TITLE: Why are the integral bounds what they are in this probability problem?
QUESTION [1 upvotes]: Harry’s magic wand breaks at a random point (location of the point is
uniform along the length of the stick, which is $40$ cm long). Suppose
the piece of the stick that Harry is left with is $X$ cm long.
Unfortunately, the next day part of the stick is accidentally burnt
while casting a spell. After this accident, the length of the stick is
reduced to $D$ cm, where $D$ is uniformly distributed between $[0,X]$.
Find $f_D(d)$.

The solution is shown here...

What I don't understand is why the bounds of integration in the final step go from $d$ to $40$, the combined regions of the wand destroyed between the two accidents. Why shouldn't the integration go, for example, from $0$ to $d$ (the portion of the wand remaining), or from $0$ to $40$ (the entire initial wand)?

REPLY [1 votes]: What I don't understand is why the bounds of integration in the final step go from d to 40

This because the joint domain $(X,D)$ is a triangle:
$$0<\underbrace{d<x<40}_{\text{X-support}}$$
In fact D is the length after the second accident and thus it cannot be longer than X.
to find $f_D(d)$ you have to integrate the joint density over all X domain, say
$x \in (d;40)$


TITLE: Do we need completeness for a weak*-convergent sequence to be bounded?
QUESTION [8 upvotes]: Let $(\phi_n)_n$ be a weak* convergent sequence in the dual of some normed space $X$ with (weak*-)limit $\phi$.
If $X$ is Banach then it follows from the uniform boundedness principle that $\sup_n \lVert \phi_n \rVert < \infty$ since $\sup_n  |\phi_n (x)| < \infty$ for every $x \in X$.
But what if $X$ is not Banach? Does this still hold true?

REPLY [12 votes]: Generally, without completeness, you can't deduce that a weak$^\ast$ convergent sequence is bounded.
Let $X = c_{00}$ be the space of sequences with only finitely many nonzero terms, endowed with the $\lVert\,\cdot\,\rVert_\infty$-norm. Its completion is $c_0$, the space of sequences converging to $0$, and its dual therefore isometrically isomorphic to $\ell^1(\mathbb{N})$. The sequence given by $x_n = n\cdot e_n$ is convergent to $0$ in $\bigl(\ell^1(\mathbb{N}),\sigma(\ell^1(\mathbb{N}),c_{00})\bigr)$ - since $\langle x_n, y\rangle = n\cdot y_n = 0$ for all sufficiently large $n$ - and unbounded.


TITLE: Regarding the role of Kolmogorov's theorem in stochastic process
QUESTION [0 upvotes]: When studying the time series using book of Time Series: Theory and Methods, I read the following theorem about distribution function.

I am not very clear how to understand this theorem, or what does this theorem tell us?

REPLY [1 votes]: Maybe one can work backwards and assume there exists some process $X$ with distribution functions $(F_t)$, that is, such that for every $t$ in $T$ and every $x=(x_s)_{s\in t}$,
$$
F_t(x)=P(\forall s\in t,X_s\leqslant x_s).
$$
Apply this to $t\cup\{r\}$ and to the vector made of $x=(x_s)_{s\in t}$ and $x_r=z$. In the limit $z\to+\infty$, the RHS involves the event
$$
\bigcup_{z}\{X_r\leqslant z,\forall s\in t,X_s\leqslant x_s\}=\{\forall s\in t,X_s\leqslant x_s\},
$$
thus one must have
$$
\lim_{z\to+\infty}F_{t\cup\{r\}}(x,z)=F_t(x).
$$
Kolmogorov theorem states that this (obviously necessary) condition on the family of distributions $(F_t)$ is also sufficient for such a stochastic process $X$ to exist.


TITLE: Use mathematical induction to prove Σ n,k=1 (1/k(k+1)) = (n/n+1) for all n in Natural numbers?
QUESTION [0 upvotes]: This is how far I can get:
p(n): nΣk=1 (1/k(k+1)) = (n/n+1)
p(1): 1Σk=1 (1/(1+1)) = (1/1+1)
            => 1/2 = 1/2
p(1) is true.
Assume that p(k) is true.
p(k) = kΣk=1, (1/k(k+1)) = k/k+1
Show that p(k+1) is true.
p(k+1): k+1 Σ k=1, (1/k+1((k+1)+1)) = (k+1/(k+1)+1)
                    => 1/(k+1)(k+2) = (k+1)/(k+2)
If this is correct, I am not sure how to finish from here.
How can I simplify p(k+1) using the induction hypothesis p(k) to show that p(k+1) is also true.
Thanks!

REPLY [1 votes]: Inductive step
$$\sum_{k=1}^{n+1}\frac1{k(k+1)}=\sum_{k=1}^{n}\frac1{k(k+1)}+\frac1{(n+1)(n+2)}=\frac n{n+1}+\frac1{(n+1)(n+2)}\\=\frac{n^2+2n+1}{(n+1)(n+2)}=\frac{n+1}{n+2}$$
Alternative proof
We can find this result by telescoping
$$\sum_{k=1}^{n+1}\frac1{k(k+1)}=\sum_{k=1}^{n+1}\frac1{k}-\frac1{k+1}=1-\frac1{n+2}=\frac{n+1}{n+2}$$

REPLY [0 votes]: Assume that it holds for $n=k$, i.e.
$$\sum_{i=1}^{k}\frac{1}{i(i+1)}=\frac{k}{k+1}.$$
Then, we have
$$\begin{align}\sum_{i=1}^{k+1}\frac{1}{i(i+1)}&=\left(\sum_{i=1}^{k}\frac{1}{i(i+1)}\right)+\frac{1}{(k+1)((k+1)+1)}\\&=\frac{k}{k+1}+\frac{1}{(k+1)(k+2)}\\&=\frac{k(k+2)+1}{(k+1)(k+2)}\\&=\frac{(k+1)^2}{(k+1)(k+2)}\\&=\frac{k+1}{(k+1)+1}.\end{align}$$
Hence, it holds for $n=k+1$.


TITLE: discrete change v. differential change in calculus
QUESTION [0 upvotes]: My understanding in univariate case is:
Suppose $f:\mathbb{R}\rightarrow\mathbb{R}$.
Then, $\frac{dy}{dx}|_{x=x_o}=f'(x)|_{x=x_o}$=$\lim_{\Delta x \to 0}\frac{\Delta f}{\Delta x}$, where $\Delta x=(x-x_o)$ and $\Delta f=f(x)-f(x_o)$.
Part of confusion is when I see something like:  $dy=f'(x)dx$.
What is happening here exactly? Why some books move around $dy$ and $dx$ as if they are discrete changes? Can somebody explain the difference between discrete changes and differential changes with clarity?
My understanding of $f'(x)$ at $x_o$ is the best affine approximation of $f$ at that point. Then, I use the term "sensitivity", because it is my understanding the derivative (or slope here) provides the information how sensitive function value at that point is to the very little change in $x$.
Then, what does $dy$ mean? Doesn't $dy$ also represent infinitesimal change? So I am confused difference between $dy$ and $\frac{dy}{dx}$ as well.
The terminologies that are used "instantaneous rate of change" or "slope" or "best affine approximation" or "how sensitive f is with respect to the argument at that point" amplifying the confusion.
Please help.

REPLY [0 votes]: The expression $\frac{dy}{dx}$ is on solid footing; it is interpretable as the slope of the function $y(x)$ or the instantaneous sensitivity of $y$ to a tiny change in $x$.
It is convenient to write $dy$ and $dx$ as if they were numeric quantities and manipulate them accordingly.  Thus for example, to solve the differential equation
$$
\frac{dy}{dx} = y^2 x
$$
you manipulate
$$
\frac{dy}{y^2} = x \, dx\\
\int\frac{dy}{y^2} =\int x \, dx\\
-\frac{1}{y} = \frac{x^2}{2}+c \\
y = -\frac{1}{\frac{x^2}{2}+c} = \frac{2}{x^2+k}
$$
And this works, but what you are really doing in the first step is a stylized shorthand for saying that taking the limits, in a suitable way, of  $x$ times the change in $x$ and of  $y^{-2}$ times the related change in $y$, those two limits are equal.
This can be made more rigourous in these simple manipulation cases by the field of hyper-reals, where infinitessimals are treated as part of the number system.  But hyperreal analysis has gone out of fashion because it does not resolve all issues of this type anyway.
In short, you need to look at equations that use $dy$ and $dx$ as if they were numbers, as shorthand for statements about limits.  This can give you peace of mind, while allowing you to take the easy shortcuts this approach enables.
A caveate:  Whatch out when partial derivatives appear!!


TITLE: Infinite Series $\sum_{n=1}^\infty\frac{H_n}{n^22^n}$
QUESTION [32 upvotes]: How can I prove that
$$\sum_{n=1}^{\infty}\frac{H_n}{n^2 2^n}=\zeta(3)-\frac{1}{2}\log(2)\zeta(2).$$
Can anyone help me please?

REPLY [4 votes]: \begin{align*}
\operatorname{Li}_3\left(\frac{1}{2}\right)&=\int _0^{\frac{1}{2}}\frac{\operatorname{Li}_2\left(x\right)}{x}\:dx=-\operatorname{Li}_2\left(\frac{1}{2}\right)\ln \left(2\right)+\int _{\frac{1}{2}}^{1}\frac{\ln \left(x\right)\ln \left(1-x\right)}{1-x}\:dx\\[2mm]
&=-\frac{1}{2}\ln \left(2\right)\zeta \left(2\right)+\frac{1}{2}\ln ^3\left(2\right)+\int _0^1\frac{\ln \left(1-x\right)\ln \left(x\right)}{x}\:dx-\int _0^{\frac{1}{2}}\frac{\ln \left(x\right)\ln \left(1-x\right)}{1-x}\:dx\\[2mm]
&=-\frac{1}{2}\ln \left(2\right)\zeta \left(2\right)+\frac{1}{2}\ln ^3\left(2\right)+\sum _{k=1}^{\infty }\frac{1}{k^3}-\frac{1}{2}\ln ^3\left(2\right)-\frac{1}{2}\int _0^{\frac{1}{2}}\frac{\ln ^2\left(1-x\right)}{x}\:dx\\[2mm]
&=-\frac{1}{2}\ln \left(2\right)\zeta \left(2\right)+\zeta \left(3\right)-\sum _{k=1}^{\infty }\frac{H_k}{k^2\:2^k}+\sum _{k=1}^{\infty }\frac{1}{k^3\:2^k}\\[2mm]
&=-\frac{1}{2}\ln \left(2\right)\zeta \left(2\right)+\zeta \left(3\right)-\sum _{k=1}^{\infty }\frac{H_k}{k^2\:2^k}+\operatorname{Li}_3\left(\frac{1}{2}\right)
\end{align*}
And magically we find the value for that sum
\begin{align*}
\sum _{k=1}^{\infty }\frac{H_k}{k^2\:2^k}=-\frac{1}{2}\ln \left(2\right)\zeta \left(2\right)+\zeta \left(3\right)
\end{align*}


TITLE: Sum of two exponential distributions with same parameter
QUESTION [0 upvotes]: Let $S, T$ be two independent random variables both with the exponential distribution and the same parameter $\lambda > 0$.
I would like to find the density function of $S+T$.
$S$ and $T$ both have the density function $f(t) = \lambda \cdot e^{-\lambda t}$ where $t>0$. How do I find the density function of $S+T$? I don't know how to begin, please help me.

REPLY [3 votes]: Suppose that $\left(X_{i}\right)_{1\leq i\leq n}$
  is independant and identically distributed according to an exponential law with a parameter $\lambda>0$
 . For all $x\in\mathbb{R}$
 , we have$$\mathbb{P}\left[X_{i}\leq x\right]=1-e^{-\lambda x}.$$
 Now, it is straightforward to use the characteristic functions : if $S_{n}=\sum_{i=1}^{n}X_{i}$
 , then we have for all $t\in\mathbb{R}$
  $$\chi_{_{S_{n}}}\left(t\right)=\prod_{i=1}^{n}\chi_{_{X_{i}}}\left(t\right)=\left(\chi_{_{X_{1}}}\left(t\right)\right)^{n}=\left(\left(1-i\lambda t\right)^{-1}\right)^{n}=\left(1-i\lambda t\right)^{-n}$$
 which is the characteristic function of a random variable that follow a gamma law of parameters $\left(n,\lambda\right)$ (I used independance and identical distribution here). Thus, the density of the law of $S_{n}$
  is the density of the gamma law (see for example http://en.wikipedia.org/wiki/Gamma_distribution).
You can also show by induction that the density of the sum of INDEPENDANT random variables is the convolution of the densities.

REPLY [2 votes]: Hint: The distribution of $Z=X+Y$ with $X$ and $Y$ independent is given by $$f_Z(z)=\int_{-\infty}^{\infty}f_X(x)f_Y(z-x)dx$$

REPLY [2 votes]: The density of the sum of two independent random variables is the convolution of their density functions.
$$f_{S+T}(z) = \int_{-\infty}^\infty f_S(x)\cdot f_T(z-x)\operatorname d x$$


TITLE: How can I compute the partial derivative under the line integral sign if the integrand is non-conservative?
QUESTION [1 upvotes]: Consider the following integral
$$I=\int_C^{\vec{y}_{0},\vec{y}} \vec{F}(\vec{x})\cdot d\vec{x}$$
where, for simplicity, $\vec{y}_{0}$ is some constant initial point on $C$ (which is an explicitly known curve) and $\vec{y}$ is some variable end point on $C$ and it is known that
$$\oint_{C_0} \vec{F}(\vec{x})\cdot d\vec{x} \neq 0 $$
for any closed curve $C_{0}$.
Does it still make sense to speak of the gradient
$$\left(\frac{\partial}{\partial y_{1}},\frac{\partial}{\partial y_{2}},\frac{\partial}{\partial y_{3}}\right)I$$
of the integral with respect to $\vec{y} = \left(y_{1},y_{2},y_{3}\right)$ ?
If so, then three questions follow.
1) If this gradient exists then doesn't that contradict the premise that $\vec{F}$ is non-conservative?
2)  Is there an equivalent of the Leibniz integral rule which specifies such a gradient?
3) How does it depend of the curve $C$?
Otherwise, how can we characterize the variational properties of this differentiation under the integral?

REPLY [1 votes]: $I$, as you have given it, is not a well-defined function of $y$. There may be two different curves ending at $y$, yielding two different line integrals. Which one should I choose to compute the value of $I$ at any given $y$?
Only when the vector field is conservative do we know that this function is well-defined. Only then may we take its gradient.


TITLE: How to see a function as a vector in a vector space
QUESTION [2 upvotes]: I know that strictly speaking my question is some sort of a duplicate of at least this previous one and I am quite sorry for that (usually I try to get the best from previous questions), but still I did not find an answer that suited my intuition, probably because not enough developed. Still, I would really like to grasp in the proper way this important point.
As a side problem, I think that a question that general can be problematic to answer, so I see it is not easy. Anyway, coming up with a specific example (or minimal working problem) could be a decent starting point. Thus, here we are.
Let's take as a building block a continuous function over $[0,1]$. When we are asked to think about this object as a vector in a vector space, is this object a vector with infinitely many elements?
Does it make sense what I wrote?
Thanks a lot to any feedback.

REPLY [3 votes]: There are two thought-patterns which I would like to encourage:

Relax your mind and stop thinking of vectors as lists of numbers! Vectors are just elements of a vector space, and that just means you can add and subtract elements since they form an additive abelian group, and you can also multiply these things with scalars. (Of course I am omitting a few connecting axioms in the interest of brevity.) Functions can be elements of sets that are vector spaces, so there is no reason to think of them any differently.
After digesting the last point, I'll tell you that you can still visualize function as vectors. It's just that this vector has a lot of entries, so many that you can't even write the entries sequentially. For each number $r\in [0,1]$, you can just think of the "$r$th coordinate" of the vector, and there is a real number there. The number at the $r$th coordinate is $f(r)$. In this way, this very long vector (with as many entries as $[0,1]$ has elements) captures the entire function. Finally, notice that vector addition and scaling  of these "function-vectors" matches the operation of pointwise addition and scaling of functions.


TITLE: Proving maclaurin polynom of a function and working with sums.
QUESTION [1 upvotes]: Let $f,g$ be a derivatible functions $n$ times.
And let:
$$
\sum_{k=0}^{n}a_kx^k, \sum_{k=0}^{n}b_kx^k
$$
Be maclaurin polynoms of $f,g$ at that order.
Prove that:
$$
(*)\sum_{k=0}^{n}(\sum_{j=0}^{k}a_jb_{k-j})x^k
$$
Is the maclaurin polynom of:
$$
f(x)g(x)
$$

What i tried:
In order to prove, i think need to show that:
$$
\lim_{x \to 0}\frac{f(x)g(x) - \sum_{k=0}^{n}(\sum_{j=0}^{k}a_jb_{k-j})x^k}{x^2} = 0
$$
Namely i show that the remainder is going to zero at $x \to 0$ (meclurian so $a = 0$)
I think its quite resonable that i will also need to use the facts (from the same logic) that:
$$
\lim_{x \to 0}\frac{f(x)- \sum_{k=0}^{n}a_kx^k}{x^2} = 0
$$
$$
\lim_{x \to 0}\frac{g(x)- \sum_{k=0}^{n}b_kx^k}{x^2} = 0
$$
Therefore i think my need is an algebric problem, working with the sums to get to identical terms.
Yet, im not good at it therefore i would like if someone can explain the conccept i need to think of in order to work and change those sums.
Thank you.

REPLY [0 votes]: Have a look at this, even though this is the infinite version the idea is in the visual provided.


text:
\begin{document}

\setcounter{page}{1}

\title[]{Some $q$-congruences for homogeneous and quasi-homogeneous multiple $q$-harmonic sums}

\author{Kh.~Hessami Pilehrood}
\address{The Fields Institute, 222 College Street, Toronto, Ontario M5T 3J1, Canada}
\email{hessamik@gmail.com}

\author{T.~Hessami Pilehrood}
\address{The Fields Institute, 222 College Street, Toronto, Ontario M5T 3J1, Canada}
\email{hessamit@gmail.com}

\author{R.~Tauraso}
\address{Dipartimento di Matematica,
Universit\`a di Roma ``Tor Vergata'',
via della Ricerca Scientifica,
00133 Roma, Italy}
\email{tauraso@mat.uniroma2.it}

\subjclass[2010]{11M32, 11B65, 05A30}
\keywords{Multiple $q$-harmonic sum, $q$-binomial identity, degenerate Bernoulli numbers, $q$-congruence,
duality relations}

\subjclass[2010]{11B65, 30E05, 11A07.}

\begin{abstract}
We show some new Wolstenholme type $q$-congruences for some classes of multiple $q$-harmonic sums of arbitrary depth with strings of indices composed of ones, twos and threes. Most of these results are $q$-extensions of the corresponding congruences for ordinary multiple harmonic sums obtained by the authors in a previous paper. Finally, we pose a conjecture concerning two kinds of cyclic sums of multiple $q$-harmonic sums.
\end{abstract}

\maketitle

\section{Introduction}

For two $l$-tuples of non-negative integers
${\bf s}:=(s_1, \ldots, s_l)$ and ${\bf t}:=(t_1,\ldots, t_l)$ and a non-negative integer $n,$ we define two classes of multiple $q$-harmonic sums
$$
H_n^q({\bf s};{\bf t})=H_n^q(s_1,\ldots,s_l;t_1,\ldots,t_l)=\sum_{1\le k_1<\cdots<k_l\le n}\frac{q^{k_1t_1+\cdots+k_lt_l}}{[k_1]_q^{s_1}\cdots[k_l]_q^{s_l}},
$$
$$
S_n^q({\bf s};{\bf t})=S_n^q(s_1,\ldots,s_l;t_1,\ldots,t_l)=\sum_{1\le k_1\le\cdots\le k_l\le n}\frac{q^{k_1t_1+\cdots+k_lt_l}}{[k_1]_q^{s_1}\cdots[k_l]_q^{s_l}},
$$
where
$$
\qq{n}=\frac{1-q^n}{1-q}=1+q+\dots+q^{n-1}
$$
is a $q$-analog of the non-negative integer $n$,
and put
$$
H_n^q({\bf s}):=H_n^q(s_1,\ldots,s_l;s_1-1,\ldots,s_l-1), \,\,\, S_n^q({\bf s}):=S_n^q(s_1,\ldots,s_l;s_1-1,\ldots,s_l-1).
$$
The number $l({\bf s}):=l$ is called the depth (or length) and $w({\bf s}):=\sum_{j=1}^ls_j$ is the weight of the multiple harmonic sum.
By convention, we put $H_n^q({\bf s};{\bf t})=S_n^q({\bf s};{\bf t})=0$ if $n<l$, and $H_n^q(\emptyset)=S_n^q(\emptyset)=1$.

Congruences for ordinary single and multiple harmonic sums have been studied since the  nineteenth century
(see \cite{G900, HPT14, Ho04, Le38, W862, Zh08} and references therein). It is well known that many multiple harmonic sums modulo a prime $p$ (or a power of $p$)
can be expressed in terms of Bernoulli numbers. The situation with $q$-analogs is much less known.
The first $q$-congruences for $q$-analogs of harmonic numbers, namely the $q$-analogs of Wolstenholme's theorem, were obtained by Andrews \cite{An99}
who proved that
for all primes $p\ge 3,$
$$
H_{p-1}^q(1;0)=\sum_{j=1}^{p-1}\frac{1}{[j]_q}\equiv \frac{p-1}{2} (1-q) \pmod{[p]_q},
$$
as well as
\begin{equation} \label{qcon-3}
H_{p-1}^q(1;1)=\sum_{j=1}^{p-1}\frac{q^j}{[j]_q}\equiv -\frac{p-1}{2}(1-q) \pmod{[p]_q}.
\end{equation}
Then Shi and Pan \cite{SP07} proved that for primes $p\ge 5$,
\begin{equation*}
\begin{split}
H_{p-1}^q(2;0)&\equiv -\frac{(p-1)(p-5)}{12} (1-q)^2 \pmod{[p]_q}, \\[3pt]
H_{p-1}^q(2;1)&\equiv -\frac{p^2-1}{12} (1-q)^2 \pmod{[p]_q}.
\end{split}
\end{equation*}
Here the above congruences are considered over the ring of polynomials with integer coefficients in variable $q.$
It is clear that $[p]_q$, as the $p$th cyclotomic polynomial, is irreducible over $\mathbb{Q}$ and therefore, the denominators of the rational functions above are coprime to $[p]_q.$

Dilcher \cite{Di08} was the first who noticed that the higher order $q$-harmonic numbers can be expressed modulo $[p]_q$ in terms of the degenerate Bernoulli numbers,
namely in terms of the sequence $K_n(p)$ defined by the generating function
\begin{equation} \label{qcon-1}
\frac{p(z-1)}{z^p-1}=\sum_{n=0}^{\infty}(-1)^{n-1}K_n(p)(z-1)^n \qquad\quad\mbox{for $|z-1|<1$.}
\end{equation}
He proved (see \cite{Di08} and also \cite{Zh13}) that for all positive integers $n$ and primes $p\ge 3,$
\begin{equation} \label{qcon-4}
H_{p-1}^q(n;1)\equiv K_n(p) (1-q)^n \pmod{[p]_q}
\end{equation}
and
\begin{equation}\label{qcon-4a}
H_{p-1}^q(n;0)\equiv (1-q)^n\Bigl(\frac{p-1}{2}+\sum_{m=2}^nK_m(p)\Bigr) \pmod{[p]_q}.
\end{equation}
The congruences above easily imply (see \cite[Theorem 1.2]{Zh13} and \cite[Theorem 2.1]{Ta13} for a similar result) that for all primes $p>3$ and integers $n\ge t\ge 1$,
\begin{equation} \label{HS}
H_{p-1}^q(n;t)\equiv (1-q)^n\sum_{m=0}^{t-1}\binom{t-1}{m}(-1)^mK_{n-m}(p) \pmod{[p]_q}.
\end{equation}
These results were further extended by Zhao \cite{Zh13} who studied generating functions of multiple $q$-harmonic sums with repeated arguments $H_{p-1}^q(\{s\}^t; \{0\}^t)$, where $\{s\}^t$ denotes $t\ge 0$ consecutive copies of the letter $s$.
In particular, he showed \cite[Corollaries 2.2, 2.3]{Zh13} that
\begin{equation} \label{111}
H_{p-1}^q(\{1\}^t)=H_{p-1}^q(\{1\}^t;\{0\}^t)\equiv \binom{p-1}{t}\frac{(1-q)^t}{t+1} \pmod{[p]_q}
\end{equation}
and
\begin{equation} \label{222}
H_{p-1}^q(\{2\}^t;\{0\}^t)\equiv (-1)^t\frac{2\cdot t!}{(2t+2)!}\binom{p-1}{t}\cdot F_{2,t}(p)\cdot(1-q)^{2t} \pmod{[p]_q},
\end{equation}
where $F_{2,t}(p)$ is a monic polynomial in $p$ of degree $t$ and $p>t$.

In this paper, we prove some new $q$-congruences modulo
$[p]_q$ for further classes of multiple $q$-harmonic sums of arbitrary depth with strings of indices composed of ones, twos and threes.
Some of these results are $q$-extensions of the corresponding congruences for ordinary multiple harmonic sums obtained by the authors in \cite{HPT14}.

We will refer to multiple $q$-harmonic sums on strings $(s,s,\ldots,s)$ as {\sl homogeneous} sums. If the vector $(s,s,\ldots,s)$
is modified by insertion of an element $u$,  we say that $H_n^q(\{s\}^a,u,\{s\}^b)$ (correspondingly, $S_n^q(\{s\}^a,u,\{s\}^b)$)
is a {\sl quasi-homogeneous} multiple $q$-harmonic (non-strict) sum.
We may summarize our results as follows.

\noindent Homogeneous sums: for any integer $t\ge 1$ and any prime $p>t$ we have modulo $[p]_q$,
\begin{align*}
S_{p-1}^q(\{1\}^t)&\equiv -K_t(p)(1-q)^t,\\
H_{p-1}^q(\{2\}^t)&\equiv (-1)^t\binom{p+t}{2t+1} \frac{(1-q)^{2t}}{p(t+1)},\\
S^q_{p-1}(\{2\}^t)&\equiv
(1-q)^{2t}\sum_{i=0}^t
\left(t\binom{t-1}{i}+(t-1)\binom{t}{i}\right)
(-1)^{i}K_{2t-i}(p), \\
H_{p-1}^q(\{3\}^t)&\equiv
\frac{(1-q)^{3t}}{(t+1)p^2}
\left(\binom{p+2t+1}{3t+2}+\binom{-p+2t+1}{3t+2}\right).
 \end{align*}
Quasi-homogeneous sums: for any non-negative integers $a,b$ and any prime $p>2t+1$ with  $t=a+b+1$, we have modulo $[p]_q$,
\begin{align*}
H^q_{p-1}(\{1\}^{a},2,\{1\}^{b})
+H^q_{p-1}(\{1\}^{b},2,\{1\}^{a})&\equiv
-\binom{p+1}{t+2}\frac{(1-q)^{t+1}}{p},\\
S_{p-1}^q(\{1\}^{a},2,\{1\}^{b})+S_{p-1}^q(\{1\}^{b},2,\{1\}^{a})
&\equiv (1-q)^{t+1}(K_{t}(p)-K_{t+1}(p)-K_{a+1}(p)K_{b+1}(p)),\\
H_{p-1}^q(\{2\}^a,3,\{2\}^b)+H_{p-1}^q(\{2\}^b,3,\{2\}^a)
&\equiv
(-1)^{t-1}\binom{p+t}{2t+1}\frac{(1-q)^{2t+1}}{p(t+1)},\\
S_{p-1}^q(\{2\}^a,3,\{2\}^b)+S_{p-1}^q(\{2\}^b,3,\{2\}^a)
&\\
\equiv(1-q)^{2t+1}\sum_{i=0}^{t}
&\left(t\binom{t-1}{i}+(t-1)\binom{t}{i}\right)
(-1)^{i+1}K_{2t-i}(p).
\end{align*}
Our derivations for homogeneous strict sums are based on application of generating functions and properties of cyclotomic and Chebyshev polynomials,
while the congruences for non-strict sums follow from some $q$-identities proved in \cite{HP13, HPZ13}.

We prove $q$-duality relations for multiple $q$-harmonic non-strict sums
by employing  $q$-binomial duality relations found by Bradley \cite{Br05}.
We also establish a kind of duality for multiple $q$-harmonic strict sums by proving some new $q$-identities which are generalizations of the divisor generating functions identities due to Dilcher \cite{Di95} and Prodinger \cite{Pr00}.

\noindent Finally, we put forward the following conjecture concerning cyclic sums of some multiple $q$-harmonic sums.
\begin{conjecture}[Cyclic-sum]
Let $d_0,d_1,\dots,d_{t}$ be non-negative integers. Then

\begin{itemize}

\item[i)] For any prime $p>r+1$, where $r=\sum_{i=0}^{t}d_i+2t$,
\begin{equation}\label{CG12}
\sum_{i=0}^{t}H^q_{p-1}\left(
\{1\}^{d_i},2,\{1\}^{d_{i+1}},2,\dots,2,\{1\}^{d_{i+t}}
\right)\equiv
(-1)^t\binom{p+t}{r+1}\frac{(1-q)^{r}}{p}
\pmod{\qq{p}},
\end{equation}

\item[ii)] For any prime $p>r+1$, where $r=\sum_{i=0}^{t}2d_i+3t$,
\begin{equation}\label{CG23}
\sum_{i=0}^{t}H^q_{p-1}\left(
\{2\}^{d_i},3,\{2\}^{d_{i+1}},3,\dots,3,\{2\}^{d_{i+t}}
\right)\equiv
N(p,r,t)\frac{(1-q)^{r}}{p}
\pmod{\qq{p}},
\end{equation}
where $N(p,r,t)$ does not depend on $q$.
\end{itemize}
In both congruences it is understood that $d_i=d_j$ if $i\equiv j$ modulo $t+1$.
\end{conjecture}
Note that the case $t=1$ is exactly Theorem \ref{T6.2} and Theorem \ref{Tx.x} below. The case of arbitrary $t$ when all $d_j$ are zero follows from Theorem \ref{T3.1} and Theorem \ref{T33.3}.

\section{Preliminary results about $K_n(p)$}

We start with recalling several known  results on the sequence $K_n(p)$ defined by (\ref{qcon-1}).
The first few values of $K_n(p)$ are as follows:
\begin{equation*}
\begin{split}
K_0(p)&=-1, \,\, K_1(p)=-\frac{p-1}{2}, \,\, K_2(p)=-\frac{p^2-1}{12}, \,\, K_3(p)=-\frac{p^2-1}{24}, \\[3pt]
K_4(p)&=\frac{(p^2-1)(p^2-19)}{720}, \,\,\,\,\, K_5(p)=\frac{(p^2-1)(p^2-9)}{480}, \,\,\ldots.
\end{split}
\end{equation*}
The properties of this sequence were studied in detail by Dilcher in \cite{Di08} (just with another notation $\widetilde{D}_n(p)=-p^nK_n(p)$).
In particular, it was shown that for $n\ge 2,$ $K_n(p)$ is a polynomial in $p^2$ of degree at most $[n/2]$ which is divisible by $p^2-1.$

The sequence $K_n(p)$ is very closely related to the degenerate Bernoulli numbers (in fact, polynomials) that were first studied by Carlitz \cite{Ca56},
and can be defined from the expansion
\begin{equation} \label{qcon-2}
\frac{x}{(1+\lambda x)^{1/\lambda}-1}=\sum_{n=0}^{\infty}\beta_n(\lambda)\frac{x^n}{n!}.
\end{equation}
By comparing generating functions (\ref{qcon-1}) and (\ref{qcon-2}) we have the relation (see \cite[Theorem 3]{Di08})
\begin{equation} \label{knp}
K_n(p)=\frac{(-1)^{n-1}}{n!} p^n \beta_n\left(\frac{1}{p}\right).
\end{equation}
Howard \cite{Ho} found explicit formulas for the coefficients of $\beta_n(\lambda)$ which together with (\ref{knp}) imply that for $n\ge 2,$
$$
K_n(p)=\frac{(-1)^{n-1}}{(n-1)!}\left(\frac{b_n}{n}
+\sum_{j=1}^{\lfloor n/2\rfloor}\frac{B_{2j}}{2j}\,s(n-1,2j-1)p^{2j}\right),
$$
where $b_n$ is the Bernoulli number of the second kind, $B_{2j}$ is the ordinary Bernoulli number, and $s(n,j)$ is the Stirling number of the first kind,
which are defined by the corresponding generating functions:
$$\frac{x}{\ln(1+x)}=\sum_{n=0}^{\infty}\frac{b_nx^n}{n!},\qquad
\frac{x}{e^x-1}=\sum_{n=0}^{\infty}\frac{B_nx^n}{n!},\qquad
\prod_{j=0}^{n-1}(x-j)=\sum_{j=0}^{n}s(n,j)x^j.
$$
From the other side, the ordinary Bernoulli numbers $B_n$
are the limit case of the degenerate Bernoulli numbers:
$$
\lim_{\lambda\to 0} \beta_n(\lambda)=B_n.
$$

\section{Preliminary results about $q$-MHS}

In this section, we describe some basic relations between multiple $q$-harmonic sums modulo $[p]_q$, which will be useful in the sequel.
These relations arise from inverting the order of summation of a nested sum and from expanding the product of two multiple $q$-harmonic sums.

\noindent Let $\overline{\bf s}=(s_l, s_{l-1}, \ldots, s_1)$ denote the reversal of ${\bf s}=(s_1, \ldots, s_{l-1}, s_l).$
Then we have the following relations.
\begin{theorem} \label{T3}
Let $p$ be a prime and ${\bf s}=(s_1,\ldots, s_l),$ ${\bf t}=(t_1,\ldots, t_l)$ be two $l$-tuples of non-negative integers. Then
\begin{align*}
H_{p-1}^q({\bf s}; {\bf t})&\equiv (-1)^{w({\bf s})}H_{p-1}^q(\overline{\bf s}; \overline{{\bf s}}-\overline{{\bf t}}) \pmod{[p]_q},\\
S_{p-1}^q({\bf s}; {\bf t})&\equiv (-1)^{w({\bf s})}S_{p-1}^q(\overline{\bf s}; \overline{{\bf s}}-\overline{{\bf t}}) \pmod{[p]_q}.
\end{align*}
In particular,
\begin{align}
H_{p-1}^q({\bf s})&\equiv (-1)^{w({\bf s})}H_{p-1}^q(\overline{\bf s}; \{1\}^l) \pmod{[p]_q},\label{EE1}\\
S_{p-1}^q({\bf s})&\equiv (-1)^{w({\bf s})}S_{p-1}^q(\overline{\bf s}; \{1\}^l) \pmod{[p]_q}.\label{EE2}
\end{align}
\end{theorem}
\begin{proof}
Reversing the order of summation, we get
\begin{equation*}
\begin{split}
H_{p-1}^q({\bf s}; {\bf t})&=\sum_{1\le k_1<\cdots<k_l<p}\frac{q^{t_1k_1+\cdots+t_lk_l}}{[k_1]_q^{s_1}\cdots [k_l]_q^{s_l}}
=\sum_{p>k_1>\cdots>k_l\ge 1}\frac{q^{t_1(p-k_1)+\cdots+t_l(p-k_l)}}{[p-k_1]_q^{s_1}\cdots[p-k_l]_q^{s_l}}\\
&=(1-q)^{w({\bf s})}\sum_{1\le k_l<\cdots<k_1<p}\frac{q^{t_1(p-k_1)+\cdots+t_l(p-k_l)+k_1s_1+\cdots+k_ls_l}}{(q^{k_1}-q^p)^{s_1}\cdots (q^{k_l}-q^p)^{s_l}} \\
&\equiv (-1)^{w({\bf s})}(1-q)^{w({\bf s})}\sum_{1\le k_l<\cdots<k_1<p}\frac{q^{k_1(s_1-t_1)+\cdots+k_l(s_l-t_l)}}{(1-q^{k_1})^{s_1}\cdots (1-q^{k_l})^{s_l}}\\
&=(-1)^{w({\bf s})}H_{p-1}^q(\overline{\bf s}; \overline{{\bf s}}-\overline{{\bf t}}) \pmod{[p]_q}.
\end{split}
\end{equation*}
Putting ${\bf t}=(s_1-1,\ldots,s_l-1),$ we get (\ref{EE1}).
The proofs for the non-strict sums are similar.
\end{proof}

For depth one $q$-harmonic sums $H_{p-1}^q(s),$ we have the following.
\begin{corollary} \label{T2}
If $p\ge 3$ is a prime, then for all positive integers  $s$, we have
\begin{equation*}
H_{p-1}^q(s)\equiv (-1)^s K_s(p) (1-q)^s \pmod{[p]_q}.
\end{equation*}
\end{corollary}
\begin{proof}
Setting $l=1$ in (\ref{EE1}) and applying (\ref{qcon-4}), we get the desired congruence.
\end{proof}

It is easy to show that the product of two multiple harmonic sums $H_n^q$ (resp.~$S_n^q$) can be expressed as a linear combination of $H_n^q$
(resp.~$S_n^q$). For example,
\begin{align*}
H_n^q(s_1)H_n^q(s_2)&=H_n^q(s_1,s_2)+H_n^q(s_2,s_1)+ H_n^q(s_1+s_2)+(1-q)H_n^q(s_1+s_2-1), \\[3pt]
S_n^q(s_1)S_n^q(s_2)&=S_n^q(s_1,s_2)+S_n^q(s_2,s_1)-S_n^q(s_1+s_2)-(1-q)S_n^q(s_1+s_2-1).
\end{align*}
From the above relations and Corollary \ref{T2} we obtain the following congruences.
\begin{lemma} \label{L1}
Let $p$ be a prime and $a, b$ be positive integers. Then we have modulo $[p]_q$,
\begin{align*}
H_{p-1}^q(a,b)+H_{p-1}^q(b,a)&\equiv (-1)^{a+b}(1-q)^{a+b}(K_a(p)K_b(p)-K_{a+b}(p)+K_{a+b-1}(p)),  \\[3pt]
S_{p-1}^q(a,b)+S_{p-1}^q(b,a)&\equiv (-1)^{a+b}(1-q)^{a+b}(K_a(p)K_b(p)+K_{a+b}(p)-K_{a+b-1}(p)).
\end{align*}
\end{lemma}

\section{Homogeneous sums $H_{p-1}^q(\{2\}^t)$ and $S_{p-1}^q(\{2\}^t)$}

In this section, we prove congruences for homogeneous strict and non-strict sums on strings composed of twos.
\begin{theorem} \label{T3.1}
For any integer $t\ge 1$ and any prime $p>t$, we have
\begin{equation}\label{CC2}
H_{p-1}^q(\{2\}^t)\equiv (-1)^t\binom{p+t}{2t+1} \frac{(1-q)^{2t}}{p(t+1)} \pmod{[p]_q}.
\end{equation}
\end{theorem}
\begin{proof}
We start by considering a generating function for the sequence $H_{p-1}^q(\{1\}^l),$ which is given by the product
\begin{equation*}
\prod_{k=1}^{p-1}\left(1+\frac{x}{1-q^k}\right)=\sum_{l=0}^{p-1}\frac{H_{p-1}^q(\{1\}^l)}{(1-q)^l}\,x^l.
\end{equation*}
By \cite[Theorem 2.1]{Zh13}, we have
\begin{equation} \label{eq:gen1}
\prod_{k=1}^{p-1}\left(1+\frac{x}{1-q^k}\right)\equiv -\frac{1}{px}\left(1-(1+x)^p\right) \pmod{[p]_q}.
\end{equation}
We are going to relate our sequence $H_{p-1}^q(\{2\}^t)$ to the finite product
\begin{equation} \label{eq:gen2}
F(z):=\prod_{k=1}^{p-1}\left(1-\frac{q^k}{(1-q^k)^2}\frac{(z-1)^2}{z}\right),
\end{equation}
the limit case of which was studied in \cite{DHP14}.
By (\ref{eq:gen2}) and (\ref{eq:gen1}), we get
\begin{equation} \label{eq:gen3}
\begin{split}
F(z)&=\prod_{k=1}^{p-1}\left(1+\frac{z-1}{1-q^k}\right)\prod_{k=1}^{p-1}\left(1+\frac{z^{-1}-1}{1-q^k}\right) \\[3pt]
&\equiv
\frac{(1-z^p)(1-z^{-p})}{p^2(z-1)(z^{-1}-1)}=\frac{(z^p-1)^2}{p^2z^{p-1}(z-1)^2} \pmod{[p]_q}.
\end{split}
\end{equation}
From the other side, for any integer $t,$ $0\le t\le p-1,$ by expanding the product in (\ref{eq:gen2}), we have
\begin{equation*}
z^tF(z)=\sum_{l=0}^{p-1}\frac{(-1)^lH_{p-1}^q(\{2\}^l)}{(1-q)^{2l}}\,(z-1)^{2l}z^{t-l}.
\end{equation*}
By expanding $z^{t-l}$ in powers of $z-1,$
\begin{equation*}
z^{t-l}=(1+(z-1))^{t-l}=\sum_{k=0}^{\infty}\binom{t-l}{k}(z-1)^k
\end{equation*}
where
\begin{equation*}
\binom{\alpha}{k}=\frac{\alpha(\alpha-1)\cdots(\alpha-k+1)}{k!},
\end{equation*}
we obtain
\begin{equation} \label{eq:gen4}
\begin{split}
z^tF(z)&=\sum_{l=0}^{p-1}\sum_{k=0}^{\infty}(z-1)^{2l+k}\binom{t-l}{k}\frac{(-1)^lH_{p-1}^q(\{2\}^l)}{(1-q)^{2l}} \\[3pt]
&=\sum_{j=0}^{\infty}(z-1)^j\sum_{l=0}^{\min([j/2],p-1)}\binom{t-l}{j-2l}\frac{(-1)^lH_{p-1}^q(\{2\}^l)}{(1-q)^{2l}}.
\end{split}
\end{equation}
To extract $H_{p-1}^q(\{2\}^t)$ from (\ref{eq:gen4}), we consider the coefficient of $(z-1)^{2t}$ on the right-hand side, which is exactly
\begin{equation} \label{eq:gen5}
\sum_{l=0}^t\binom{t-l}{2t-2l}\frac{(-1)^lH_{p-1}^q(\{2\}^l)}{(1-q)^{2l}}=\frac{(-1)^tH_{p-1}^q(\{2\}^t)}{(1-q)^{2t}},
\end{equation}
since $\binom{t-l}{2t-2l}$ is distinct from zero for $l=0,1,\ldots, t$ if and only if $l=t.$

Summarizing (\ref{eq:gen3}), (\ref{eq:gen4}) and (\ref{eq:gen5}), we get
\begin{equation} \label{eq:gen6}
\frac{(-1)^tH_{p-1}^q(\{2\}^t)}{(1-q)^{2t}}\equiv [(z-1)^{2t}]\frac{(z^p-1)^2z^{t-p+1}}{p^2(z-1)^2} \pmod{[p]_q},
\end{equation}
where $[(z-a)^j]f(z)$ is the coefficient of $(z-a)^j$ in Taylor's expansion of $f(z)$ centered at~$a$.
By expanding $\frac{(z^p-1)^2 z^{t-p+1}}{p^2(z-1)^2}$ into powers of $z-1,$ we have
\begin{equation*}
\begin{split}
\frac{(z^p-1)^2 z^{t-p+1}}{p^2(z-1)^2}&=\frac{1}{p^2(z-1)^2}(z^{p+t+1}-2z^{t+1}+z^{t-p+1}) \\
&=\frac{1}{p^2(z-1)^2}\left((1+(z-1))^{p+t+1}-2(1+(z-1))^{t+1}+(1+(z-1))^{t-p+1}\right) \\
&=\frac{1}{p^2(z-1)^2}\left(\sum_{k=0}^{p+t+1}\binom{p+t+1}{k}(z-1)^k-2\sum_{k=0}^{t+1}\binom{t+1}{k}(z-1)^k \right.\\
&+
\left.\sum_{k=0}^{\infty}\binom{t-p+1}{k}(z-1)^k\right),
\end{split}
\end{equation*}
and therefore by (\ref{eq:gen6}),
\begin{equation*}
\frac{(-1)^tH_{p-1}^q(\{2\}^t)}{(1-q)^{2t}}\equiv \frac{1}{p^2}\left(\binom{p+t+1}{2t+2}+\binom{t-p+1}{2t+2}\right)
=\frac{1}{p(t+1)}\binom{p+t}{2t+1} \pmod{[p]_q},
\end{equation*}
that completes the proof.
\end{proof}
By applying a similar argument as above, we are able to find the right-hand side of congruence (\ref{222}) explicitly.
\begin{theorem}
For any positive integer $t$ and any prime $p>t,$ we have
\begin{equation}\label{qcon-13.5}
H_{p-1}^q(\{2\}^t;\{0\}^t)\equiv \left((-1)^t\binom{p-1}{2t+1}+\binom{p-1}{t}\right) \frac{(1-q)^{2t}}{(t+1)p} \pmod{[p]_q}.
\end{equation}
\end{theorem}
\begin{proof}
 Consider the product
\begin{equation} \label{qcon-14}
F_1(z)=\prod_{k=1}^{p-1}\left(1-\frac{z^2}{(1-q^k)^2}\right)=\sum_{t=0}^{p-1}\frac{(-1)^tH_{p-1}^q(\{2\}^t;\{0\}^t)}{(1-q)^{2t}}z^{2t}.
\end{equation}
Similarly as in the previous proof, we have
\begin{equation} \label{qcon-15}
\begin{split}
F_1(z)&=\prod_{k=1}^{p-1}\left(1-\frac{z}{1-q^k}\right)\left(1+\frac{z}{1-q^k}\right)
\equiv \frac{-1}{p^2z^2}(1-(1-z)^p)(1-(1+z)^p) \\
&=\frac{-1}{p^2z^2}(1-(1+z)^p-(1-z)^p+(1-z^2)^p)
 \pmod{[p]_q}.
\end{split}
\end{equation}
Now by comparing coefficients of $z^{2t}$ on the right-hand sides of (\ref{qcon-14}) and (\ref{qcon-15}), we easily get
$$
\frac{(-1)^tH_{p-1}^q(\{2\}^t;\{0\}^t)}{(1-q)^{2t}}\equiv \frac{-1}{p^2}\left(-\binom{p}{2t+2}-\binom{p}{2t+2}+(-1)^{t+1}\binom{p}{t+1}\right)
$$
which implies (\ref{qcon-13.5}).
\end{proof}
\begin{theorem} \label{T3.3}
For any  integer $t\ge 0$  and
for any prime $p>2t+1$,
\begin{equation*}
S^q_{p-1}(\{2\}^t)\equiv
(1-q)^{2t}\sum_{i=0}^t
\left(t\binom{t-1}{i}+(t-1)\binom{t}{i}\right)
(-1)^{i}K_{2t-i}(p)
\pmod{\qq{p}}.
\end{equation*}
\end{theorem}
\begin{proof}
By (23) in \cite{HP13}, we have
$$S^q_n(\{ 2\}^t)=\sum_{k=1}^n
\qbin{n}{k}\qbin{n+k}{k}^{-1}\frac{(-1)^{k-1}(1+q^k)q^{\binom{k}{2}+tk}}{\qq{k}^{2t}},
$$
where
$$\qbin{n}{k}=\prod_{j=1}^k\frac{1-q^{n-k+j}}{1-q^j}$$
is the {\sl Gaussian $q$-binomial coefficient}.
By (17) and (18) in \cite{Ta13}, for $k=1,\dots,p-1$,
$$\qq{p}\qbin{p-1}{k}\qbin{p-1+k}{k}^{-1}\equiv
(-1)^{k}q^{-\binom{k}{2}-k}
\left(\qq{k}-\qq{p}
-\qq{p}\qq{k}\sum_{j=1}^{k-1}\frac{1+q^j}{\qq{j}}
\right)\!\!\pmod{\qq{p}^2}.$$
Hence, for $n=p-1$ with $p>2t+1$, we obtain
\begin{align*}
S^q_{p-1}(\{ 2\}^t)
&\equiv\frac{1}{\qq{p}}\sum_{k=1}^{p-1}
\frac{(1+q^k)q^{(t-1)k}}{\qq{k}^{2t}}
\left(-\qq{k}+\qq{p}
+\qq{p}\qq{k}\sum_{j=1}^{k-1}\frac{1+q^j}{\qq{j}}\right)\\
&\equiv-\frac{1}{\qq{p}}\sum_{k=1}^{p-1}
\frac{q^{(t-1)k}+q^{tk}}{\qq{k}^{2t-1}}
 +\sum_{k=1}^{p-1}
\frac{q^{(t-1)k}+q^{tk}}{\qq{k}^{2t}}\\
&\qquad +\sum_{1\leq j<k\leq p-1}
\frac{(q^{tk}+q^{j+(t-1)k})+(q^{(t-1)k}+q^{j+tk})}{\qq{j}\qq{k}^{2t-1}}
\pmod{\qq{p}}.
\end{align*}
By (13) in \cite{Ta13}, we have
$$\sum_{1\leq j<k\leq p-1}
\frac{q^{tk}+q^{j+(t-1)k}}{\qq{j}\qq{k}^{2t-1}}
\equiv
\sum_{k=1}^{p-1}\frac{1}{\qq{j}}
\sum_{k=1}^{p-1}\frac{q^{tk}}{\qq{k}^{2t-1}}
-\sum_{k=1}^{p-1}\frac{q^{tk}}{\qq{k}^{2t}}
\pmod{\qq{p}},$$
and
$$\sum_{1\leq j<k\leq p-1}
\frac{q^{(t-1)k}+q^{j+tk}}{\qq{j}\qq{k}^{2t-1}}
\equiv
\sum_{k=1}^{p-1}\frac{1}{\qq{j}}
\sum_{k=1}^{p-1}\frac{q^{(t-1)k}}{\qq{k}^{2t-1}}
-\sum_{k=1}^{p-1}\frac{q^{(t-1)k}}{\qq{k}^{2t}}
\pmod{\qq{p}}.$$
Hence
\begin{align*}
S^q_{p-1}(\{ 2\}^t)
&\equiv
\left(-\frac{1}{\qq{p}}+\sum_{k=1}^{p-1}\frac{1}{\qq{j}}\right)
\sum_{k=1}^{p-1}
\frac{q^{(t-1)k}+q^{tk}}{\qq{k}^{2t-1}}\\
&\equiv
(2t-1)\sum_{k=1}^{p-1}\frac{q^{tk}}{\qq{k}^{2t}}
-(t-1)(1-q)\sum_{k=1}^{p-1}\frac{q^{tk}}{\qq{k}^{2t-1}}
\pmod{\qq{p}},
\end{align*}
where we used (11) in \cite{Ta13}.
Finally, we apply \eqref{HS}, which completes the proof.
\end{proof}

\section{Homogeneous sums $H_{p-1}^q(\{3\}^t)$}

In this section, we consider homogeneous sums $H_{p-1}^q(\{s\}^t)$ when $s=3$. Before proving the main result of this section, we state the following
combinatorial lemma.
We owe the proof of this lemma to Robin Chapman \cite{RC15}.
\begin{lemma} \label{CL}
Let $p, s$ be positive integers, $\delta\in\{0, 1\}$ and
\begin{align*}
T(p,s,\delta)&:=[x^{3s}]\sum_{k=0}^{p-\delta}(-1)^k\binom{p-k-\delta}{k-\delta}(2-x)^{p-2k}(1+x)^{s-k}, \\
\widetilde{T}(p,s,\delta)&:=[x^{3s}]\sum_{k=0}^{p-\delta}(-1)^k\binom{p-k-\delta}{k-\delta}(2-x)^{p-2k}(1+x)^{p+s-k}.
\end{align*}
Then
$$
T(p,s,\delta)=(-1)^{\delta}\binom{s+p-2\delta+1}{3s+1}, \quad
\widetilde{T}(p,s,\delta)=(-1)^{s+\delta}\binom{2s+p-2\delta+1}{3s+1}.
$$
\end{lemma}
\begin{proof}
Consider  generating functions of both sequences
$$
G_{s,\delta}(y)=(-1)^{\delta}\sum_{p=1}^{\infty}T(p,s,\delta)y^p \quad\text{and}\quad
\widetilde{G}_{s,\delta}(y)=(-1)^{\delta}\sum_{p=1}^{\infty}\widetilde{T}(p,s,\delta)y^p.
$$
Then we have
\begin{align*}
G_{s,\delta}(y)
&=
[x^{3s}]\sum_{p=1}^{\infty}\sum_{k=0}^{p-2\delta}(-1)^{k}\binom{p-k-2\delta}{k}
(2-x)^{p-2k-2\delta}(1+x)^{s-k-\delta}y^p\\
&=
y^{2\delta}[x^{3s}](1+x)^{s-\delta}\sum_{r\ge 0}\sum_{k=0}^r(-1)^k\binom{r}{k}
(2-x)^{r-k}(1+x)^{-k}y^{k+r}\\
&=y^{2\delta}[x^{3s}](1+x)^{s-\delta}
\sum_{r\ge 0}y^r(2-x)^r\left(1-\frac{y}{(2-x)(1+x)}\right)^{r}\\
&=y^{2\delta}[x^{3s}]\frac{(1+x)^{s-\delta}}{
1-y(2-x)\left(1-\frac{y}{(2-x)(1+x)}\right)}\\
&=y^{2\delta}[x^{3s}]\frac{(1+x)^{s-\delta+1}}{
1+x-y(2-x)(1+x)+y^2}\\
&=y^{2\delta}[x^{3s}]\frac{(1+x)^{s-\delta+1}}{
(1-y)^2+(1-y)x+yx^2}.
\end{align*}
Hence, by letting $u=1/(1-y)$, we easily obtain
\begin{align*}
(-1)^{\delta}T(p,s,\delta)
&=[y^{p-2\delta}][x^{3s}]
\frac{(1+x)^{s-\delta+1}}{
(1-y)^2+(1-y)x+yx^2}\\
&=[y^{p-2\delta}]\frac{[x^{3s}]}{(1-y)^2}
\frac{(1+x)^{s-\delta+1}}{
1+ux+(u^2-u)x^2}\\
&=[y^{p-2\delta}]\frac{[x^{3s}]}{(1-y)^2}
(1+x)^{s-\delta+1}\sum_{j=0}^{\infty}a_j x^j\\
&=[y^{p-2\delta}]\frac{1}{(1-y)^2}
\sum_{j=0}^{s-\delta+1}\binom{s-\delta+1}{j}
a_{2s+\delta-1+j} \\
&=[y^{p-2\delta}]\frac{1}{(1-y)^2}
\sum_{j=0}^{s-\delta+1}\binom{s-\delta+1}{j}
\frac{\alpha^{2s+\delta+j}-\beta^{2s+\delta+j}}{\alpha-\beta} \\
&=[y^{p-2\delta}]\frac{1}{(1-y)^2}
\frac{\alpha^{2s+\delta}(1+\alpha)^{s-\delta+1}-\beta^{2s+\delta}(1+\beta)^{s-\delta+1}}{\alpha-\beta} \\
&=[y^{p-2\delta}]\frac{y^{2s}}{(1-y)^{3s+2}}
\frac{\alpha^{\delta}(1+\alpha)^{1-\delta}-\beta^{\delta}(1+\beta)^{1-\delta}}{\alpha-\beta} \\
&=[y^{p-2\delta-2s}](1-y)^{-(3s+2)}
=\binom{s+p-2\delta+1}{3s+1}
\end{align*}
where we used the fact that $\alpha^2(1+\alpha)=\beta^2(1+\beta)=u(u-1)^2=y^2/(1-y)^3$.

Similarly, for the second generating function, we have
\begin{align*}
\widetilde{G}_{s,\delta}(y)
&=
[x^{3s}]\sum_{p=1}^{\infty}\sum_{k=0}^{p-2\delta}(-1)^{k}\binom{p-k-2\delta}{k}
(2-x)^{p-2k-2\delta}(1+x)^{p+s-k-\delta}y^p\\
&=
y^{2\delta}[x^{3s}](1+x)^{s+\delta}\sum_{r\ge 0}\sum_{k=0}^r(-1)^k\binom{r}{k}
(2-x)^{r-k}(1+x)^{r}y^{k+r}\\
&=y^{2\delta}[x^{3s}](1+x)^{s+\delta}
\sum_{r\ge 0}y^r(2-x)^r(1+x)^r\left(1-\frac{y}{2-x}\right)^{r}\\
&=y^{2\delta}[x^{3s}]\frac{(1+x)^{s+\delta}}{
1-y(1+x)(2-x-y)}\\
&=y^{2\delta}[x^{3s}]\frac{(1+x)^{s+\delta}}{
(1-y)^2-y(1-y)x+yx^2}.
\end{align*}
By comparing the coefficients of the powers of $y$ on both sides, we get
\begin{align*}
(-1)^{\delta}\widetilde{T}(p,s,\delta)
&=[y^{p-2\delta}][x^{3s}]
\frac{(1+x)^{s+\delta}}{
(1-y)^2-y(1-y)x+yx^2}\\
&=[y^{p-2\delta}]\frac{[x^{3s}]}{(1-y)^2}
\frac{(1+x)^{s+\delta}}{
1-(u-1)x+(u^2-u)x^2}\\
&=[y^{p-2\delta}]\frac{[x^{3s}]}{(1-y)^2}
(1+x)^{s+\delta}\sum_{j=0}^{\infty}\widetilde{a}_j x^j\\
&=[y^{p-2\delta}]\frac{1}{(1-y)^2}
\sum_{j=0}^{s+\delta}\binom{s+\delta}{j}
\widetilde{a}_{2s-\delta+j} \\
&=[y^{p-2\delta}]\frac{1}{(1-y)^2}
\sum_{j=0}^{s+\delta}\binom{s+\delta}{j}
\frac{\widetilde{\alpha}^{2s-\delta+j+1}-\widetilde{\beta}^{2s-\delta+j+1}}{\widetilde{\alpha}-\widetilde{\beta}} \\
&=[y^{p-2\delta}]\frac{1}{(1-y)^2}
\frac{\widetilde{\alpha}^{2s-\delta+1}(1+\widetilde{\alpha})^{s+\delta}-
\widetilde{\beta}^{2s-\delta+1}(1+\widetilde{\beta})^{s+\delta}}{\widetilde{\alpha}-\widetilde{\beta}} \\
&=[y^{p-2\delta}]\frac{(-1)^sy^{s}}{(1-y)^{3s+2}}
\frac{\widetilde{\alpha}^{1-\delta}(1+\widetilde{\alpha})^{\delta}-\widetilde{\beta}^{1-\delta}(1+\widetilde{\beta})^{\delta}}{\widetilde{\alpha}-\widetilde{\beta}} \\
&=(-1)^s[y^{p-2\delta-s}](1-y)^{-(3s+2)}
=(-1)^s\binom{2s+p-2\delta+1}{3s+1}
\end{align*}
where we used the fact that $\widetilde{\alpha}^2(1+\widetilde{\alpha})=\widetilde{\beta}^2(1+\widetilde{\beta})=u^2(1-u)=-y/(1-y)^3$.
\end{proof}

\begin{theorem} \label{T33.3}
For any positive integer $t$ and for any prime $p>t$,
\begin{equation}\label{33}
\frac{H_{p-1}^q(\{3\}^t)}{(1-q)^{3t}}\equiv
\frac{1}{(t+1)p^2}
\left(\binom{p+2t+1}{3t+2}+\binom{-p+2t+1}{3t+2}\right)
 \pmod{[p]_q}.
\end{equation}
\end{theorem}

\begin{proof}
Since by (\ref{EE1}), $H_{p-1}^q(\{3\}^t)\equiv (-1)^t H_{p-1}^q(\{3\}^t; \{1\}^t)
\pmod{[p]_q}$, we may deal with $H_{p-1}^q(\{3\}^t; \{1\}^t)$.
A generating function for the sequence $H_{p-1}^q(\{3\}^t; \{1\}^t)$
can be derived from the product
$$
F(z)=\prod_{k=1}^{p-1}\left(1+\frac{q^k}{(1-q^k)^3}\frac{(z-1)^3}{z}\right).
$$
We have
$$
z^tF(z)=\sum_{l=0}^{p-1}\frac{H_{p-1}^q(\{3\}^l; \{1\}^l)}{(1-q)^{3l}}\,(z-1)^{3l}z^{t-l}
$$
and since
$$
z^{t-l}=(1+(z-1))^{t-l}=\sum_{k=0}^{\infty}\binom{t-l}{k}(z-1)^k,
$$
we get
\begin{equation*}
\begin{split}
z^tF(z)&=\sum_{l=0}^{p-1}\sum_{k=0}^{\infty}
(z-1)^{3l+k}\binom{t-l}{k}\frac{H_{p-1}^q(\{3\}^l; \{1\}^l)}{(1-q)^{3l}} \\
&=\sum_{j=0}^{\infty}(z-1)^j
\sum_{l=0}^{\min([j/3],p-1)}\binom{t-l}{j-3l}\frac{H_{p-1}^q(\{3\}^l; \{1\}^l)}{(1-q)^{3l}}.
\end{split}
\end{equation*}
To extract $H_{p-1}^q(\{3\}^t; \{1\}^t),$ we consider
\begin{equation} \label{11}
[(z-1)^{3t}]z^tF(z)=\sum_{l=0}^{t}\binom{t-l}{3t-3l}
\frac{H_{p-1}^q(\{3\}^l; \{1\}^l)}{(1-q)^{3l}}=
\frac{H_{p-1}^q(\{3\}^t; \{1\}^t)}{(1-q)^{3t}}.
\end{equation}
From the other side, we can factor $F(z)$ as
$$
F(z)=\prod_{k=1}^{p-1}\left(1+\frac{z-1}{1-q^k}\right)
\left(1-\frac{\frac{(z-1)}{2}(1+\sqrt{1-4/z})}{1-q^k}
\right)\left(1-\frac{\frac{(z-1)}{2}(1-\sqrt{1-4/z})}{1-q^k}
\right).
$$
Then by (\ref{eq:gen1}) we have
\begin{align*}
F(z)\equiv -\frac{(1-z^p)}{p(z-1)}\cdot&
\frac{\bigl(1-(1-\frac{(z-1)}{2}(1+\sqrt{1-4/z}))^p\bigr)}
{p(z-1)(1+\sqrt{1-4/z})}\\
&\qquad\quad\times \frac{\bigl(1-(1-\frac{(z-1)}{2}(1-\sqrt{1-4/z}))^p\bigr)}
{p(z-1)(1-\sqrt{1-4/z})} \pmod{[p]_q},
\end{align*}
or
$$F(z)\equiv
\frac{z(z^p-1)}{p^3(z-1)^3}\left(1+\frac{1}{z^p}-
\frac{2}{z^{p/2}}T_p\left(\frac{\sqrt{z}(3-z)}{2}\right)\right)
 \pmod{[p]_q}.
$$
where $T_n(x)$ denotes the
Chebyshev polynomial of the first kind.

Hence, by (\ref{11}), we conclude that
\begin{equation} \label{oo}
\frac{H_{p-1}^q(\{3\}^t; \{1\}^t)}{(1-q)^{3t}}
\equiv
[(z-1)^{3t}]\frac{z^{t+1}(z^p-1)}{p^3(z-1)^3}
\left(1+\frac{1}{z^p}-
\frac{2}{z^{p/2}}T_p\left(\frac{\sqrt{z}(3-z)}{2}\right)\right) \pmod{[p]_q}.
\end{equation}
By Kummer's  formula
$$
a^n+b^n=\sum_{k=0}^{[n/2]}(-1)^k\frac{n}{n-k}\binom{n-k}{k}(ab)^k
(a+b)^{n-2k},
$$
we have
\begin{equation} \label{oa}
\frac{2}{z^{p/2}}T_p\left(\frac{\sqrt{z}(3-z)}{2}\right)=\sum_{k=0}^{(p-1)/2}(-1)^k\frac{p}{p-k}\binom{p-k}{k}\frac{1}{z^k}
(3-z)^{p-2k}.
\end{equation}
Taking into account that
$$\frac{p}{p-k}\binom{p-k}{k}=\binom{p-k}{k}+\binom{p-k-1}{k-1}$$
and by substituting (\ref{oa}) into (\ref{oo}), we get modulo $[p]_q$,
\begin{equation*}
\begin{split}
\frac{H_{p-1}^q(\{3\}^t; \{1\}^t)}{(1-q)^{3t}}
&\equiv \frac{1}{p^3}
[(z-1)^{3t+3}](z^{t+p+1}-z^{t+1-p}) \\
&-\frac{1}{p^3}(\widetilde{T}(p,t+1,0)-T(p,t+1,0)+\widetilde{T}(p,t+1,1)-T(p,t+1,1)).
\end{split}
\end{equation*}
Since
$$
[(z-1)^{3t+3}](z^{t+p+1}-z^{t+1-p})=\binom{p+t+1}{3t+3}+(-1)^t\binom{p+2t+1}{3t+3},
$$
by Lemma \ref{CL}, we readily get the desired congruence.
\end{proof}

\section{Quasi-homogeneous sums $H_{p-1}^q(\{2\}^a,3,\{2\}^b)$ and $S_{p-1}^q(\{2\}^a,3,\{2\}^b)$}

The aim of this section is to prove first congruences for quasi-homogeneous $q$-harmonic sums.

\begin{theorem} \label{Tx.x}
For any integers $a,b\geq 0$  and
for any prime $p>2t+1$ with $t=a+b+1$,
\begin{equation*}
\begin{split}
&H_{p-1}^q(\{2\}^a,3,\{2\}^b)+H_{p-1}^q(\{2\}^b,3,\{2\}^a)
\equiv
(-1)^{t-1}\binom{p+t}{2t+1}
\frac{(1-q)^{2t+1}}{p(t+1)}
\pmod{\qq{p}},\\
&S_{p-1}^q(\{2\}^a,3,\{2\}^b)+S_{p-1}^q(\{2\}^b,3,\{2\}^a) \\
&\qquad\qquad\quad\equiv
-(1-q)^{2t+1}\sum_{i=0}^t
\left(t\binom{t-1}{i}+(t-1)\binom{t}{i}\right)
(-1)^{i}K_{2t-i}(p)
\pmod{\qq{p}}.
\end{split}
\end{equation*}
\end{theorem}
\begin{proof}
We have
\begin{equation*}
\begin{split}
&\quad H_{p-1}^q(\{2\}^a,3,\{2\}^b)+(1-q)H_{p-1}^q(\{2\}^t) \\[3pt]
&=\sum_{k_1<k_2<\ldots<k_a}
\frac{q^{k_1+\dots+k_a}}{[k_1]_q^2\dots[k_a]_q^2}
\sum_{k_a<n<k_{a+1}}\frac{q^{2n}+q^n(1-q^n)}{[n]_q^3}
\sum_{k_{a+1}<k_{a+2}<\dots<k_{a+b}}
\frac{q^{k_{a+1}+\dots+k_{a+b}}}{[k_{a+1}]^2\dots[k_{a+b}]^2} \\[3pt]
&=H_{p-1}^q(\{2\}^a,3,\{2\}^b; \{1\}^{a+b+1})
\equiv -H_{p-1}^q(\{2\}^b,3,\{2\}^a) \pmod{\qq{p}},
\end{split}
\end{equation*}
where in the last congruence we used \eqref{EE1}.
Hence
$$H_{p-1}^q(\{2\}^a,3,\{2\}^b)+H_{p-1}^q(\{2\}^b,3,\{2\}^a)
\equiv
-(1-q)H_{p-1}^q(\{2\}^t)
\pmod{\qq{p}}$$
which, by \eqref{CC2}, implies the first  congruence of the theorem.
In a similar way, we show that
$$S_{p-1}^q(\{2\}^a,3,\{2\}^b)+S_{p-1}^q(\{2\}^b,3,\{2\}^a)
\equiv
-(1-q)S_{p-1}^q(\{2\}^t)
\pmod{\qq{p}}$$
and then by Theorem \ref{T3.3}, we conclude the proof.
\end{proof}

\section{Duality for multiple $q$-harmonic (non-strict) sums}

\noindent In \cite{Di95}, Dilcher established the following identity: for any pair
of positive integers $n$, $s$,
\begin{equation}\label{D}
\sum_{k=1}^n \qbin{n}{k}\frac{(-1)^{k}
q^{\binom{k}{2}+sk}}{\qq{k}^{s} }
=-\sum_{1\leq j_1\leq j_2\leq\cdots\leq j_s\le n}
\prod_{i=1}^s\frac{q^{j_i}}{\qq{j_i}}.
\end{equation}
Identity \eqref{D} is a generalization of the case $s=1$ due to Van Hamme \cite{VH82} and by
taking the limit as $n$ goes to infinity one obtains a remarkable $q$-series
related to overpartitions and divisor generating functions. For example, if $s=1$ and $|q|<1$ then
$$
\sum_{k=1}^{\infty} \frac{(-1)^{k-1}
q^{\binom{k+1}{2}}}{(1-q)(1-q^2)\cdots(1-q^{k-1})(1-q^{k})^2 }
=\sum_{j=1}^{\infty}\frac{q^j}{1-q^{j}}=\sum_{j=1}^{\infty}d(j)q^j.
$$
where $d(j)$ is the number of divisors of $j$ (see the pioneering paper of Uchimura \cite{Uc81}).

\noindent With this motivation, several authors have recently investigated \eqref{D}
and they extended it along several directions: see for example
\cite{An13, Br05, FL03, FL05, GZ11, GZ13, IS12, MSS13, Pr01, Pr04, Xu14, Ze05}.
In~\cite{Pr00}, Prodinger shows the inversion of \eqref{D},
\begin{equation}\label{H}
\sum_{k=1}^n \qbin{n}{k}(-1)^{k}
q^{\binom{k}{2}-nk}\!\!\!\!\!\sum_{1\leq j_1\leq j_2\leq\dots\leq j_s=k}
\prod_{i=1}^s\frac{q^{j_i}}{\qq{j_i}}
=-\sum_{k=1}^n\frac{q^{(s-1)k}}{\qq{k}^{s}},
\end{equation}
which is the $q$-analog of a formula of Hernandez \cite{He99}.

In \cite{Br05}, Bradley extended identity (\ref{D}) to multiple $q$-binomial sums
$$
A_n(s_1, \ldots, s_l)=
\sum_{k=1}^n\qbin{n}{k}(-1)^{k-1}
q^{\binom{k}{2}+k}\sum_{1\le k_1\le k_2\le\cdots \le k_l=k}
\prod_{j=1}^l\frac{q^{(s_j-1)k_j}}{[k_j]_q^{s_j}},
$$
(note that we changed the order of summation here in comparison with Bradley's definition)
and proved a duality identity between  $q$-harmonic non-strict sums $S_{n}^q$ and $A_n$:
for positive integers $r, a_1, b_1, \ldots, a_r, b_r,$
\begin{equation} \label{BrD}
S_n^q\Bigl(a_1, \{1\}^{b_1-1}, \overset{r}{\underset{j=2}{{\bf Cat}}}\{a_j+1, \{1\}^{b_j-1}\};{\bf 1}\Bigr)=
A_n\Bigl(\overset{r-1}{\underset{j=1}{{\bf Cat}}}\{\{1\}^{a_j-1}, b_j+1\}, \{1\}^{a_r-1}, b_r\Bigr)
\end{equation}
where ${\bf Cat}_{j=1}^l\{s_j\}$ stands for the concatenated argument sequence $s_1, \ldots, s_l$
and ${\bf 1}$ denotes a vector whose all components are equal to $1$.
From (\ref{BrD}) we get a duality relation for multiple $q$-harmonic non-strict sums modulo $[p]_q$.
\begin{theorem} \label{T4.1}
Let $p$ be a prime. Then for positive integers $r, a_1, b_1, \ldots, a_r, b_r,$ we have
\begin{equation*}
\begin{split}
S_{p-1}^q&\Bigl(\overset{r-1}{\underset{j=1}{{\bf Cat}}}\{\{1\}^{a_j-1}, b_j+1\}, \{1\}^{a_r-1}, b_r\Bigr) \\
&\equiv (-1)^{a_1+b_1+\cdots+a_r+b_r}
S_{p-1}^q\Bigl(\overset{r-1}{\underset{j=1}{{\bf Cat}}}\{\{1\}^{b_{r-j+1}-1}, a_{r-j+1}+1\}, \{1\}^{b_1-1}, a_1\Bigr) \pmod{[p]_q}.
\end{split}
\end{equation*}
\end{theorem}
\begin{proof} For $1\leq j <p$, we have $\qq{p-j}=q^{-j}(\qq{p}-\qq{j})$, which implies that
\begin{equation} \label{nk}
\qbin{p-1}{k}
=\prod_{j=1}^{k}\frac{\qq{p-j}}{\qq{j}}
\equiv\prod_{j=1}^{k}(-q^{-j})
=(-1)^{k}q^{-\binom{k+1}{2}}
\pmod{\qq{p}}.
\end{equation}
Now by Theorem \ref{T3} and (\ref{BrD}), we readily get the congruence.
\end{proof}
In particular, for $r=1$ we obtain the following result.
\begin{corollary} \label{C3}
Let $p$ be a prime. Then for positive integers $a, b$ we have
$$
S_{p-1}^q(\{1\}^{a-1},b)\equiv(-1)^{a+b}\,S_{p-1}^q(\{1\}^{b-1},a) \pmod{[p]_q}.
$$
\end{corollary}

\begin{theorem}\label{T4.2}
Let $a, b$ be non-negative integers and $t$ be a positive integer. Then
\begin{equation*}
\begin{split}
&S_{p-1}^q(\{1\}^t)\equiv -K_t(p)(1-q)^t \pmod{[p]_q},\\
&S_{p-1}^q(1,\{2\}^{t-1},1)\equiv (1-q)^{2t}\sum_{j=0}^t\Bigl(t\binom{t-1}{j}+(t-1)\binom{t}{j}\Bigr)(-1)^{j+1}K_{2t-j}(p) \pmod{[p]_q}, \\
&S_{p-1}^q(\{1\}^{a},2,\{1\}^{b})+S_{p-1}^q(\{1\}^{b},2,\{1\}^{a}) \\
&\qquad\qquad\equiv (1-q)^{a+b+2}(K_{a+b+1}(p)-K_{a+b+2}(p)-K_{a+1}(p)K_{b+1}(p)) \pmod{[p]_q}.
\end{split}
\end{equation*}
\end{theorem}
\begin{proof} Setting $b=1,$ $a=t$ in Corollary \ref{C3}, we get the first congruence by Theorem \ref{T2}.
Setting $r=t,$ $a_1=\ldots=a_t=1,$ $b_1=\ldots=b_{t-1}=1,$ $b_t=2$ in Theorem \ref{T4.1}, we get
$$
S_{p-1}^q(\{2\}^t)\equiv -S_{p-1}^q(1,\{2\}^{t-1},1) \pmod{[p]_q},
$$
and the result follows immediately from Theorem \ref{T3.3}.
If we put $r=2,$ $a_1=a+1,$ $b_1=b_2=1,$ $a_2=b,$ we get
\begin{equation} \label{S121}
S_{p-1}^q(\{1\}^a,2,\{1\}^b)\equiv (-1)^{a+b+1}S_{p-1}^q(b+1,a+1) \pmod{[p]_q}.
\end{equation}
Now by Lemma \ref{L1}, we easily derive the last congruence of the theorem.
\end{proof}

\section{Duality for multiple $q$-harmonic (strict) sums}

\noindent In this section, we provide  further extensions of \eqref{D} and \eqref{H} and then consider application of the new identities to $q$-congruences.
\begin{theorem} \label{TwoId}
Let $n,s_1,s_2,\dots, s_l$ be positive integers. Then
\begin{align}\label{DE}
&\sum_{k=1}^n \qbin{n}{k}(-1)^{k}
q^{\binom{k}{2}+k}\!\!\!\!\sum_{1\leq k_1<k_2<\cdots<k_l=k}
\prod_{i=1}^l\frac{q^{(s_i-1)k_i}}{\qq{k_i}^{s_i}}
=(-1)^l\!\!\!
\underset{j_i<j_{i+1},\; i\in I}{\sum_{1\leq j_1\leq j_2\leq\dots\leq j_w\leq n}}
\prod_{i=1}^w\frac{q^{j_i}}{\qq{j_i}},\\
\label{HE}
&\sum_{k=1}^n \qbin{n}{k}(-1)^{k}
q^{\binom{k}{2}-nk}\!\!\!
\underset{j_i<j_{i+1},\; i\in I}{\sum_{1\leq j_1\leq j_2\leq\dots\leq j_w=k}}
\prod_{i=1}^w\frac{q^{j_i}}{\qq{j_i}}
=(-1)^l\!\!\!
\sum_{1\leq k_1<k_2<\cdots<k_l\leq n}
\prod_{i=1}^l\frac{q^{(s_i-1)k_i}}{\qq{k_i}^{s_i}}
\end{align}
where $w=\sum_{i=1}^ls_i$ and
$I=\{s_1,s_1+s_2,\dots,s_1+s_2+\dots+s_{l-1}\}$.
\end{theorem}

\noindent The proof of Theorem \ref{TwoId} will be given in the next section.

\noindent  From \eqref{DE} with $n$ replaced by $p-1$, where $p$ is a prime, with the help of congruence \eqref{nk} we get a kind of duality for multiple $q$-harmonic $H_n^q$ sums.
\begin{theorem} \label{T6.1}
Let $p$ be a prime and $s_1, \ldots, s_l$ be positive integers. Then
$$
H_{p-1}^q(s_1,\ldots,s_l)\equiv (-1)^l \underset{j_i<j_{i+1},\; i\in I}{\sum_{1\leq j_1\leq j_2\leq\dots\leq j_w\leq p-1}}
\prod_{i=1}^w\frac{q^{j_i}}{\qq{j_i}} \pmod{[p]_q},
$$
where $w=\sum_{i=1}^ls_i$ and
$I=\{s_1,s_1+s_2,\dots,s_1+s_2+\dots+s_{l-1}\}$.
\end{theorem}
\noindent
As an application of the duality relation above, we prove the following congruence for quasi-homogeneous sums.
\begin{theorem} \label{T6.2}
For any integers $a,b\geq 0$  and
for any prime $p>t$, where $t=a+b+1$,
\begin{equation}\label{CC1}
H^q_{p-1}(\{1\}^{a},2,\{1\}^{b})
+H^q_{p-1}(\{1\}^{b},2,\{1\}^{a})\equiv
-\binom{p+1}{t+2}\frac{(1-q)^{t+1}}{p}
\pmod{\qq{p}}.
\end{equation}
\end{theorem}
\begin{proof}
By Theorem \ref{T6.1}, we easily  obtain the following
congruences modulo $\qq{p}$,
\begin{align*}
-H^q_{p-1}(\{1\}^{a},2,\{1\}^{b})
&\equiv(-1)^{a+b}
\!\!\!
\sum_{1\leq j_1<\dots<j_{a}<j_{a+1}\leq  j_{a+2}<j_{a+3}
<\dots<j_{t+1}\leq p-1}
\prod_{i=1}^{t+1}\frac{q^{j_i}}{\qq{j_i}}\\
&\equiv(-1)^{a+b}
\!\!\!
\sum_{1\leq j_{t+1}<\dots<j_{a+3}<j_{a+2}\leq  j_{a+1}<j_{a}
<\dots<j_{1}\leq p-1}
\prod_{i=1}^{t+1}\frac{q^{p-j_i}}{\qq{p-j_i}}\\
&\equiv
\!\!\!
\sum_{1\leq j_{t+1}<\dots<j_{a+3}<j_{a+2}\leq  j_{a+1}<j_{a}
<\dots<j_{1}\leq p-1}
\prod_{i=1}^{t+1}\frac{1}{\qq{j_i}}.
\end{align*}
The right-hand side of the last congruence can be decomposed as
$$
\sum_{1\leq j_{t+1}<\dots<j_{a+3}<j_{a+2}\leq  j_{a+1}<j_{a}
<\dots<j_{1}\leq p-1}
\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!
\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!
((1-q)\qq{j_{a+1}}+q^{j_{a+1}})\prod_{i=1}^{r}\frac{1}{\qq{j_i}}
\;\;\;\;\;\;\;\;+\!\!\!\!\!\!\!\!\!\!\!\!
\sum_{1\leq j_{t+1}<\dots<j_{1}\leq p-1}
\prod_{i=1}^{t+1}\frac{1}{\qq{j_i}}.
$$
Therefore
$$-H^q_{p-1}(\{1\}^{a},2,\{1\}^{b})\equiv
(1-q)H^q_{p-1}(\{1\}^{t})
+H^q_{p-1}(\{1\}^{b},2,\{1\}^{a})+H^q_{p-1}(\{1\}^{t+1}).
$$
Finally,  by employing congruence (\ref{111}), we obtain
\begin{align*}
H^q_{p-1}(\{1\}^{a},2,\{1\}^{b})
&+H^q_{p-1}(\{1\}^{b},2,\{1\}^{a})
\equiv
-(1-q)H^q_{p-1}(\{1\}^{t})-H^q_{p-1}(\{1\}^{t+1})\\
&\equiv
-(1-q)\binom{p-1}{t}\frac{(1-q)^{t}}{r}
-\binom{p-1}{t+1}\frac{(1-q)^{t+1}}{t+2}\\
&\equiv
-\binom{p+1}{t+2}\frac{(1-q)^{t+1}}{p}
\pmod{\qq{p}}.
\end{align*}
\end{proof}

\section{Proof of Theorem \ref{TwoId}}

\noindent In order to prove identities \eqref{DE} and \eqref{HE}, we need first some auxiliary statements.
\begin{lemma} Let $n,j$ be positive integers with $j\leq n$. Then
\begin{align}\label{C1}
\sum_{k=j}^n \qbin{k-1}{j-1}q^k&=\qbin{n}{j}q^j,\\
\label{C2}
\sum_{k=j}^n\qbin{n}{k}(-1)^{k}q^{\binom{k}{2}}
&=\qbin{n-1}{j-1}(-1)^{j}q^{\binom{j}{2}}.
\end{align}
\end{lemma}
\begin{proof} Since
$$
\qbin{k}{j}=\qbin{k-1}{j-1}q^{k-j}+\qbin{k-1}{j},
$$
it follows that
$$\sum_{k=j}^n\qbin{k-1}{j-1}q^k=
q^j\sum_{k=j}^n\qbin{k-1}{j-1}q^{k-j}
=q^j\sum_{k=j}^n\left(\qbin{k}{j}-\qbin{k-1}{j}\right)=
\qbin{n}{j}q^j.$$
As regards \eqref{C2}, since
$$
\qbin{n}{k}=\qbin{n-1}{k-1}+q^k\qbin{n-1}{k},
$$
we have
\begin{align*}
\sum_{k=j}^n \qbin{n}{k}(-1)^{k}q^{\binom{k}{2}}&=
\sum_{k=j}^n\left(\qbin{n-1}{k-1}+q^k\qbin{n-1}{k}\right)(-1)^{k}q^{\binom{k}{2}}\\
&=\sum_{k=j}^n\left(\qbin{n-1}{k-1}(-1)^{k}q^{\binom{k}{2}}-
\qbin{n-1}{k}(-1)^{k+1}q^{\binom{k+1}{2}}\right)\\
&=\qbin{n-1}{j-1}(-1)^{j}q^{\binom{j}{2}}.
\end{align*}
\end{proof}

\noindent The following lemma provides some simple properties of the $q$-binomial transform of
$\{a_n\}_{n\geq 1}$ given by
\begin{equation}\label{BT}
b_n=\sum_{k=1}^n \qbin{n}{k}(-1)^{k}q^{\binom{k}{2}}a_k,
\end{equation}
and its inverse
\begin{equation}\label{IBT}
a_n=\sum_{k=1}^n \qbin{n}{k}(-1)^{k}q^{\binom{n-k}{2}}b_k.
\end{equation}

\begin{lemma} Let $\{a_n\}_{n\geq 1}$ and $\{b_n\}_{n\geq 1}$ be two sequences which satisfy \eqref{BT}.
Then for any positive integer $r$,
\begin{equation}\label{TC}
\sum_{k=1}^n \qbin{n}{k}\frac{(-1)^{k}
q^{\binom{k}{2}+rk}}{\qq{k}^{r}}\cdot a_{k}
=\sum_{1\leq k_1\leq k_2\leq\cdots\leq k_r\le n}
\prod_{i=1}^r\frac{q^{k_i}}{\qq{k_i}}\cdot b_{k_1},
\end{equation}
and
\begin{equation}\label{TCD}
\sum_{k=1}^n \qbin{n}{k}(-1)^{k}q^{\binom{k}{2}-nk}\!\!\!\!
\sum_{1\leq k_1\leq k_2\leq\cdots\leq k_r=k}
\prod_{i=1}^r\frac{q^{k_i}}{\qq{k_i}}\cdot b_{k_1}
=\sum_{k=1}^n
\frac{q^{(r-1)k}}{\qq{k}^r}\cdot a_{k}.
\end{equation}
\end{lemma}
\begin{proof}
For $i\geq 0$, let
$$b_n(i)=\sum_{k=1}^n \qbin{n}{k}(-1)^{k}q^{\binom{k}{2}}a_k(i)\\
\quad\mbox{with}\quad a_k(i)=\frac{q^{ik}a_k}{\qq{k}^i}.$$
Hence, by \eqref{C1},
\begin{align*}
\sum_{k=1}^n \frac{q^k b_k(i)}{\qq{k}}&=
\sum_{k=1}^n \frac{q^k}{\qq{k}}
\sum_{j=1}^k \qbin{k}{j}(-1)^{j}q^{\binom{j}{2}}a_j(i)
=\sum_{j=1}^n (-1)^{j}q^{\binom{j}{2}}a_j(i)
\sum_{k=j}^n \qbin{k}{j} \frac{q^k}{\qq{k}}\\
&=
\sum_{j=1}^n
\frac{(-1)^{j}q^{\binom{j}{2}}a_j(i)}{\qq{j}}
\sum_{k=j}^n \qbin{k-1}{j-1}q^k
=
\sum_{j=1}^n \qbin{n}{j}\frac{(-1)^{j}q^{\binom{j}{2}+j}a_j(i)}{\qq{j}}\\
&=\sum_{j=1}^n \qbin{n}{j} (-1)^{j}q^{\binom{j}{2}}a_j(i+1)=b_n(i+1).
\end{align*}
Therefore
$$
b_n(r)=
\sum_{k_r=1}^n \frac{q^{k_r} b_{k_r}(r-1)}{\qq{k_r}}
=\sum_{k_r=1}^n \frac{q^{k_r}}{\qq{k_r}}
\sum_{k_{r-1}=1}^{k_r} \frac{q^{k_{r-1}} b_{k_{r-1}}(r-2)}{\qq{k_{r-1}}}
=\!\!\!\!\!\sum_{1\leq k_1\leq k_2\leq\cdots\leq k_r\le n}
\prod_{i=1}^r\frac{q^{k_i}}{\qq{k_i}}\cdot b_{k_1}
$$
and \eqref{TC} is proved.

\noindent Finally, by Lemma 1 in \cite{Pr00} (or by the use of the inverse relation \eqref{IBT}),
it follows that
$$\sum_{k=1}^n \qbin{n}{k}(-1)^{k}q^{\binom{k}{2}-nk}(b_k(r)-b_{k-1}(r))=\sum_{k=1}^n q^{-k} a_k(r),$$
which yields \eqref{TCD}.
\end{proof}

\subsection{Proof of identities \eqref{DE} and \eqref{HE}}
We begin with the following result.

\begin{lemma}
Let ${\bf s}\in ({\mathbb N}^{*})^l$ and
let $r$ be a positive integer. Then
\begin{align} \label{I2}
\sum_{k=1}^n \qbin{n}{k}(-1)^{k}q^{\binom{k}{2}}
H^q_{k-1}({\bf s},r)&=
\frac{q^n}{\qq{n}}\sum_{k=1}^n \qbin{n}{k}\frac{(-1)^{k}q^{\binom{k}{2}+(r-1)k}H^q_{k-1}({\bf s})}{\qq{k}^{r-1}}\nonumber\\
&\qquad\qquad\qquad
-\sum_{k=1}^n \qbin{n}{k}
\frac{(-1)^{k}q^{\binom{k}{2}+rk}H^q_{k-1}({\bf s})}{\qq{k}^{r}}.
\end{align}
\end{lemma}
\begin{proof} By using \eqref{C2}, and the recursive relation
\begin{align*}
H^q_{n}({\bf s},r)&=
\sum_{k=1}^{n}
\frac{q^{(r-1)k}H^q_{k-1}({\bf s})}{\qq{k}^r},
\end{align*}
we have
\begin{align*}
\sum_{k=1}^n \qbin{n}{k}(-1)^{k}q^{\binom{k}{2}}
H^q_{k-1}({\bf s},r)&=
\sum_{k=1}^n \qbin{n}{k}(-1)^{k}q^{\binom{k}{2}}
\sum_{j=1}^{k-1}
\frac{q^{(r-1)j}H^q_{j-1}({\bf s})}{\qq{j}^r}\nonumber\\
&=
\sum_{j=1}^n \frac{q^{(r-1)j}H^q_{j-1}({\bf s})}{\qq{j}^r}
\sum_{k=j+1}^n\qbin{n}{k}(-1)^{k}q^{\binom{k}{2}}\nonumber\\
&=
\sum_{j=1}^n \frac{q^{(r-1)j}H^q_{j-1}({\bf s})}{\qq{j}^r}
\left(\qbin{n-1}{j}(-1)^{j+1}q^{\binom{j}{2}+j}\right)\nonumber\\
&=\sum_{j=1}^n \frac{(-q^j)\qq{n-j}}{\qq{n}\qq{j}}\qbin{n}{j}
\frac{(-1)^{j}q^{\binom{j}{2}+(r-1)j}H^q_{j-1}({\bf s})}{\qq{j}^{r-1}},
\end{align*}
and the proof is complete as soon as we note that
$$
\frac{q^n}{\qq{n}}-\frac{q^j}{\qq{j}}
=\frac{q^n(1-q^j)-q^j(1-q^n)}{\qq{n}\qq{j}(1-q)}
=\frac{(-q^j)\qq{n-j}}{\qq{n}\qq{j}}.
$$
\end{proof}

\noindent In order to reduce the use of symbols indicating multiple nested sums,
we introduce
$$
T^q_{n}({\bf s}):=\sum_{}
\prod_{i=1}^{w({\bf s})} \frac{q^{j_i}}{\qq{j_i}}
$$
where the sum is intended to be taken over all integers
satisfying the conditions:
$$\mbox{
$1\leq j_i\leq n$, $j_i<j_{i+1}$ for $i\in\{s_1,s_1+s_2,\dots,s_1+s_2+\dots+s_{l-1}\}$ and
$j_i\leq j_{i+1}$ otherwise.}$$
Note that the following recursive relations hold
\begin{align}\label{TT}
T^q_{n}({\bf s},1)=\sum_{k=1}^n \frac{q^k T^q_{k-1}({\bf s})}{\qq{k}}
\quad\mbox{and}\quad
T^q_{n}({\bf s},r)=\sum_{k=1}^n \frac{q^kT^q_{k}({\bf s},r-1)}{\qq{k}}\quad\mbox{for $r>1$}.
\end{align}

\noindent In the next theorem, the identities \eqref{I4}
and \eqref{I5} are equivalent to \eqref{DE} and \eqref{HE} respectively.

\begin{theorem}
Let ${\bf s}\in ({\mathbb N}^{*})^l$ and
let $r$ be a positive integer. Then
\begin{align}\label{I3}
&\sum_{k=1}^n \qbin{n}{k}(-1)^{k}
q^{\binom{k}{2}}H^q_{k-1}({\bf s})=
(-1)^{l+1} T^q_{n-1}({\bf s}),\\
&\label{I4}
\sum_{k=1}^n \qbin{n}{k}\frac{(-1)^{k}
q^{\binom{k}{2}+rk}H^q_{k-1}({\bf s})}{\qq{k}^{r}}=
(-1)^{l+1} T^q_{n}({\bf s},r),\\
&\label{I5}
\sum_{k=1}^n \qbin{n}{k}(-1)^{k}q^{\binom{k}{2}-nk}\!\!\!\!
\sum_{1\leq k_1\leq k_2\leq\cdots\leq k_r=k}
\prod_{i=1}^r\frac{q^{k_i}}{\qq{k_i}}\cdot T^q_{k_1-1}({\bf s})
=(-1)^{l+1} H^q_{n}({\bf s},r).
\end{align}
\end{theorem}
\begin{proof}
We proceed by induction on the length $l$.

\noindent The base case $l=0$ of \eqref{I3}
is true because by \eqref{C2},
$$\sum_{k=1}^n \qbin{n}{k}(-1)^{k} q^{\binom{k}{2}}=-1.$$
The base case $l=0$ of \eqref{I4} follows from \eqref{TC} and the previous equation, and reads
$$\sum_{k=1}^n \qbin{n}{k}\frac{(-1)^{k}
q^{\binom{k}{2}+rk}}{\qq{k}^{r} }=-T^q_n(r).$$
In a similar way, for $l=0$, \eqref{TCD} implies \eqref{I5}.

\noindent Now assume that $l>0$ and let ${\bf s}=({\bf t},s)$. By \eqref{I2}, we have
\begin{align*}
\sum_{k=1}^n \qbin{n}{k}(-1)^{k}
q^{\binom{k}{2}}H^q_{k-1}({\bf t}, s)
&=\frac{q^n}{\qq{n}}\sum_{k=1}^n \qbin{n}{k}\frac{(-1)^{k}q^{\binom{k}{2}+(s-1)k}H^q_{k-1}({\bf t})}{\qq{k}^{s-1}}\nonumber\\
&\qquad\qquad\qquad
-\sum_{k=1}^n \qbin{n}{k}
\frac{(-1)^{k}q^{\binom{k}{2}+sk}H^q_{k-1}({\bf t})}{\qq{k}^{s}}.
\end{align*}
If $s=1$ then by the inductive step and \eqref{TT}, we get
\begin{align*}
\sum_{k=1}^n \qbin{n}{k}(-1)^{k}
q^{\binom{k}{2}}H^q_{k-1}({\bf t},1)
&=\frac{q^n }{\qq{n}}(-1)^{l({\bf t})+1}T^q_{n-1}({\bf t})
-(-1)^{l({\bf t})+1}T^q_{n}({\bf t},1)\\
&=(-1)^{l+1}\left(
T^q_{n}({\bf t},1)-\frac{q^nT^q_{n-1}({\bf t})}{\qq{n}}
\right)
=(-1)^{l+1} T^q_{n-1}({\bf s}).
\end{align*}
On the other hand, if $s>1$ then by the inductive step and \eqref{TT}, we obtain
\begin{align*}
\sum_{k=1}^n \qbin{n}{k}(-1)^{k}
q^{\binom{k}{2}}H^q_{k-1}({\bf t},s)
&=\frac{q^n}{\qq{n}}(-1)^{l({\bf t})+1}T^q_{n}({\bf t},s-1)
-(-1)^{l({\bf t})+1}T^q_{n}({\bf t},s)\\
&=(-1)^{l+1}\left(
T^q_{n}({\bf t},s)-\frac{q^nT^q_{n}({\bf t},s-1)}{\qq{n}}
\right)
=(-1)^{l+1} T^q_{n-1}({\bf s}),
\end{align*}
and therefore \eqref{I3} holds.

\noindent Putting
$a_n=H_{n-1}^q({\bf s})$ and
$b_n=(-1)^{l+1} T^q_{n-1}({\bf s})$
in \eqref{TC} and using \eqref{I3}, we immediately get
\begin{align*}
\sum_{k=1}^n \qbin{n}{k}\frac{(-1)^{k}
q^{\binom{k}{2}+kr}H^q_{k-1}({\bf s})}{\qq{k}^{r}}
&=\sum_{1\leq k_1\leq k_2\leq\cdots\leq k_r\le n}
\prod_{j=1}^r\frac{q^{k_j}}{\qq{k_j}}\cdot (-1)^{l+1} T^q_{k_1-1}({\bf s})\\
&=(-1)^{l+1} T^q_{n}({\bf s},r)
\end{align*}
and the proof of \eqref{I4} is complete.
Similarly, by \eqref{I3} and \eqref{TCD}, we obtain \eqref{I5}.
\end{proof}

\vspace{0.1cm}

\centerline{{\bf Acknowledgment.}}

\vspace{0.2cm}

 Kh.~Hessami Pilehrood and T.~Hessami Pilehrood acknowledge support from the Fields Institute Research Immersion Fellowships.

config: arxiv


TITLE: Is $\sum_{n=1}^\infty {1\over 3^{\sqrt{n}}}$ convergent?
QUESTION [2 upvotes]: Is   $\sum_{n=1}^\infty {1\over 3^{\sqrt{n}}}$ convergent ?
I use it to compare with $1/n^2$, and then I used LHôpitals rule multiple times. Finally , I can solve it. However,I think we have other methods!  somebody can help me?

REPLY [1 votes]: Note that if $$k\le\sqrt{n}< k+1$$ then, since the function $x\mapsto 3^x$ is increasing, we have $$3^k\le3^{\sqrt{n}}<3^{k+1},\;\;\text{and}$$ $$\frac{1}{3^k}\ge\frac{1}{3^{\sqrt{n}}}>\frac{1}{3^{k+1}}$$for every $n\in [k^2,(k+1)^2[$ then
\begin{align}
\sum_{n=1}^{N^2-1}{\frac{1}{3^{\sqrt{n}}}} & = \frac{1}{3^1}+\frac{1}{3^{\sqrt{2}}}+\frac{1}{3^{\sqrt{3}}}+\frac{1}{3^2}+\ldots+\underbrace{\frac{1}{3^{\sqrt{(N-1)^2}}}+\frac{1}{3^{\sqrt{(N-1)^2+1}}}+\ldots+\frac{1}{3^{\sqrt{N^2-1}}}}_{2N-1\,\text{terms}}\\
& \le \frac{1}{3^1}+\frac{1}{3^1}+\frac{1}{3^1}+\frac{1}{3^2}+\ldots+\underbrace{\frac{1}{3^{N-1}}+\ldots+\frac{1}{3^{N-1}}}_{2N-1\;\text{terms}}\\
& = \sum_{k=1}^{N-1}{\frac{2k+1}{3^k}}
\end{align}
Then, for comparison test, $\sum_{n=1}^{\infty}{\frac{1}{3^{\sqrt{n}}}}$ converges if $\sum_{k=1}^{\infty}{\frac{2k+1}{3^k}}$ does, and this happens due the ratio test.


TITLE: Point within triangle maximally distant from four points
QUESTION [1 upvotes]: Suppose I have a right triangle $ABC$ where the midpoint of the hypotenuse $AB$ is $M$, and the side $AC$ is longer than $BC$. How to determine the point $X$ within the triangle such that the distance from $X$ to any of $A$, $B$, $C$, $M$ is as great as possible? That is, the minimum of $XA$, $XB$, $XC$, $XM$ is maximal? I'm speculating it could be the centroid of $AMC$, but I'm really not sure. Thanks.

REPLY [1 votes]: First let's solve a simpler problem. Given an isosceles triangle $ABC$ with vertex $C$, the point $X$ inside the triangle that maximizes the minimal distance from the three vertices is:

the circumcenter, if it lies within the triangle (that is, if the triangle is acute);
the intersection of $AB$ with the perpendicular bisector of $BC$, if the triangle is obtuse (that is, if angle $ACB$ is obtuse). We can also use the perpendicular bisector of $AC$ instead, by symmetry.

(For right isosceles triangles, take the limit in either case - we get the center of the hypotenuse $AB$.)
Now for the problem given: the point $X$ might be inside the acute isosceles triangle $BCM$, or it might be inside the obtuse isosceles triangle $ACM$.

In the first case, any point inside $BCM$ is farther from $A$ than it is from $M$. So the best we can do is take $X$ to be the circumcenter of $BCM$.
In the second case, let $N$ be the midpoint of $AC$. Any point in triangle $ANM$ has an analogous point in triangle $CNM$ (its reflection in the line $NM$0; but the point in $ANM$ is farther from $B$ than the analogous point in $CNM$. Therefore the best we can do is to take $X$ to be the intersection of $AN$ with the perpendicular bisector of $AM$.

These are the only two choices for $X$, and I believe one can show that the latter choice is always optimal.


TITLE: Vector space structure on $(-1,1) \subset \mathbb{R}$ (or: möbius strip as vector bundle)
QUESTION [1 upvotes]: I'm first putting the question into it's context, so probably you can see if i'm asking the wrong question to get what i want.
The Task is to show that the Möbius (Moebius) strip is a Vector bundle with base space $S^1$ and typical fiber $\mathbb{R}$.
The strip is defined as follows: Let $R$ be the open square $R = (-\pi, 2 \pi) \times (-1,1) \subset \mathbb{R}^2$. On $R$ we define the equivalence relation $\sim$ to be: $(x_1,y_1) \sim (x_2,y_2)$ if $\{x_1=x_2, y_1=y_2\}$ or $\{|x_1-x_2|=2\pi,y_1=-y_2\}$. The Möbius strip is now defined as $E = R / \sim$.
As projection $\pi: E \to S^1$ i chose simply the projection on the first component. The fiber over every point of $S^1$ is therefore the interval $(-1,1)$ which needs to be isomorphic to $\mathbb{R}$ as vectorspace.
My question is now: What is the vector space structure on the fiber $(-1,1)$?
Or:
How can i turn (-1,1) into a vector space?
My first try was to define vector addition as $x \oplus y  = \tanh\left( \tanh^{-1}(x) + \tanh^{-1}(y)\right)$ but then i couldn't find a sensible scalar multiplication with $\mathbb{R}$.
Any help and comments are appreciated.

REPLY [2 votes]: Choose your favorite homeomorphism $f\colon\mathbb R \to (-1, 1)$, for example $\tanh$, and define
$$v + w = f(f^{-1}(v) + f^{-1}(w))$$
$$a\cdot v = f(a\cdot f^{-1}(v))$$
All you're doing here is transplanting the vector space structure of $\mathbb R$ onto $(-1, 1)$ so there isn't really anything to check as far as it being a well defined vector space.
You will, on the other hand, have to check that your transition maps are linear.  This should hold as long as you don't vary the homeomorphism $f$ from fiber to fiber.