{n} f(\frac{k}{n})\binom{n}{k} x^k (1-x)^{n-k}=\sum_{k=0}^{n} f(\frac{k}{n})B_{k,n}(x),$$
for $k, \, n \in \Bbb Z_+$, where $B_{k, n} (x)=\binom{n}{k}x^k (1-x)^{n-k}$ is called the Bernstein polynomial of degree $n$
(see [1, 9, 10]).

In [9], Phillips introduced the $q$-extension of Bernstein polynomials and
Kim-Jang-Yi proposed the modified $q$-Bernstein polynomials of degree $n$,
which are different $q$-Berstein polynomials of Phillips (see [1]).

Let $q$ be regarded as either a complex number $q\in\Bbb C$ or a $p$-adic number
$q\in\Bbb C_p$. If $q\in\Bbb C$, then we always assume that $|q|< 1$.
If $q\in\Bbb C_p$, we normally assume that $|1-q|_p < p^{- \frac{1}{p-1}}$,
which yields the relation $q^x = \exp(x \log q)$ for $|x|_p \le 1$ (see [1-11]).

Here, the symbol $| \cdot |_p$ stands for the $p$-adic absolute on $\Bbb C_p$
with $|p|_p = 1/p$. The $q$-bosonic natural numbers are defined by
$[n]_q=\frac{1-q^n}{1-q} =1+q+ \cdots + q^{n-1} \, (n \in \Bbb N)$, and the $q$-factorial is defined by
$[n]_q!=[n]_q[n-1]_q\cdots[2]_q[1]_q$.  In this paper we use the notation
for the Gaussian binomial coefficients in the form of
$${\binom{n}{k}}_q=\frac{[n]_q!}{[n-k]_q![k]_q!}=\frac{[n]_q[n-1]_q\cdots[n-k+1]_q}{[k]_q!}, \quad (\text{see [3, 8]}).$$
Note that $\underset{q \rightarrow 1}{\lim} {\binom{n}{k}}_q ={\binom{n}{k}} = \frac{n(n-1)\cdots(n-k+1)}{n!}.$

Let $p$ be a fixed prime number. Throughout this paper, the symbol
$\Bbb Z_p$, $\Bbb Q_p$, $\Bbb C$ and $\Bbb C_p$ denote the ring of $p$-adic integers, the field of
$p$-adic rational numbers, the complex number field, and the completion of algebraic closure
of $\Bbb Q_p$, respectively. Let $\mathbb{N}$ be the set of natural
numbers and $\Bbb Z_+ = \Bbb N \cup \{ 0 \}$.

Let $UD(\Bbb Z_p)$ be the space of uniformly differentiable function
on $\Bbb Z_p$. For $f \in UD(\Bbb Z_p)$, the $p$-adic $q$-integral on $\Bbb Z_p$ is defined by
$$I_q (f)=\int_{\Bbb Z_p
} f(x) d\mu_q (x) = \lim_{N \rightarrow \infty} \frac{1}{[p^N]_q} \sum_{x=0}^{p^N
-1} f(x)q^x, \quad (\text{see [4, 5, 6]}).
$$
Carlitz' $q$-Bernoulli number $\beta_{k,q}$ can be defined respectively by
$\beta_{0,q}=1$ and by the rule that $q(q \beta +1)^k -  \beta_{k,q}$ is equal to $1$ if $k=1$ and to $0$ if $k>1$
with the usual convention of replacing $\beta^i$ by $\beta_{i, q}$ (see [5]). As was shown in [5],
Carlitz's $q$-Bernoulli numbers can be represented by $p$-adic $q$-integral on $\Bbb Z_p$ as follows :

\begin{eqnarray}\int_{\Bbb Z_p
} [x]_q^n d\mu_q (x) = \lim_{N \rightarrow \infty} \frac{1}{[p^N]_q} \sum_{x=0}^{p^N
-1} [x]_q^n q^x = \beta_{n, q}, \quad n \in \Bbb Z_+.
\end{eqnarray}

The $k$-th order factorial of the $q$-number $[x]_q$, which is defined by
$$[x]_{k,q} =[x]_q [x-1]_q \cdots [x-k+1]_q = \frac{(1-q^x)(1-q^{x-1})\cdots(1-q^{x-k+1})}{(1-q)^k}, $$
is called the $q$-factorial of $x$ of order $k$ (see [3]). Thus, we note that $\binom{x}{k}_q = \frac{[x]_{k,q}}{[k]_q !}.$

In this paper, we consider $q$-Bernstein polynomials on $\Bbb Z_p$ and we investigate some interesting properties of
$q$-Bernstein polynomials related $q$-Stirling numbers and Carlitz's $q$-Bernoulli numbers.

\section{$q$-Bernstein polynomials related to $q$-Stirling numbers and $q$-Bernoulli numbers}

In this section, we assume that $q \in \Bbb C_p$ with $|1-q|_p < p^{- \frac{1}{p-1}}$.
For $f \in UD(\Bbb Z_p)$, we consider $q$-Bernstein type operator on $\Bbb Z_p$ as  follows :
\begin{eqnarray} \mathbb{B}_{n,q}(f)(x)&=& \sum_{k=0}^{n} f(\frac{k}{n})\binom{n}{k} [x]_q^k [1-x]_q^{n-k}\\
&=&\sum_{k=0}^{n} f(\frac{k}{n})B_{k,n}(x, q),\notag \end{eqnarray}
for $k, \, n \in \Bbb Z_+$, where $B_{k, n} (x, q)=\binom{n}{k}[x]_q^k [1-x]_q^{n-k}$ is called q-Bernstein type polynomials of degree $n$
(see [1]).

Let $(Eh)(x)=h(x+1)$ be the shift operator. Consider the $q$-difference operator as follows :
\begin{eqnarray}
\Delta_q^n = \prod_{i=1}^n (E- q^{i-1}I),
\end{eqnarray}
where $(Ih)(x)=h(x)$. From (3), we note that
\begin{eqnarray}
f(x)=\sum_{n \ge 0} \binom{x}{n}_q \Delta_q^n f(0), \quad (\text{see [3]}),
\end{eqnarray}
where
\begin{eqnarray}
\Delta_q^n f(0)=\sum_{k=0}^{n}  \binom{n}{k}_q (-1)^k q^{\binom{k}{2}} f(n-k).
\end{eqnarray}
The $q$-Stiring number of the first kind is defined by
\begin{eqnarray}
\prod_{k=1}^n (1+[k]_q z) = \sum_{k=0}^{n} S_1 (n, k : q)z^k,
\end{eqnarray}
and the $q$-Stiring number of the second kind is also defined by
\begin{eqnarray}
\prod_{k=1}^n (\frac{1}{1+[k]_q z}) = \sum_{k=0}^{n} S_2 (n, k : q)z^k, \quad (\text{see [3]}).
\end{eqnarray}
By (3), (4), (5), (6) and (7), we see that
\begin{eqnarray}
S_2 (n, k : q)z^k = \frac{q^{- \binom{k}{2}}}{[k]_q !} \sum_{j=0}^{k}(-1)^j q^{\binom{j}{2}}\binom{k}{j}_q [k-j]_q^n
= \frac{q^{- \binom{k}{2}}}{[k]_q !} \Delta_q^k 0^n, \quad (\text{see [1]}).
\end{eqnarray}

For $q \in \Bbb C_p$ with $|1-q|_p < p^{- \frac{1}{p-1}}$, we have that for $k, n \in \Bbb{Z_+}$,
\begin{eqnarray*}
F_q^{(k)}(t,x)&=& \frac{t^k e^{[1-x]_q t} [x]_q^k}{k!}=[x]_q^k \sum_{n=0}^{\infty}
\binom{n+k}{k} [1-x]_q^n \frac{t^{n+k}}{(n+k)!} \\
&=&\sum_{n=k}^{\infty}
\binom{n}{k}[x]_q^k [1-x]_q^{n-k} \frac{t^n}{n!}\\
&=&\sum_{n=0}^{\infty} B_{k,n}(x,q) \frac{t^n}{n!}, \quad \text{(see [1])}.
\end{eqnarray*}
 Thus, we note that
$\frac{t^k e^{[1-x]_q t} [x]_q^k}{k!}$ is the generating function of $q$-Berstein polynomials (see [1]).
It is easy to show that
\begin{eqnarray}
[1-x]_q^{n-k} = \sum_{m=0}^{\infty} \sum_{l=0}^{n-k} \binom{l+m-1}{m} \binom{n-k}{l} (-1)^{l+m} q^l [x]_q^{l+m}(q-1)^m.
\end{eqnarray}
By (1), (2) and (9), we obtain the following theorem.

\begin{theorem}
For $k, n \in \Bbb{Z_+}$ with $n \ge k$, we have
\begin{eqnarray*}
& & \int_{\Bbb Z_p} B_{k, n}(x, q)d \mu_q (x) \\ & & \qquad=\sum_{m=0}^{\infty} \sum_{l=0}^{n-k}\binom{l+m-1}{m} \binom{n-k}{l}
(-1)^{l+m} q^l (q-1)^m \beta_{l+m+k, q},
\end{eqnarray*}
where $\beta_{n,q}$ are the $n$-th Carlitz $q$-Bernoulli numbers.
\end{theorem}

\medskip

In [3], it is known that
\begin{eqnarray}
[x]_q^n = \sum_{k=0}^{n} q^{\binom{k}{2}} \binom{x}{k}_q [k]_q !S_2 (n, k : q),
\end{eqnarray}
and
\begin{eqnarray}
\frac{\sum_{k=i-1}^{n} \frac{\binom{k}{i}}{\binom{n}{i}}B_{k,n} (x,q)}{([x]_q +[1-x]_q)^{n-i}}
=[x]_q^i, \quad \text{for} \,\, i \in \Bbb N, \quad (\text{see [1]}).
\end{eqnarray}
By (2) and (11), we see that for $i \in \Bbb N$,
\begin{eqnarray}
&&[x]_q^i  = \sum_{m=0}^{\infty} \sum_{k=i-1}^n \sum_{l=0}^{m+n-k} \sum_{p=0}^{\infty}
\frac{\binom{k}{i}\binom{n}{k}}{\binom{n}{i}}\binom{l+p-1}{p}  \\
& & \quad \times
\binom{m+n-k}{l} \binom{n-i+m-1}{m}(-1)^{l+p+m}q^l(q-1)^p [x]_q^{n-i-m+k+p+l}. \notag
\end{eqnarray}
By (12), we obtain the following theorem.

\begin{theorem}
For $k, n \in \Bbb{Z_+}$ and $i \in \Bbb N$, we have
\begin{eqnarray*}
&&\beta_{i, q}  = \sum_{m=0}^{\infty} \sum_{k=i-1}^n \sum_{l=0}^{m+n-k} \sum_{p=0}^{\infty}
\frac{\binom{k}{i}\binom{n}{k}}{\binom{n}{i}}\binom{l+p-1}{p}  \\
& & \quad \times
\binom{m+n-k}{l} \binom{n-i+m-1}{m}(-1)^{l+p+m}q^l(q-1)^p \beta_{n-i-m+k+p+l, q}. \notag
\end{eqnarray*}
\end{theorem}

\medskip

From (10) and (11),  we note that
\begin{eqnarray}
\frac{\sum_{k=i-1}^{n} \frac{\binom{k}{i}}{\binom{n}{i}}B_{k,n} (x,q)}{([x]_q +[1-x]_q)^{n-i}}
=\sum_{k=0}^{i} q^{\binom{k}{2}} \binom{x}{k}_q [k]_q !S_2 (i, k : q).
\end{eqnarray}
In (3), it is known that
\begin{eqnarray}
\int_{\Bbb Z_p} \binom{x}{n}_q d \mu_q (x) = \frac{(-1)^n}{[n+1]_q}q^{(n+1)-\binom{n+1}{2}}.
\end{eqnarray}
By (13), (14) and Theorem 2, we have
\begin{eqnarray*}
\beta_{n,q}=q \sum_{k=0}^{m} \frac{[k]_q !}{[k+1]_q} (-1)^k S_2 (k, n-k,:q).
\end{eqnarray*}
For $S_2 (n, k :q)$, we see that
\begin{eqnarray}
S_2 (n, k:q)= \frac{1}{(1-q)^k}\sum_{j=0}^{k}(-1)^{k-j}\binom{k+n}{k-j} \binom{j+n}{j}_q, \quad \text{(see [3])},
\end{eqnarray}
and
\begin{eqnarray*}
\binom{n}{k}_q= \sum_{j=0}^{n}\binom{n}{j}(q-1)^{j-k}S_2 (k, j-k :q).
\end{eqnarray*}
By simple calculation, we show that
\begin{eqnarray}
q^{nx}&=& \sum_{k=0}^{n}(q-1)^k q^{\binom{k}{2}} \binom{n}{k}_q [x]_{k, q} \\
&=&\sum_{m=0}^{n}\{\sum_{k=m}^{n}(q-1)^k  \binom{n}{k}_q S_1 (k, m :q)\}[x]_q^m, \notag
\end{eqnarray}
and
\begin{eqnarray}
\int_{\Bbb Z_p} q^{nx} d \mu_q (x) = \sum_{m=0}^{\infty} \binom{n}{m}(q-1)^m
 \beta_{m,q}.\end{eqnarray}
By (16) and (17), we see that
\begin{eqnarray*}
\binom{n}{m}_q= \sum_{k=m}^{n}(q-1)^{-m+k}\binom{n}{k}_q S_1 (k, m :q).
\end{eqnarray*}
Therefore, we obtain the following theorem.

\begin{theorem}
For $k, n \in \Bbb{Z_+}$, we have
\begin{eqnarray*}
B_{k, n}(x,q)=\sum_{m=k}^{n}(q-1)^{-k+m}\binom{n}{m}_q S_1 (m, k :q)[x]_q^k [1-x]_q^{n-k}.
\end{eqnarray*}
\end{theorem}

\medskip

From the definition of the $q$-Stirling numbers of the first kind, we derive
\begin{eqnarray}
q^{\binom{n}{2}}\binom{x}{n}_q [n]_q ! = [x]_{n,q}q^{\binom{n}{2}}=\sum_{k=0}^n S_1 (n, k:q)[x]_q^k.
\end{eqnarray}
By (13) and (18), we obtain the following theorem.

\begin{theorem}
For $k, n \in \Bbb{Z_+}$ and $i \in \Bbb N$, we have
\begin{eqnarray*}
\frac{\sum_{k=i-1}^{n} \frac{\binom{k}{i}}{\binom{n}{i}}B_{k,n} (x,q)}{([1-x]_q +[x]_q)^{n-i}}
=\sum_{k=0}^{i} (\sum_{l=0}^{k}S_1 (n, l : q)  S_2 (i, k : q) [x]_q^l).
\end{eqnarray*}
\end{theorem}

\medskip

By Theorem 2 and Theorem 4, we obtain the following result.
\begin{corollary}
For $i \in \Bbb N$, we have
\begin{eqnarray*}
\beta_{i, q}= \sum_{k=0}^{i}(\sum_{l=0}^{k}S_1(n, l :q )S_2(i, k :q) \beta_{l, q}),
\end{eqnarray*}
where $\beta_{i,q}$ are the $i$-th Carlitz $q$-Bernoulli numbers.
\end{corollary}

\medskip

In [3], the $q$-Bernoulli polynomials of order $k \, (\in \Bbb Z_+)$ are defined by
\begin{eqnarray}
\beta_{n, q}^{(k)}(x)= \frac{1}{(1-q)^n}\sum_{i=0}^{n}(-1)^{i}\binom{n}{i}q^{ix} \underbrace{\int_{\Bbb Z_p}
\cdots \int_{\Bbb Z_p}}_{k-\text{times}}q^{\sum_{l=1}^k (k-l+i)x_l} d \mu_q (x_1) \cdots
d \mu_q (x_k).
\end{eqnarray}
From (19), we note that
\begin{eqnarray*}
\beta_{n, q}^{(k)}= \frac{1}{(1-q)^n}\sum_{i=0}^{n}(-1)^{i}\binom{n}{i}
\frac{(i+k)\cdots(i+1)}{[i+k]_q \cdots [i+1]_q} q^{ix}, \quad (\text{see [3]}).
\end{eqnarray*}
The inverse $q$-Bernoulli polynomial of order $k$ are also defined by
\begin{eqnarray}
\beta_{n, q}^{(-k)}(x)= \frac{1}{(1-q)^n}\sum_{i=0}^{n} \frac{(-1)^{n}\binom{n}{i} q^{xi} }
{\int_{\Bbb Z_p}
\cdots \int_{\Bbb Z_p} q^{\sum_{l=1}^k (k-l+i)x_l} d \mu_q (x_1) \cdots
d \mu_q (x_k)},
\end{eqnarray}
(see [3]). In the special case $x=0$, $\beta_{n, q}^{(k)}=\beta_{n, q}^{(k)}(0)$ are called
the $n$-th $q$-Bernoulli numbers of order $k$ and $\beta_{n, q}^{(-k)}=\beta_{n, q}^{(-k)}(0)$ are called
the $n$-th inverse $q$-Bernoulli numbers of order $k$.

By (20), we see that
\begin{eqnarray}
\beta_{k, q}^{(-n)}&=& \frac{1}{(1-q)^k}\sum_{j=0}^{k}(-1)^{j}\binom{k}{j}
\frac{[j+n]_q \cdots [j+1]_q}{(j+n) \cdots (j+1)} \notag \\
&=& \frac{1}{(1-q)^k}\sum_{j=0}^{k}(-1)^{j} \frac{\binom{k}{j}}{(j+n) \cdots (j+1)}
[n]_q! \binom{j+n}{n}_q \\
&=& \frac{1}{(1-q)^k}\sum_{j=0}^{k}(-1)^{j} \frac{\binom{k+n}{n-j}}{\binom{k+n}{n}}\binom{j+n}{n}_q \frac{[n]_q!}{n!} \notag\\
&=& \frac{[n]_q!}{\binom{k+n}{n} n!} \{ \frac{1}{(1-q)^k} \sum_{j=0}^{k}(-1)^{j}\binom{k+n}{k-j}\binom{j+n}{n}_q \notag \}.
\end{eqnarray}
From (15) and (21), we note that
\begin{eqnarray}
S_2 (n, k :q)=\binom{k+n}{n} \frac{[n]_q !}{n!} \beta_{k,q}^{(-n)}.
\end{eqnarray}
By (13) and (22), we obtain the following theorem.

\begin{theorem}
For $k, n \in \Bbb{Z_+}$ and $i \in \Bbb N$, we have
\begin{eqnarray*}
\frac{\sum_{k=i-1}^{n} \frac{\binom{k}{i}}{\binom{n}{i}}B_{k,n} (x,q)}{([1-x]_q +[x]_q)^{n-i}}
=\sum_{k=0}^{i}q^{\binom{k}{2}} \binom{x}{k}_q [k]_q ! \binom{k+i}{i} \frac{[i]_q !}{i!}
\beta_{k,q}^{(-i)}.
\end{eqnarray*}
\end{theorem}

\medskip

It is not difficult to show that
\begin{eqnarray}
q^{\binom{n}{2}} \binom{x}{n}_q &=& \frac{1}{[n]_q!} \prod_{k=0}^{n-1} ([x]_q - [k]_q) \\
&=& \frac{1}{[n]_q!}  \sum_{k=0}^n (-1)^k [x]_q^{n-k} S_1 (n-1, k :q).
\end{eqnarray}
By Theorem 4 and (23), we obtain the following result.

\begin{corollary}
For $k, n \in \Bbb{Z_+}$ and $i \in \Bbb N$, we have
\begin{eqnarray*}
\frac{\sum_{k=i-1}^{n} \frac{\binom{k}{i}}{\binom{n}{i}}B_{k,n} (x,q)}{([1-x]_q +[x]_q)^{n-i}}
=\sum_{k=0}^{i}\sum_{j=0}^{k}(-1)^j [x]_q^{n-j}S_1 (k-1, j : q)
\binom{k+i}{i} \frac{[i]_q !}{i!}\beta_{k,q}^{(-i)}.
\end{eqnarray*}
\end{corollary}

\bigskip

config: arxiv


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Edit:
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TITLE: Eigenvalue problem with Robin Boundary Conditions
QUESTION [1 upvotes]: Solve the eigenvalue problem with Robin boundary conditions: $$\phi''(x)+\lambda\phi(x)=0,$$ $$\phi(0)+\phi'(0)=0,$$ $$\phi(\pi)+\phi'(\pi)=0$$
I worked out that both $\lambda=0$ and $\lambda<0$ aren't eigenvalues, and for $\lambda>0$ with $\lambda=+p^2$ I found that $c_1+c_2=0$ and $c_1\cos(\pi p)+c_2p\cos(\pi p)=0$. Something cool was supposed to happen and I'm not sure if I'm doing it right. $$ $$
So continuing with this I got that $p=1$ and therefore $\lambda=1$.

REPLY [1 votes]: Every non-zero solution of
$$
              \phi''+\lambda\phi =0, \\ \phi(0)+\phi'(0)=0
$$
can be normalized so that $\phi(0)-\phi'(0)=1$ by scaling the solution. If you could not do that, then the solution $\phi$ would satisfy $\phi(0)=\phi'(0)=0$, which is satisfied only by the $0$ function. So, start by solving $\phi''+\lambda \phi=0$ subject to
$$
                \phi(0)+\phi'(0)=0,\;\;\; \phi(0)-\phi'(0)=1, \\
               \implies \phi(0)=1/2,\;\; \phi'(0)=-1/2.
$$
This has the unique solution
$$
                \phi_{\lambda}(x) = \frac{1}{2}\cos(\sqrt{\lambda}x)-\frac{1}{2\sqrt{\lambda}}\sin(\sqrt{\lambda}x).
$$
In order for $\lambda$ to be an eigenvalue it will necessary for $\phi_{\lambda}(\pi)+\phi_{\lambda}'(\pi)=0$, which gives the eigenvalue equation
$$
          \frac{1}{2}\cos(\sqrt{\lambda}\pi)-\frac{1}{2\sqrt{\lambda}}\sin(\sqrt{\lambda}\pi)-\frac{\sqrt{\lambda}}{2}\sin(\sqrt{\lambda}\pi)-\frac{1}{2}\cos(\sqrt{\lambda}\pi) = 0 \\
        \left(1+\lambda\right) \frac{\sin(\sqrt{\lambda}\pi)}{\sqrt{\lambda}}=0.
$$
The eigenvalues are
$$
             \lambda=-1 \mbox{ and } \lambda = 1^2,2^2,3^2,\cdots .
$$
The corresponding eigenfunctions are
$$
           \phi_{-1}=\frac{1}{2}\cosh(x)-\frac{1}{2}\sinh(x)=e^{-x},\\
           \phi_{n^2}=\frac{1}{2}\cos(nx)-\frac{1}{2n}\sin(nx),\;\; n=1,2,3,\cdots.
$$


TITLE: Tangential behavior of Riemannian exponential
QUESTION [2 upvotes]: I wonder if the following holds in an arbitrary Riemannian manifold $M$:
assume $x\in M$, $h\in T_x M$, do we have for $u\in T_x M$ exponentiable (if necessary of small enough norm) that:
$$\lim_{t\to 0} \frac{d(\exp_{\exp_x u}t\tau(h), \exp_x (u+th))}{t}=0$$
where $\tau(h)$ is the parallel transport of $h$ along $[0,1]\rightarrow M, t\mapsto \exp_x tu$.
If it doesn't hold in general, then which geometrical conditions are sufficient for it to hold, and do we have at least in general that for any $x\in M$, $h\in T_x M$:
$$\lim_{t\to 0, u\to 0} \frac{d(\exp_{\exp_x u}t\tau(h), \exp_x (u+th))}{t}=0$$

REPLY [4 votes]: The second identity is always true because both arguments of $d$ are smooth functions of $u$ and $t$ and they coincide when $u=0$.
The first one holds true for all $u$ and $h$ only if the metric is flat. Indeed, the l.h.s. is the length of the difference of the initial velocity vectors of two curves $t\mapsto \exp_{\exp_x u} t\tau(h)$ and $t\mapsto\exp_x(u+th)$. The first velocity vector is $\tau(h)$, so its length equals $|h|$. The second one is the derivative of $\exp_x$ at $u$ along $h$. If its length equals $|h|$ for all $u$ and $h$, then $\exp_p$ is a Riemannian isometry (by definition), so the metric is flat.


TITLE: When to extrapolate?
QUESTION [1 upvotes]: I know extrapolating is probably dependent on the question being asked. But are there some guidelines on when extrapolation is "okay"? If you can't be as certain that a set of data will maintain a pattern, how can you say extrapolation is reliable? If it's not reliable how can you say it's useful? Do businesses make decisions off extrapolation?
Sorry that was a lot of questions.

REPLY [0 votes]: In statistics, the word extrapolation is perhaps most often used in the
context of regression. I will confine my Answer to extrapolation in simple regression. (Similar comments hold for multiple regression.)
Simple linear regression. Suppose you have $n$ pairs $(x_i, Y_i)$ modeled by the equation
$$Y_i = \beta_0 + \beta_1 x_i + e_i,$$
where 'errors' $e_i \stackrel{iid}{\sim} \mathsf{Norm}(0, \sigma).$ Also suppose that
data $(x_i, Y_i)$ have been used to make a least-squares line, estimating
coefficients of the linear model $\beta_0$ by $\hat \beta_0$ and $\beta_1$ by
$\hat \beta_1$ and the error
variance $\sigma^2$ by $\hat \sigma^2.$
Prediction. Then, given a value $x_p,$ a corresponding value $\hat Y_p$ can be 'predicted'
as $\hat Y_p = \hat \beta_0 + \hat \beta_1 x_p,$ and a 95% prediction interval (PI)
can be found using also $\hat \sigma^2$ as in your textbook or on slide 11 of these notes, using slightly different notation.
If the value $x_p$ lies within the span of the $n$ values $x_i$ of the data,
then the predicted value $\hat Y_p$ is called an interpolation. The prediction interval is shortest when $x_p \approx \bar x$ and becomes wider as $x_p$
becomes farther from $\bar x.$ [Notice the factor $(x_p - \bar x)^2$ in the
third term under the square root in the formula for the prediction interval.]
However, if the value $x_p$ lies beyond the span of the $n$ values $x_i$ of the data (far to the right or far to the left),
then the predicted value $\hat Y_p$ is called an extrapolation. Extrapolation
is always somewhat risky. Of course, the formula for the PI gives much longer interval (indicating risk that is taken into account by the regression procedure.)
Added risks of extrapolation. But there is additional risk in extrapolation because the formulas for $\hat \beta_0, \hat \beta_1, \hat \sigma^2,$ and for the PI assume
a linear model within the span of the data's $x_i.$ The true model may
be a gently bent non-linear function that curves away from the estimated
model as $x$-values move beyond the span of the data. This risk is not
accounted for in any of the formulas.
A small 'chunk' of a gently bent curve may be 'almost-linear' so that
regression diagnostics (such as plots of residuals) show no difficulty.
But as we move away from that 'chunk' we may need a different linear model
or a model that is distinctly non-linear. As you suggested in your
Question, the 'pattern' may change if you extrapolate.
Extrapolation to the future. By their very nature, commercial, climatic, and economic forecasts of future events
always involve extrapolation. Data for $x$'s in 2000-2017 may fit a linear model
that no longer applies in $x_p = 2018.$ Wars, environmental events, elections,
and so on may change the business or economic situation in unexpected ways.
It is futile to criticize such extrapolation to the near future because
(short of crystal balls and tarot cards) there is no alternative. But
extrapolations must always be viewed with skepticism.
Famous quote (due to ancient Norse sages, Niels Bohr, Albert Einstein, or Yogi Berra, depending
on your reading of the Internet): "Prediction is difficult, especially about
the future."


TITLE: When will velocity and acceleration vectors be perpendicular?
QUESTION [3 upvotes]: Suppose a particle is moving in the $xy$ plane with
  $$x=at, \quad y=at(1-bt),$$
  where $a$ and $b$ are positive constants. When will the velocity vector and acceleration vector be perpendicular?

I know that these vectors are perpendicular in circular motion. Should I then use the circle equation
$$ x^2+y^2 = r^2 $$
then substitute $r$ with centripetal acceleration
$$ a_c = v^2/r $$
and then substitute for $v$? But I don't see how I will get an equation that has something to do with vectors.

REPLY [3 votes]: Another approach uses $v\cdot a=\frac{d}{dt}(v^2/2)$, so orthogonality is equivalent to $v^2=\dot{x}^2+\dot{y}^2=a^2+(a-2abt)^2$ being constant. This is equivalent to $a=0$ or $b=0$.


TITLE: Why is the use of an implication arrow correct in the definition of injective functions: $ \forall a\forall b(f(a) = f(b) \to a = b) $?
QUESTION [1 upvotes]: According to an excerpt, from a textbook, about injective functions:

We can express that $f$ is one-to-one using quantifiers as
  $$ \forall a\forall b(f(a) = f(b) \to a = b) $$
  where the universe of discourse is the domain

I am having trouble understanding why the use of an implication arrow is correct for the definition of injective functions?
My counter example to the definition of injective functions:

The proposition $ f(a) \neq f(b) \to a = b $ would hold true to the definition of injective functions above, as $ \mathbf F \to \mathbf T $ holds true for conditional propositions.
Because $a = b$ this means that the single input $a$ gives out 2 different output values, $f(a)$ and $f(b)$
And this goes against the notion of one-to-one, as it becomes one-to-two. And it even goes against the definition of a function in general.

Is my line of reasoning correct? If not, where did I go wrong? Please, explain in detail my fault in the counter example.
And isn't the use of a biconditional arrow the true correct way of defining an injective function?
$$ \forall a\forall b(f(a) = f(b) \iff a = b) $$

REPLY [0 votes]: "The proposition f(a)≠f(b)→a=b would hold true to the definition above, as F→T
holds true for conditional propositions."
$f(a) \ne f(b)$ isn't a false statement.
The definition $f(a) \ne f(b) \implies a = b$ would describe constant functions and only constant functions.
This could be a working definition of "constant function" although it would be convoluted, confusing, and, to anyone who doesn't live, eat and breath predicate logic, paradoxical.  ("Y" is lot more straightforward than "not Y $\implies$ something imcompatible" although, logically, they are equivalent.)
"Because a=b
this means that the single input a gives out 2 different output values, f(a) and f(b)"
$f(a)$ and $f(b)$ are not two different output values.  THey are one output value written two different ways.
∀a∀b(f(a)=f(b)⟺a=b)
This is a fine way to define injectivity but as $a=b \implies f(a) = f(b)$ is always trivially true the bidirection is not nescessary.


TITLE: Sample the solution space of an inequality system $A x \leq b$
QUESTION [0 upvotes]: Given the following inequality system:
$A x \leq b, A \in \Re^{m \times n}$
I would like to sample the solution space. My first approach to solve this problem was by brute force, e.g. randomly generate $k$ solutions and test whether they satisfy the system, but as the system gets larger this process takes time. Do you know of any existing algorithm that can sample the solution space efficiently?

REPLY [0 votes]: I'm assuming that your polytope is bounded and that you want to sample uniformly from the polytope.
There are Markov Chain Monte Carlo sampling methods for this problem.  In particular, an algorithm called "Hit-and-Run" is widely used. See this survey paper by Santosh Vempala:
https://www.cc.gatech.edu/~vempala/papers/survey.pdf


TITLE: What is Venus's core made of?
QUESTION [2 upvotes]: As we all know Venus's surface is so hot that it can probably melt lead.
What would be in it's in core?
Is it in the liquid or solid state?
What would be it's temperature?
How many cores does it have?

REPLY [0 votes]: Venus core is probably made of iron and nickel. With a sprinkle of other elements. Also some radioactive elements might provide some heat in Venus' core. Its inner core might be solid but its outer core might be liquid. However the outer core will not be as fluid as here on Earth because of Venus's slow rotation. This is why Venus does not have a magnetic field. Venus's slow rotation is not able to make Venus's core create a magnetic field.