TITLE: Circle homeomorphism
QUESTION [0 upvotes]: Simple problem, searchig for a solution.
Let $A$ be a proper, closed and path connected subset of the $S^1$, then $A \cong [0,1]$.
So currently I am trying too use the fact that $f:[0,1] \rightarrow S^1$ by $f(t)=(cos(2\pi t),sin(2\pi t) )$ is a continuous surjective map. However, this leads nowhere and I have become stuck.
I am looking for an elementary proof.
Thanks

REPLY [1 votes]: This only holds if $A$ is a proper and non-singleton subset of $S^1$. Note that $A$ being closed in $S^1$ is compact. Pick $p \in S^1\setminus A$ and note that $S^1 \setminus \{p\} \simeq \Bbb R$ (stereographic projection) and a (path-)connected compact subset of $\Bbb R$ is a singleton or of the form $[a,b], a < b$ and thus homeomorphic to $[0,1]$.


TITLE: Parallelogram law functional equation: $ f ( x + y ) + f ( x - y ) = 2 \big( f ( x ) + f ( y ) \big) $
QUESTION [8 upvotes]: I'm trying to solve the following functional equation, called the parallelogram law:
$$ f ( x + y ) + f ( x - y ) = 2 \big( f ( x ) + f ( y ) \big) $$
It can be easily proven that we have the family of solutions $ f ( x ) = c x ^ 2 $. Is there other type of solution apart from this?
I've proven that this are all the solutions in the case $ f : \mathbb Q \to \mathbb Q $ this is the only solution. To prove this, first let $ x = y = 0 $. We then get
$$ 2 f ( 0 ) = 4 f ( 0 ) \implies f ( 0 ) = 0 $$
Now will prove by induction that $ f ( n x ) = n ^ 2 f ( x ) $, $ \forall n \in \mathbb N $. Statement hold for $ n = 1 $. Now, supposing the induction hypothesis holds, let $ x = n x $ and $ y = x $. Then
$$ f \big( ( n + 1 ) x \big) + f \big( ( n - 1 ) x \big) = 2 \big( f ( n x ) + f ( x ) \big) $$
$$ f \big( ( n + 1 ) x \big) + ( n - 1 ) ^ 2 f ( x ) = 2 \left( n ^ 2 + 1 \right) f ( x ) $$
$$ f \big( ( n + 1 ) x \big) = ( n + 1 ) ^ 2 f ( x ) $$
Now letting $ x = 0 $, we get
$$ f ( y ) + f ( - y ) = 2 f ( y ) \implies f ( - y ) = f ( y ) $$
$$ f ( n x ) = n ^ 2 f ( x ) \text , \ \forall n \in \mathbb Z $$
Also notice that
$$ f ( x ) = f \left( n \frac x n \right) = n ^ 2 f \left( \frac x n \right) $$
$$ f \left( \frac x n \right) = \frac { f ( x ) } { n ^ 2 } \text , \ \forall n \in \mathbb Z \setminus \{ 0 \} $$
We can conclude that $ f ( r x ) = r ^ 2 f ( x ) \forall r \in \mathbb Q $.
The above reasoning is valid also when $ f : \mathbb R \to \mathbb R $. If we assume continuity  in some interval or monotonicity in some interval $ [ a , b ] $, the solution is of form of a quadratic polynomial. If the solution is not unique, what are the minimal assumptions one has to make to ensure the statement hold (injectivity, positivity, Riemann integrability, measurability, not being dense in a region of $\mathbb R ^ 2 $)?

REPLY [7 votes]: $ \def \Q {\mathbb Q}
\def \R {\mathbb R}
\def \Rp {\mathbb R _ +}
\def \B {\tilde B}
\DeclareMathOperator \sgn {sgn} $
Assume that $ B : \R ^ 2 \to \R $ is biadditive (and hence bilinear over $ \Q $). If you define $ f : \R \to \R $ with $ f ( x ) = B ( x , x ) $ for all $ x \in \R $, then for any $ x , y \in \R $ you can verify that
$$ \begin {align*}
f ( x + y ) + f ( x - y ) & = B ( x + y , x + y ) + B ( x - y , x - y ) \\
& = B ( x , x + y ) + B ( y , x + y ) + B ( x , x - y ) - B ( y , x - y ) \\
& = { B ( x , x ) + B ( x , y ) + B ( y , x ) + B ( y , y ) \\
+ B ( x , x ) - B ( x , y ) - B ( y , x ) + B ( y , y ) } \\
& = 2 B ( x , x ) + 2 B ( y , y ) \\
& = 2 f ( x ) + 2 f ( y ) \text .
\end{align*} $$
Conversely, you can show that for any $ f : \R \to \R $ satisfying
$$ f ( x + y ) + f ( x - y ) = 2 f ( x ) + 2 f ( y ) \tag 0 \label 0 $$
for all $ x , y \in \R $ (also called a quadratic function satisfying the quadratic functional equation), there is a symmetric biadditive $ B : \R ^ 2 \to \R $ such that $ f ( x ) = B ( x , x ) $ for all $ x \in \R $. To see this, define $ B : \R ^ 2 \to \R $ with
$$ B ( x , y ) = \frac { f ( x + y ) - f ( x - y ) } 4 \tag 1 \label 1 $$
for any $ x , y \in \R $. You've managed to show that \eqref{0} implies
$$ f ( r x ) = r ^ 2 f ( x ) \tag 2 \label 2 $$
for all $ r \in \Q $ and all $ x \in \R $. For $ r = 0 $, \eqref{2} gives $ f ( 0 ) = 0 $, and for $ r = 2 $, it gives $ f ( 2 x ) = 4 f ( x ) $ for all $ x \in \R $. These, together with letting $ y = x $ in \eqref{1}, show that $ f ( x ) = B ( x , x ) $ for all $ x \in \R $. For $ r = - 1 $, \eqref{2} shows that $ f $ is even, which by swapping $ x $ and $ y $ in \eqref{1} shows that $ B $ is symmetric. Thus, it's sufficient to prove additivity of $ B $ in the first variable, and biadditivity will follow. By evenness of $ f $ and substituting $ - x $ for $ x $ in \eqref{1}, we can see that $ B $ is odd in the first variable. By \eqref{0}, for any $ x , y , x \in \R $ we have
$$ \begin {align*}
4 B ( x + y , z ) + 4 B ( x - y , z ) & = { f ( x + y + z ) + f ( x - y + z ) \\ - f ( x + y - z ) - f ( x - y - z ) } \\
& = 2 f ( x + z ) + 2 f ( y ) - 2 f ( x - z ) - 2 f ( y ) \\
& = 2 f ( x + z ) - 2 f ( x - z ) \\
& = 8 B ( x , z ) \text ,
\end {align*} $$
or equivalently
$$ B ( x + y , z ) + B ( x - y , z ) = 2 B ( x , z ) \text . $$
Substituting $ - y $ for $ x $ and $ x $ for $ y $, we get
$$ B ( - y + x , z ) + B ( - y - x , z ) = 2 B ( - y , z ) \text . $$
for all $ x , y , x \in \R $. Subtracting the last two equations, we have
$$ B ( x + y , z ) - B ( - x - y , z ) = 2 B ( x , z ) - 2 B ( - y , z ) $$
for all $ x , y , x \in \R $, which by oddness of $ B $ in the first variable, proves additivity of $ B $ in the first variable.

Now, note that given any Hamel basis $ ( e _ i ) _ { i \in I } $ of $ \R $ over $ \Q $, and any "matrix" $ ( \alpha _ { i j } ) _ { i , j \in I } $ of real numbers, there is a biadditive $ B : \R ^ 2 \to \R $ such that for any $ x , y \in \R $,
$$ B ( x , y ) = \sum _ { k = 1 } ^ n \sum _ { l = 1 } ^ m s _ { i _ k } \alpha _ { { i _ k } { j _ l } } t _ { j _ l } \text , $$
where $ x = \sum _ { k = 1 } ^ n s _ { i _ k } e _ { i _ k } $ and $ y = \sum _ { l = 1 } ^ m s _ { j _ l } e _ { j _ l } $. As you've noted, for any additive $ A , C : \R \to \R $, defining $ B : \R ^ 2 \to \R $ with $ B ( x , y ) = A ( x ) C ( y ) $ gives a biadditive map. But those are not the only possibilities; for example, you can add finitely many different maps of this form, and end up with another biadditive map. Even those don't cover all the possibilities; you can find a relation between the "entries" of the "matrix" of $ A ( x ) C ( y ) $ and the values of $ A $ and $ C $ at the basis points, and "construct" a "matrix" that is not a finite sum of "matrices" of that form.
You can see that the only regular biadditive functions are those of the form $ B ( x , y ) = k x y $, and consequently the only regular quadratic functions are those of the form $ f ( x ) = k x ^ 2 $, for some constant $ k \in \R $. Similar to the well-known case of additive functions, regularity here can have order theoretical, topological or measure-theoretical meanings. Any Lebesgue measurable biadditive/quadratic function must be of the mentioned form. Also, the graph of any biadditive/quadratic function not of the above form is dense. Therefore, any property implying that the graph can't be dense, will force the function to be of the mentioned form. Such properties include continuity at a point, being bounded above or below on a nontrivial compact subset of the domain, and others, much similar to the familiar ones for the case of Cauchy's functional equation. In the following, I provide proofs for these facts. I only give the proofs in terms of biadditive functions, and it should be straightforward to conclude the corresponding result for quadratic functions, from what we've already discussed.

Measurability:
Let $ B : \R ^ 2 \to \R $ be biadditive and Lebesgue measurable. For any $ y \in \R $, define $ A ^ y : \R \to \R $ with $ A ^ y ( x ) = B ( x , y ) $ for all $ x \in \R $. $ A ^ y $ is additive for every $ y \in \R $, and by Tonelli's theorem, $ A ^ y $ is Lebesgue measurable for almost every $ y \in \R $. Similarly, for any $ x \in \R $, we can define $ A _ x : \R \to \R $ with $ A _ x ( y ) = B ( x , y ) $, and $ A _ x $ will be additive for every $ x \in \R $ and Lebesgue measurable for almost every $ x \in \R $. As any Lebesgue measurable additive function on $ \R $ must be linear, there is a full subset $ E $ of $ \R $ and functions $ g , h : E \to \R $ such that $ B ( x , y ) = g ( x ) y = x h ( y ) $ for all $ x , y \in E $. Fixing $ x _ * \in E \setminus \{ 0 \} $, for any $ x \in E \setminus \{ 0 \} $ we have
$$ \frac { g ( x ) } x = \frac { h ( x _ * ) } { x _ * } = \frac { g ( x _ * ) } { x _ * } = \frac { h ( x ) } x \text . $$
Therefore, setting $ k = \frac { g ( x _ * ) } { x _ * } $, we have $ B ( x , y ) = k x y $ for all $ x , y \in E \setminus \{ 0 \} $. As $ E \setminus \{ 0 \} $ is a full subset of $ \R $, for any $ x ,y \in \R $, there are $ x _ 0 , x _ 1 , y _ 0 , y _ 1 \in E \setminus \{ 0 \} $ such that $ x = x _ 0 + x _ 1 $ and $ y = y _ 0 + y _ 1 $, for which we have
$$ \begin {align*}
B ( x , y ) & = B ( x _ 0 + x _ 1 , y _ 0 + y _ 1 ) \\
& = B ( x _ 0 , y _ 0 ) + B ( x _ 0 , y _ 1 ) + B ( x _ 1 , y _ 0 ) + B ( x _ 1 , y _ 1 ) \\
& = k x _ 0 y _ 0 + k x _ 0 y _ 1 + k x _ 1 y _ 0 + k x _ 1 y _ 1 \\
& = k ( x _ 0 + y _ 0 ) ( x _ 1 + y _ 1 ) \\
& = k x y \text .
\end {align*} $$

Denseness of Graphs:
Let $ B : \R ^ 2 \to \R $ be biadditive and $ k = B ( 1 , 1 ) $. Given that $ B ( x _ * , y _ * ) \ne k x _ * y _ * $ for some $ x _ * , y _ * \in \R $, you can show that the graph of $ B $ is a dense subset of $ \R ^ 3 $. It's sufficient to prove this for the case $ k = 0 $; that's because defining $ \B : \R ^ 2 \to \R $ with $ \B ( x , y ) = B ( x , y ) - k x y $ for all $ x , y \in \R $, $ \B $ is biadditive, we have $ \B ( 1 , 1 ) = 0 $, and since $ ( x , y ) \mapsto k x y $ is continuous, denseness of the graph of $ \B $ in $ \R ^ 3 $ implies that of $ B $. Again, it's sufficient to prove that the graph of $ B $ is dense in the upper half-space $ \R ^ 2 \times \Rp $; that's because denseness of the graph of $ B $ in the lower half-space follows from denseness of the graph of the biadditive function $ - B $ in the upper half-space, and the union of the upper and lower half-spaces is dense in $ \R ^ 3 $. Finally, letting $ z _ * = B ( x _ * , y _ * ) $, it's sufficient to consider the case $ z _ * > k x _ * y _ * $, since if $ z _ * < k x _ * y _ * $, we have $ B ( x _ * , - y _ * ) = - z _ * > - k x _ * y _ * $, and we can replace $ y _ * $ with $ - y _ * $. So, from now on, without loss of generality, we assume that $ k = 0 $ and $ z _ * > 0 $, and only prove density of the graph of $ B $ in the upper half-space.
If $ B ( x _ * , 1 ) \ne 0 $, then defining $ A : \R \to \R $ with $ A ( x ) = B ( x , 1 ) $ for all $ x \in \R $, $ A $ will be an additive function with $ A ( 1 ) = 0 $ and $ A ( x _ * ) \ne A ( 1 ) x _ * $. Therefore, the graph of $ A $ must be dense in $ \R ^ 2 $. Hence, for any $ p \in \Q $, the graph of $ B | _ { \R \times \{ p \} } $ is dense in $ \R \times \{ p \} \times \R $. As $ \Q $ is dense in $ \R $, it follows that the graph of $ B $ is dense in $ \R ^ 3 $. Similarly, the graph of $ B $ must be dense in $ \R ^ 3 $ in case $ B ( 1 , y _ * ) \ne 0 $. So, it remains to investigate the case $ B ( x _ * , 1 ) = B ( 1 , y _ * ) = 0 $. Consider a ball in $ \R ^ 3 $ centered at $ ( x , y , z ) $ with radius $ r \in \Rp $, where $ x , y , z \in \Q $ and $ z > 0 $. Choose $ a , b , c \in \Q $ such that $ c \ne 0 $, $ \left| \frac z { z _ * } - c ^ 2 \right| < \frac r { \sqrt 3 z _ * } $, $ | x _ * - a | < \frac r { \sqrt 3 | c | } $ and $ | y _ * - b | < \frac r { \sqrt 3 | c | } $. Define $ X , Y , Z \in \R $ with $ X = x + c ( x _ * - a ) $, $ Y = y + c ( y _ * - b ) $ and $ Z = B ( X , Y ) $. Then, by bilinearity of $ B $ over $ \Q $,
$$ \begin {align*}
Z & = B \big( x + c ( x _ * - a ) , y + c ( y _ * - b ) \big) \\
& = { B ( x , y ) + c \big( B ( x _ * , y ) - B ( a , y ) + B ( x , y _ * ) - B ( x , b ) \big) \\
+ c ^ 2 \big( B ( x _ * , y _ * ) - B ( a , y _  * ) - B ( x _ * , b ) + B ( a , b ) \big) } \\
& = c ^ 2 z _ * \text .
\end {align*} $$
This implies that $ | X - x | ^ 2 + | Y - y | ^ 2 + | Z - z | ^ 2 < r ^ 2 $, and therefore we've found a point on the graph of $ B $ in the intended ball. As every ball in the upper half-space contains a ball centered at a point with rational coordinates, the result follows.
Furthermore, letting $ \Delta = \left\{ ( x , x ) \big| x \in \R ^ 2 \right\} $ (the diagonal of $ \R ^ 2 $), you can prove that if $ B ( x _ * , x _ * ) \ne k x _ * ^ 2 $, then the graph of $ B | _ \Delta $ is dense in $ \Delta \times \R $. Without loss of generality, we make all the assumptions mentioned before, and it will be sufficient to find suitable $ a , c \in \Q $ like before (given $ x _ * \in \R $ and $ x , z \in \Q $ with $ z > 0 $) and define $ X $ and $ Z $ like before, and note that this time it's like we're given $ y _ * = x _ * $ and $ y = x $, and required to let $ b = a $ so that $ Y $ becomes equal to $ X $. Without loss of generality, we can also assume that $ B $ is symmetric; that's because defining $ \B : \R ^ 2 \to \R $ with $ \B ( x , y ) = \frac { B ( x , y ) + B ( y , x ) } 2 $ for all $ x , y \in \R $, we can see that $ \B $ agrees with $ B $ on the diagonal. In particular, this means that we can assume $ B ( x _ * , 1 ) = B ( 1 , x _ * ) $, and we denote this value by $ w _ * $. We will also assume that $ x \ne 0 $, as this won't affect the denseness results.
If $ w _ * = 0 $, then choosing $ a $ and $ c $ like above (and setting $ b = a $) does the job. In case $ w _ * \ne 0 $, first note that no matter how we choose $ a $ and $ c $, we have
$$ \begin {align*}
Z & = B \big( x + c ( x _ * - a ) , x + c ( x _ * - a ) \big) \\
& = { B ( x , x ) + c \big( B ( x _ * , x ) - B ( a , x ) + B ( x , x _ * ) - B ( x , a ) \big) \\
+ c ^ 2 \big( B ( x _ * , x _ * ) - B ( a , x _  * ) - B ( x _ * , a ) + B ( a , a ) \big) } \\
& = ( z _ * - 2 a w _ * ) c ^ 2 + 2 x w _ * c \\
& = 2 w _ * c ^ 2 ( x _ * - a ) + ( z _ * - 2 x _ * w _ * ) c ^ 2 + 2 x w _ * c \text .
\end {align*} $$
If $ z _ * - 2 x _ * w _ * = 0 $, then it's sufficient to choose $ c $ and $ a $ so that $ c \ne 0 $, $ \left| \frac z { 2 x w _ * } - c \right| < \frac r { 4 \sqrt 3 | x w _ * | } $ and $ | x _ * - a | < \min \left( \frac r { \sqrt 3 | c | } , \frac r { 4 \sqrt 3 | w_ * | c ^ 2 } \right) $, and we'll have $ 2 | X - x | ^ 2 + | Z - z | ^ 2 < r ^ 2 $. In case $  z _ * - 2 x _ * w _ * \ne 0 $, let
$$ d _ * = w _ * x ^ 2 + ( z _ * - 2 x _ * w _ * ) \left( z + \frac r { 2 \sqrt 3 } \right) \text , $$
and assume that $ d _ * > 0 $. It's then sufficient to choose $ c $ and $ a $ such that $ c \ne 0 $, $ \frac { x w _ * - \sqrt { d _ * } } { z _ * - 2 x _ * w _ * } < c < \frac { x w _ * + \sqrt { d _ * } } { z _ * - 2 x _ * w _ * } $ and $ | x _ * - a | < \min \left( \frac r { \sqrt 3 | c | } , \frac r { 4 \sqrt 3 | w_ * | c ^ 2 } \right) $. This will prove denseness of the graph of $ B | _ \Delta $ in a subregion of $ \Delta\times \R $ described with $ d _ * ( x , z ) > 0 $, when $ x $ and $ z $ are seen as real variables. Now, note that we can argue the same starting with the point $ q x _ * $ instead of $ x _ * $, where $ q \in \Q $ and $ q $ has the same sign as $ w _ * $. Then we will have $ B ( q x _ * , q x _ * ) = q ^ 2 z _ * $ and $ B ( q x _ * , 1 ) = q  w _ * $, and we will end up proving denseness of the graph of $ B | _ \Delta $ in the subregion of $ \Delta \times \R $ described with
$$ q w _ * x ^ 2 + q ^ 2 ( z _ * - 2 x _ * w _ * ) \left( z + \frac r { 2 \sqrt 3 } \right) > 0 \text , $$
or equivalently
$$ x ^ 2 + q \left( \frac { z _ * } { w _ * } - 2 x _ * \right) \left( z + \frac r { 2 \sqrt 3 } \right) > 0 \text . $$
As we can take $ q $ as small as we want, this means that we have the result for $ x ^ 2 > 0 $, or equivalently $ x \ne 0 $.


TITLE: Finite Series Calculation
QUESTION [1 upvotes]: I am a totally math newbie.  I would like to calculate the series below,
$1 + 1.05 + 1.1025 + 1.157625 + \ldots$
So every new element is the 5% more of the previous one.

REPLY [1 votes]: A number 5% larger than $x$ is $1.05x$, so the terms of your series is $1.05^n$. Such a series is called a geometric series, and the sum of a number of terms from such a series is given by the "formula":
$$
\frac{\text{last term not included}-\text{first term}}{\text{ratio}-1}
$$
or if we fill in your values:
$$
\frac{1.05^{29+1}-1}{1.05-1}
$$
(29 because the first term is $1.05^0$)
If we put that into a calculator (or computer) we get the value $66.438847503$.


TITLE: Asymptotic expansion as $N \rightarrow \infty$ of $\sum_{k=1}^{\left\lfloor{N/2}\right\rfloor} k \sum_{e \mid 2k}\frac{\Lambda \left({e}\right)}{e}$
QUESTION [1 upvotes]: This expression comes from the asymptotic expansion of
$$\sum_{k=1}^{\left\lfloor{N/2}\right\rfloor} \sum_{i=1}^{k} \tau \left({i \left({2\, k - i}\right)}\right)$$
From Adrian W. Dudek, "Note on the number of divisors of reducible quadratic polynomials", arXiv:1806.01404v1 [math.NT] 4 Jun 2018.  Here $\tau \left({x}\right)$ us the number of divisors of $x$ and $\left\{{x}\right\}$ is the fractional part of $x$.
This expansion is $$\sum_{n \le x} \tau \left({n \left({n + v}\right)}\right) = \frac{6}{{\pi}^{2}} x \left({{\log}^{2} \left({x}\right) + {A}_{1} \left({v}\right) \log \left({x}\right) + {A}_{2} \left({v}\right)}\right) + O \left({{x}^{2/3 + \epsilon}}\right)$$
where $v$ is a positive integer (in this case I have $v = - 2\, k$ which may be valid as long as there is no $\tau \left({0}\right)$ case) and $${A}_{1} = 4\, \gamma- 2 - 4 \frac{{\zeta}^{\prime} \left({2}\right)}{\zeta \left({2}\right)} - 2\sum_{e \mid v} \frac{{\Lambda} \left({e}\right)}{e}$$ and $${A}_{2} = \left({2\, \gamma - 1 - 2 \frac{{\zeta}^{\prime} \left({2}\right)}{\zeta \left({2}\right)}}\right)^{2} + 1 - 4\frac{{\zeta}^{\prime \prime} \left({2}\right)}{\zeta \left({2}\right)} + 4 \left({\frac{{\zeta}^{\prime} \left({2}\right)}{\zeta \left({2}\right)}}\right)^{2} - \left({4\, \gamma- 2 - 4 \frac{{\zeta}^{\prime} \left({2}\right)}{\zeta \left({2}\right)}}\right) \sum_{e \mid v} \frac{{\Lambda} \left({e}\right)}{e} + 2 \sum_{e \mid v} \frac{{\Lambda} \left({e}\right) \log \left({E}\right)}{e} + \sum_{e \mid v} \frac{{\Lambda}_{2} \left({e}\right)}{e}$$ where ${\Lambda}_{k} \left({n}\right) = \sum_{d \mid n} \mu \left({d}\right) \left({\log \left({\frac{n}{d}}\right)}\right)^{k}$ which is the generalized Von Mangoldt function and $\Lambda \left({n}\right) = {\Lambda}_{1} \left({n}\right)$.
In general I am looking for the asymptotic expansions as $N \rightarrow \infty$ of $$\sum_{k=1}^{\left\lfloor{N/2}\right\rfloor} k \sum_{e \mid 2k} \frac{\Lambda \left({e}\right)}{e}$$ $$\sum_{k=1}^{\left\lfloor{N/2}\right\rfloor} k \log \left({k}\right) \sum_{e \mid 2k} \frac{\Lambda \left({e}\right)}{e}$$ $$\sum_{k=1}^{\left\lfloor{N/2}\right\rfloor} k \sum_{e \mid 2k} \frac{\Lambda \left({e}\right) \log \left({k}\right)}{e}$$ and $$\sum_{k=1}^{\left\lfloor{N/2}\right\rfloor} k \sum_{e \mid 2k} \frac{{\Lambda}_{2} \left({e}\right)}{e}$$.

REPLY [3 votes]: $$\sum_n (2n)^{-s} 2n \sum_{2d+1|2n} \frac{\Lambda(d)}d= (\frac{-\zeta'(s)}{\zeta(s)}-\frac{2^{-s}\log 2}{1-2^{-s}}) 2^{1-s}\zeta(s-1)$$
Following the same Tauberian theorem as in the proof of the PNT we get
$$\sum_{2n\le x}2n \sum_{2d+1|2n}\frac{\Lambda(d)}d\sim Res((\frac{-\zeta'(s)}{\zeta(s)}-\frac{2^{-s}\log 2}{1-2^{-s}}) 2^{1-s}\zeta(s-1)\frac{x^s}{s},2)$$ $$\sim \frac{x^2}{4}(\frac{-\zeta'(2)}{\zeta(2)}-\frac{2^{-2}\log 2}{1-2^{-2}})$$
The error term is $O(\frac{x^2}{\log^k x})$, under the RH it can be improved to $O(x^{3/2+\epsilon})$.


TITLE: Confusion regarding stopping times
QUESTION [2 upvotes]: I am having a hard time with "stopping time" definitions, notation and reasoning behind it.
$\tau \in \{0,1,2,...;+\infty\}$

What formaly does $\{ \tau \leq n \}$ mean? Does it represent an event? Particularly in  $ \{ \tau \leq n \} = \{ w:\tau(w) \leq n \} \in \mathscr{F_n} $. What is $\tau(w)$ then?
Can you please explain meaning of the $
\{ \tau = n \} =
\{ \tau \leq n \} \setminus \{ \tau \leq n - 1 \} \in \mathscr{F_n}
$? I unfortunately have no intuitive understandring of this statement.
Can you please refer to clear construction examples of sigma algebra / filtration needed for stopping time to be measurable / adapted.
Can you please recommend good book/article with clear and intuituve guidance to grasp a motivation and understanding of stopping times?

REPLY [3 votes]: To start $\tau$ is a random variable. Therefore many authors use a capital $T$ to denote stopping times. So, everything you know about random raviables applies. Now,

Yes, $\{\tau\le n\}$ is an event. It is the event, that your process will stop before time $n$.
The event, that the stopping criterion will occur at point $n$ is the event, that the stopping criterion occured before $n$ but after $n-1$. The information is obtained only at point $n$ not earlier, hence $\mathcal F_n$.
The most common example, is an infinite sequence of coin tosses which induces a random walk. Formally, let $$P(X_i=1)=P(X_i=-1)=\frac12$$ for any $i\in\mathbb N$ and let $S_n=\sum_{i=1}^n X_i$ with $S_0=0$ (this is an arbitrary starting condition). You can define the stopping time $$\tau=\inf{\{n\in N: S_n\ge 5\}}$$ or in words, the first point in time that your random walk, equals or exceeds $5$. Now, e.g. $P(\tau=1)=0$, because there is no chance to reach $5$ starting from $0$ and adding at most $1$ at each step. But $P(\tau=5)>0$.
There are many, many, so hmm no, I cannot pick one.


TITLE: Finite morphisms from $\mathbb{P}_k^1$ to a curve?
QUESTION [3 upvotes]: I use the following definition for a curve over an algebraically closed field $k$ : a scheme $X$ over $k$ is a curve if it is integral of dimension $1$, and separated and of finite type over $k$.
Now, it is a well known fact of the theory of curves that any nonsingular curve $X$ can be embedded in $\mathbb{P}^3$. However, can we say anything about the existence of finite maps of the form $\mathbb{P}^1\to X$? If $f:\mathbb{P}^1\to X$ is finite, then can we conclude that it is an isomorphism and therefore that $X$ is rational?

REPLY [6 votes]: The existence of a finite morphism $\mathbb{P}^1 \to X$ implies, by pulling back to $\mathbb{P}^1$ the rational functions on $X$, the existence of a finite field extension $k(X) \subseteq k(\mathbb{P}^1)$.
Now $k(\mathbb{P}^1) \simeq k(t)$, the pure transcendental extension of degree $1$ of $k$, so $k(X)$ is an intermediate extension between $k$ and $k(t)$. Then, by Lüroth's theorem, there exists a rational function $\phi(t) \in k(t)$ such that $k(X) = k(\phi(t))$.
In other words, $k(X)$ is a simple transcendental extension of $k$, and this means that $X$ is birational, and hence isomorphic, to $\mathbb{P}^1$.


TITLE: Find the mistake.
QUESTION [0 upvotes]: We all know the product rule of differentiation
That, $(uv)'=u'v+v'u$
I tried proving the same and what I have done I'm showing you, you have to find the mistake in my proof (if you are interested )
$$
(g(x).h(x))'=f'(x)
$$
$$
={\frac{d}{dx}[f(x)]}
$$
$$
={\frac{d}{dx}[g(x).h(x)]}
$$
$$
=\lim_{\delta x\to 0}{\frac{g(x+\delta x).h(x+\delta x)-g(x).h(x)}{\delta x}}
$$
$$
=\lim_{\delta x\to 0}{\frac{g(x+\delta x).h(x+\delta x)}{\delta x}}-\lim_{\delta x\to 0}{\frac{g(x).h(x)}{\delta x}}
$$
$$
=\lim_{\delta x\to 0}{\frac{g(x+\delta x)}{\delta x}}.\lim_{\delta x\to 0}h(x+\delta x)-\lim_{\delta x\to 0}{\frac{g(x)}{\delta x}}.\lim_{\delta x\to 0}h(x)
$$
$$
=(\lim_{\delta x\to 0}{\frac{g(x+\delta x)}{\delta x}}).h(x)-(\lim_{\delta x\to 0}{\frac{g(x)}{\delta x}}).h(x)
$$
$$
=(\lim_{\delta x\to 0}{\frac{g(x+\delta x)}{\delta x}}-\lim_{\delta x\to 0}{\frac{g(x)}{\delta x}}).h(x)
$$
$$
=(\lim_{\delta x\to 0}{\frac{g(x+\delta x)-g(x)}{\delta x}}).h(x)
$$
$$
=g'(x).h(x)
$$
What I'm getting from my proof is definitely wrong but what I have used in my proof is all correct so where I'm doing the mistake.
Please help me!!!

REPLY [1 votes]: your problem is $$\lim_{\delta x\to 0}{\frac{g(x+\delta x).h(x+\delta x)}{\delta x}}=\lim_{\delta x\to 0}{\frac{g(x+\delta x)}{\delta x}}.\lim_{\delta x\to 0}h(x+\delta x)$$
and
$$\lim_{\delta x\to 0}{\frac{g(x).h(x)}{\delta x}}=\lim_{\delta x\to 0}{\frac{g(x)}{\delta x}}.\lim_{\delta x\to 0}h(x)$$
you can't seperate them by this way.


text:
\begin{document}
\title[Generalized inverses and polar decomposition]
{Generalized inverses and polar decomposition of unbounded
regular operators on Hilbert $C^*$-modules}
\author{M. Frank}
\address{Michael Frank, \newline Hochschule f\"{u}r Technik,
Wirtschaft und Kultur (HTWK) Leipzig,
Fachbereich IMN, Gustav-Freytag-Strasse 42A, D-04277 Leipzig,
Germany } \email{mfrank@imn.htwk-leipzig.de}
\author{K. Sharifi}
\address{Kamran Sharifi, \newline Department of Mathematics,
Shahrood University of Technology, P. O. Box 3619995161-316,
Shahrood, Iran} \email{sharifi.kamran@gmail.com and
sharifi@shahroodut.ac.ir}

\subjclass{Primary 46L08; Secondary 47L60, 46C05}
\keywords{Hilbert $C^*$-module, unbounded operators, polar
          decomposition, generalized inverses,
          $C^*$-algebras of compact operators}

\begin{abstract}
In this note we show that an unbounded regular operator $t$ on
Hilbert $C^*$-modules over an arbitrary $C^*$ algebra $
\mathcal{A}$ has polar decomposition if and only if the closures
of the ranges of $t$ and $|t|$ are orthogonally complemented, if
and only if the operators $t$ and $t^*$ have unbounded regular
generalized inverses. For a given $C^*$-algebra $ \mathcal{A}$ any
densely defined $\mathcal A$-linear closed operator $t$ between
Hilbert $C^*$-modules has polar decomposition, if and only if any
densely defined $\mathcal A$-linear closed operator $t$ between
Hilbert $C^*$-modules has generalized inverse, if and only if
$\mathcal A$ is a $C^*$-algebra of compact operators.
\end{abstract}
\maketitle

\section{Introduction.}
In the theory of $C^*$-algebras, an important role is played by
the spaces which are modules over a $C^*$-algebra and are
equipped with a structure which is like an inner product but which,
instead of being scalar-valued as in the case of Hilbert spaces,
takes its values in the $C^*$-algebra. Such modules
are called (pre-)Hilbert $C^*$-modules. Let us quickly recall the
definition of a Hilbert $C^*$-module.

A (left) {\it pre-Hilbert $C^*$-module} over a (not necessarily
unital) $C^*$-algebra $\mathcal{A}$ is a left $\mathcal{A}$-module
$E$ equipped with an $\mathcal{A}$-valued inner product $\langle
\cdot , \cdot \rangle : E \times E \to \mathcal{A}$,
which is $\mathcal A$-linear in the first variable and has the
properties:
$$ \langle x,y \rangle=\langle y,x \rangle ^{*},
   \ \ \ \langle x,x \rangle \geq 0 \ \ {\rm with} \
   {\rm equality} \ {\rm if} \ {\rm and} \ {\rm only} \
   {\rm if} \ x=0.
$$
We always suppose that the linear structures of $\mathcal A$
and $E$ are compatible.

A pre-Hilbert $\mathcal{A}$-module $E$ is called a {\it Hilbert $
\mathcal{A}$-module} if $E$ is a Banach space with respect to the
norm $\| x \|=\|\langle x,x\rangle \|_{\mathcal A}^{1/2}$. If $E$,
$F$ are two Hilbert $\mathcal{A}$-modules then the set of all
ordered pairs of elements $E \oplus F$ from $E$ and $F$ is a
Hilbert $\mathcal{A}$-module with respect to the $\mathcal
A$-valued inner product $\langle (x_{1},y_{1}),(x_{2},y_{2})
\rangle= \langle x_{1},x_{2}\rangle_{E}+\langle y_{1},y_{2}
\rangle _{F}$. It is called the {\it orthogonal sum of $E$ and
$F$}. A pre-Hilbert $\mathcal A$-module $E$ of a pre-Hilbert
$\mathcal A$-module $F$ is an orthogonal summand if $E \oplus
E^\bot = F$, where $E^\bot$ denotes the orthogonal complement of
$E$ in $F$. If $F$ is a Hilbert $ \mathcal{A}$-module and $F=E
\oplus E^\bot$ then $E$ and $E^\bot$ are necessarily  Hilbert $
\mathcal{A}$-submodules (cf. \cite{WEG}, Lemma 15.3.4). Some
interesting results about orthogonally complemented submodules can
be found in \cite{FR2}, \cite{FR3}, \cite{MAG}, \cite{SCH}. For
the basic theory of Hilbert $C^*$-modules we refer to the books
\cite{LAN}, \cite{M-T} and to some chapters of \cite{WEG}.

As a convention, throughout the present paper we assume
$\mathcal{A}$ to be an arbitrary $C^*$-algebra (i.e. not necessarily
unital). Since we deal with bounded and unbounded operators at the
same time we simply denote bounded operators by capital letters and
unbounded operators by lower case letters. We use the denotations
$Dom(.)$, $Ker(.)$ and $Ran(.)$ for domain, kernel and range of
operators, respectively.

Suppose $E,\ F$ are Hilbert $\mathcal{A}$-modules. We denote the
set of all bounded $\mathcal{A}$-linear maps $T: E \to F$ for which
there is a map $T^*: F \to E$  such that the equality $ \langle Tx,y
\rangle _{F} = \langle x,T^*y \rangle _{E}$ holds for any $ x \in
E,\ y \in F$ by $B(E,F)$. The operator $T^*$ is called the {\it
adjoint operator} of $T$.

The polar decomposition is a useful tool that represents an
operator as a product of a partial isometry and a positive
element. It is well known that every bounded operator on Hilbert
spaces has polar decomposition. In general bounded adjointable
operators on Hilbert $C^*$-modules do not have polar composition,
but Wegge-Olsen has given a necessary and sufficient
condition for bounded adjointable operators to admit polar
decomposition. He has proved that a bounded adjointable operator
$T$ has polar decomposition if and only if $\overline{Ran(T)}$ and
$\overline{Ran(|T|)}$ are orthogonal direct summands (cf.
\cite{WEG}, Theorem 15.3.7).

Let us review the polar decomposition of densely defined closed
operators on Hilbert spaces. Suppose $H$ and $H^{'}$ are Hilbert
spaces and $t : Dom(t) \subseteq H \to H^{'}$ is a densely
defined closed operator, then there exists a partial isometry $
\mathcal{V} \in B(H,H^{'})$ such that
$$
    t=\mathcal{V}|t|, \ \ Ker(t)=Ker( \mathcal{V})
$$
where $|t|:=(t^*t)^{1/2}$ (cf. \cite{KAT}, VI. Section 2.7).
Furthermore, every densely defined closed operator $t : Dom(t)
\subseteq H \to H^{'}$ has a densely defined, closed generalized
inverse, i.e. there exists a densely defined closed operator $s$
such that $tst=t$, $sts=s$, $(ts)^*= \overline{ts}$ and $(st)^*=
\overline{st}$ (cf. \cite{PYT}, Lemma 12).

In \cite{GUL} Gulja\v{s} lifts the above facts to the densely
defined closed operators on Hilbert $C^*$-modules over arbitrary
$C^*$-algebras $K(H)$ of all compact operators on a Hilbert space
$H$ of arbitrary cardinality. In fact he has found a bijective
operation-preserving map between the space of all densely defined
closed operators on Hilbert $K(H)$-modules and the space of all
densely defined closed operators on a suitable Hilbert space,
so he may lift certain properties of operators from Hilbert spaces
to Hilbert $K(H)$-modules (cf. \cite{GUL}, Theorems 1, 2, 3).

In the present note we give a necessary and sufficient condition
for unbounded regular operators to admit polar decomposition. In
fact we will prove that an unbounded regular operator $t$ on
Hilbert $C^*$-modules over an arbitrary $C^*$ algebra
$\mathcal{A}$ has polar decomposition if and only if
$\overline{Ran(t)}$ and $\overline{Ran(|t|)}$ are orthogonally
complemented, if and only if the operators $t$ and $t^*$ have
unbounded regular generalized inverses.

Some interesting characterizations of an arbitrary $C^*$-algebra
of compact operators (i.e. of a $C^*$-algebra that admits a
faithful $*$-representation in the set of all compact operators on
a certain Hilbert space) have been given in \cite{ARV},
\cite{FR1}, \cite{F-S}, \cite{MAG}, \cite{SCH}. Beside the work
of these authors we give other descriptions of the $C^*$-algebra
of compact operators via the above properties.

\newcounter{cou001}

\section{Preliminaries}

In this section we recall some definitions and basic facts about
regular operators on Hilbert $\mathcal{A}$-modules. These
operators were first introduced by Baaj and Julg in \cite{B-J}.
More details and properties can be found in chapters 9 and 10 of
\cite{LAN}, and in the papers \cite{F-S},  \cite{KK}, \cite{PAL},
\cite{KUS}, \cite{WOR}.

Let $E, F$  be Hilbert $\mathcal{A}$-modules, we will use the
notation $t : Dom(t) \subseteq E \to F$ to indicate that $t$ is
an $\mathcal{A}$-linear operator whose domain $Dom(t)$ is a dense
submodule of $E$ (not necessarily identical with $E$) and whose
range is in $F$. Given $t,s:Dom(t),Dom(s) \subseteq E \to F$, we
write $s \subseteq t$ if $Dom(s)\subseteq Dom(t)$ and $s(x)=t(x)$
for all $x \in Dom(s)$. A densely defined operator $t: Dom(t)
\subseteq E \to F$ is called {\it closed} if its graph
$G(t)=\{(x,t(x)): \ x \in Dom(t)\}$ is a closed submodule of the
Hilbert $\mathcal{A}$-module $E \oplus F$. If $t$ is closable,
the operator $s : Dom(s) \subseteq E \to F$ with the property
$G(s)= \overline{G(t)}$ is called the {\it closure} of $t$ denoted
by $s= \overline{t}$.
The operator $ \overline{t}$ is the smallest closed operator that
contains $t$.

A densely defined operator $t: Dom(t) \subseteq E \to F$
is called {\it adjointable} if it possesses a densely defined map
$t^*: Dom(t^*) \subseteq F \to E$ with the domain
$$
   Dom(t^*)= \{y \in F : {\rm there} \ {\rm exists} \ z \in E \
   {\rm such} \ {\rm that} \ \langle t(x),y \rangle _{F} = \langle
   x , z \rangle _{E} \ {\rm for} \ {\rm any} \ x \in Dom(t) \}
$$
which satisfies the property $\langle t(x),y \rangle_{F} = \langle
x,t^*(y) \rangle_{E}$,  for any $x \in Dom(t), \ y \in Dom(t^*)$.
This property implies that $t^*$ is a closed $\mathcal{A}$-linear
map.

\begin{remark}
Recall that the composition of two densely defined operators
$t,s$ is the unbounded operator $ts$ with $Dom(ts)=\{x\in Dom(s)
: \ s(x)\in Dom(t) \}$ given by $(ts)(x)=t(s(x))$ for all  $x\in
Dom(ts)$. The operator $ts$ is not necessarily densely defined.
Suppose two densely defined operators $t,s$ are adjointable, then
$s^*t^* \subseteq (ts)^*$. If $T$ is a bounded adjointable
operator, then $s^*T^* = (Ts)^*$.
\end{remark}

A densely defined closed $\mathcal{A}$-linear map $t: Dom(t)
\subseteq E \to F$ is called {\it regular} if it is adjointable
and the operator $1+t^*t$ has a dense range. We denote the set of
all regular operators from $E$ to $F$ by $R(E,F)$. A criterion of
regularity via the graph of densely defined operators has been
given in \cite{F-S}. In fact a densely defined operator $t$ is
regular if and only if its graph is orthogonally complemented in
$E \oplus F$ (cf. \cite{F-S}, Corollary 3.2). If $t$ is regular
then $t^*$ is regular and $t=t ^{**}$, moreover $t^*t$ is regular
and selfadjoint (cf.~\cite{LAN}, Corollaries 9.4, 9.6 and
Proposition 9.9). Define $Q_{t}=(1+t^*t)^{-1/2}$ and
$F_{t}=tQ_{t}$, then $Ran(Q_{t})=Dom(t)$,  $0 \leq Q_{t} \leq 1$
in $B(E,E)$ and $F_{t}\in B(E,F)$ (cf.~\cite{LAN}, chapter 9). The
bounded operator $F_{t}$ is called the bounded transform (or
$z$-transform) of the regular operator $t$. The map $t\to F_{t}$
defines a bijection $$ R(E,F) \to \{ T \in B(E,F):\| T \|\leq 1 \
\, {\rm and} \ \ Ran(1- T^* T ) \ {\rm is} \ {\rm dense} \ in \ F
\}, $$ (cf. \cite{LAN}, Theorem 10.4). This map is
adjoint-preserving, i.e. $F_{t}^*=F_{t^*}$, and for the bounded
transform $F_{t}=tQ_{t}=t(1+t^*t)^{-1/2}$ we have $\|F_{t}\|\leq
1$ and $$ t=F_{t}(1-F_{t}^*F_{t})^{-1/2} \ {\rm and} \
Q_{t}=(1-F_{t}^*F_{t})^{1/2} \, . $$

For a regular operator $t \in R(E):=R(E,E)$ some usual
properties may be defined. A regular operator $t$ is called
{\it normal} iff $Dom(t)=Dom(t^*)$ and $\langle t(x),t(x)
\rangle=\langle t^*(x),t^*(x) \rangle$ for any $x \in Dom(t)$.
The operator $t$ is called {\it selfadjoint} iff $t^*=t$, and
$t$ is called {\it positive} iff $t$ is normal and
$\langle t(x),x \rangle \geq 0$ for any $x \in Dom(t)$.
Remarkably, a regular operator $t$ is selfadjoint (resp.,
positive) iff its bounded transform $F_t$ is selfadjoint
(resp., positive), cf. \cite{KUS,LAN}. Moreover, both
$t$ and $F_t$ have the same range and the same kernel.
A regular operator $t$ has closed range if and only if
its adjoint operator $t^*$ has closed range, and then for
$|t|:=(t^*t)^{1/2}$ the orthogonal sum decompositions $E=
Ker(t) \oplus Ran(t^*)= Ker(|t|) \oplus \overline{Ran(|t|)}$,
$F= Ker(t^*) \oplus Ran(t) = Ker(|t^*|) \oplus
\overline{Ran(|t^*|)}$ exist, cf. Proposition 2.2 of
\cite{F-S} and Result 7.19 of \cite{KUS}.

\begin{remark}
Let $t$ be a regular operator on an arbitrary Hilbert
$\mathcal{A}$-module $E$, and $F _{t} $ and $Q_{t}$ be as above
then one can see that $F_{t} \cdot p(F^*_{t}F_{t})=p(F_{t}F^*_{t})
\cdot F_{t}$ \ for any polynomial $p$ and, hence, by continuity
for any $p$ in $\mathbf{C}([0,1])$. In particular,
$F_{t}(1-F^*_{t}F_{t}) ^{1/2}=(1-F_{t}F^*_{t}) ^{1/2}F_{t}$ and so
by the equalities $Q_{t}=(1-F^*_{t}F_{t})^{1/2}$ and
$Q_{t^*}=(1-F_{t}F^*_{t}) ^{1/2}$ we have
$tQ_{t}^2=Q_{t^*}tQ_{t}$.
\end{remark}

Before closing this section we would like to define the concept of
generalized (or pseudo-) inverses of unbounded regular operators,
which is motivated by the definitions of densely defined closed
operators in \cite{GUL} and \cite{PYT}.

\begin{definition}
Let $t \in R(E,F)$ be a regular operator between two Hilbert
${\mathcal{A}}$-modules $E,F$ over some fixed $C^*$-algebra
${\mathcal{A}}$. A regular operator $s \in R(F,E)$ is called
the generalized inverse of $t$ if $tst=t$, $sts=s$, $(ts)^*=
\overline{ts}$ and $(st)^*= \overline{st}$.
\end{definition}

If a regular operator $t$ has a generalized inverse $s$, then
the above definition implies that $Ran(t) \subseteq Dom(s)$
and $Ran(s) \subseteq Dom(t)$. Note, that bounded
$\mathcal{A}$-linear operators may admit generalized inverses
in the set of regular operators even if they do not admit any
bounded generalized inverse operator. For examples, consider
contractive operators on Hilbert spaces with dense, but
non-closed range. Moreover, for bounded linear operators on
Hilbert spaces, for example, the property to admit polar
decomposition does not imply the property to admit a bounded
generalized inverse. More surprising are the results for
unbounded operators described in the next section.

\section{The polar decomposition and generalized inverses}

\begin{theorem} \label{thm_polar_decomp}
If $E,F$ are arbitrary Hilbert $\mathcal{A}$-modules over a
$C^*$-algebra of coefficients $\mathcal{A}$ and $t \in R(E,F)$
denotes a regular operator then the following conditions are
equivalent:

\begin{list}{(\roman{cou001})}{\usecounter{cou001}}
\item $t$ has a unique polar decomposition $t= \mathcal{V}|t|$,
where $\mathcal{V} \in B(E,F)$ is a partial isometry for which
$Ker(\mathcal{V})=Ker(t)$, $Ker(\mathcal{V^*})=Ker(t^*)$,
$Ran(\mathcal{V})=\overline{Ran(t)}$,
$Ran(\mathcal{V^*})=\overline{Ran(|t|)}$. That is
$\overline{Ran(t)}$ and $\overline{Ran(|t|)}=\overline{Ran(t^*)}$
are final and initial submodules of the partial isometry
$\mathcal{V}$, respectively.

\item $E=Ker(|t|) \oplus \overline{Ran(|t|)}$ and $F=Ker(t^*) \oplus
\overline{Ran(t)}$.

\item $t$ and $t^*$ have unique generalized inverses which are
adjoint to each other, $s$ and $s^*$.
\end{list}

\noindent
In this situation, $\mathcal{V^*}\mathcal{V}=\overline{t^*s^*}$ is
the projection onto $\overline{Ran(|t|)} = \overline{Ran(t^*)}$,
$\mathcal{V}\mathcal{V^*}=\overline{ts}$ is the projection onto
$\overline{Ran(t)}$ and $\mathcal{V^*}\mathcal{V}|t|=|t|$,
$\mathcal{V^*}t=|t|$ and $\mathcal{V}\mathcal{V^*}t=t$.
\end{theorem}

\begin{proof}

(i) $\Rightarrow$ (ii) Let $\mathcal{V} \in B(E,F)$ be a partial
isometry satisfying condition (i). Then the identity map
of $E$ can be written as the sum of two orthogonal projections
$I-\mathcal{V^*}\mathcal{V}$ and $\mathcal{V^*}\mathcal{V}$. By
Result 7.19 of \cite{KUS} we have
$$
   Ran(I-\mathcal{V^*}\mathcal{V})=Ker(
   \mathcal{V})=Ker(t)=Ker(|t|) \, ,
$$
$$
   Ran( \mathcal{V^*}\mathcal{V})=Ran(
   \mathcal{V^*})=\overline{Ran(|t|)} \, .
$$
So we get $E=Ran(I-\mathcal{V^*}\mathcal{V})\oplus
Ran(\mathcal{V^*}\mathcal{V})=Ker(|t|)\oplus\overline{Ran(|t|)}$.
Similarly, $F$ can be decomposed as $F=Ker(t^*) \oplus
\overline{Ran(t)}$.

(ii) $\Rightarrow$ (i) Let $t \in R(E,F)$ be a regular operator
and $F_{t}$ be its bounded transform. Then Proposition 2.2 of
\cite{F-S}, Result 7.19 of \cite{KUS} and Proposition 3.7 of
\cite{LAN} imply that
$$
  Ker(t)=Ker(F_{t})=Ker(|F_{t}|)=Ker(|t|),
  Ker(t^*)=Ker(F_{t^*})
$$
$$
  \overline{Ran(|F_{t}|)}=\overline{Ran(F_{t}^*)}= \overline{Ran(t^*)}
  =\overline{Ran(|t|)}, \overline{Ran(t)}=\overline{Ran(F_{t})} \, .
$$
From the above equalities and (ii) we have $E=Ker(|F_{t}|)
\oplus \overline{Ran(|F_{t}|)}$, $F=Ker(F_{t}^*) \oplus
\overline{Ran(F_{t})}$. Now Theorem 15.3.7 of \cite{WEG} implies
that there exists a unique partial isometry $ \mathcal{V} \in
B(E,F)$ such that $F_{t}=\mathcal{V}|F_{t}|$ and
$$
Ker(
\mathcal{V})=Ker(F_{t}), \ \ Ker( \mathcal{V^*})=Ker(F_{t}^*),
$$
$$
Ran(\mathcal{V})=\overline{Ran(F_{t})},
Ran(\mathcal{V^*})=\overline{Ran(|F_{t}|)}.
$$

Therefore
$$
  Ker( \mathcal{V})=Ker(t),  Ker(\mathcal{V^*})=
  Ker(t^*) \, ,
$$
$$
  Ran( \mathcal{V})=\overline{Ran(t)},
  Ran(\mathcal{V^*})=\overline{Ran(|t|)} \, .
$$
Furthermore, by Remark 2.2 we have $F_{t}=
\mathcal{V}|F_{t}|=\mathcal{V}(t^*Q_{t^*}tQ_{t})^{
1/2}=\mathcal{V}(t^*tQ_{t}^{2}) ^{1/2}$ that is
$tQ_{t}=\mathcal{V}(t^*t)^{1/2} Q_{t}.$ But $Q_{t}:E
\longrightarrow Ran(Q_{t})=Dom(t)$ is invertible, so
$t=\mathcal{V}(t^*t)^{1/2}= \mathcal{V}|t|$.

(ii) $\Rightarrow$ (iii) Recall, that $Ker(|t|) = Ker(t)$ and
$Ran(|t|)=Ran(t^*)$. We set $Dom(s):=Ran(t) \oplus Ker(t^*)$
and define $s: Dom(s)\subseteq F \longrightarrow E$ by
$s(t(x_{1}+x_{2})+x_{3})=x_{1}$, for all $x_{1} \in Dom(t) \cap
\overline{Ran(t^*)}$, $x_{2} \in Dom(t) \cap Ker(t)$ and $x_{3}
\in Ker(t^*)$. This definition is correct since $E = Ker(t)
\oplus \overline{Ran(t^*)}$ by supposition. Then $s$ is an
$\mathcal{A}$-linear module map the domain of which
is a dense $\mathcal{A}$-submodule of $F$, since
$F=\overline{Ran(t)}\oplus Ker(t^*)$.

For each $x \in Dom(t)$ with $x=x_{1}+x_{2}$, $x_{1} \in
Dom(t) \cap  \overline{Ran(t^*)}$, $x_{2} \in Dom(t) \cap Ker(t)$
we have $tst(x)=ts(t(x_1+x_2)+0)=t(x_1)=t(x_1+x_2)$, i.e.
$tst=t$. Similarly, for each $x=t(x_{1}+x_{2})+x_{3} \in Dom(s)$
such that $x_{1} \in Dom(t) \cap  \overline{Ran(t^*)}$, $x_{2} \in
Dom(t) \cap Ker(t)$ and $x_{3} \in Ker(t^*)$, we have $sts(x)=
st(x_{1}+x_{2})=x_{1}=s(x)$, and so $sts=s$. Now we are going
to derive the properties of Definition 2.3 to demonstrate that
$s$ is a regular operator and the generalized inverse of
the operator $t$.

By the definition of $s$, the equality $ts(t(x_1+x_2)+x_3) =
t(x_1)=t(x_1+x_2)$ holds. Consequently, the operator $ts$ acts on
$Ran(t)$ as the identity operator, and on the orthogonal
complement $Ran(t)^\bot$ as the zero operator. By continuity, the
closure $\overline{ts}$ of $ts$ is the projection onto the
orthogonal summand $\overline{Ran(t)}$ of $F$. So,
$(ts)^*=\overline{(ts)^*}=(\overline{ts})^*=\overline{ts}$.
Analogously, the operator equality $(st)^*=\overline{st}$ can be
derived, and $(st)^*$ can be shown to be the projection onto the
orthogonal summand $\overline{Ran(s)}$ of $E$.

We set $Dom( \tilde{s} ):=Ran(t^*) \oplus Ker(t)$ and define the
module map $\tilde{s}:Dom( \tilde{s} )\subseteq E \longrightarrow
F $ by $ \tilde{s}(t^*(y_{1}+y_{2})+y_{3})=y_{1}$, for any $y_{1}
\in Dom(t^*) \cap \overline{Ran(t)}$, $y_{2} \in Dom(t^*) \cap
Ker(t^*)$ and $y_{3} \in Ker(t)$. Then $\tilde{s}$ is an
$\mathcal{A}$-linear module map which domain $Dom(\tilde{s})$ is
a dense $\mathcal{A}$-submodule of $E$ since
$E=\overline{Ran(|t|)}\oplus Ker(|t|)=\overline{Ran(t^*)}\oplus
Ker(t)$. We also have $t^* \tilde{s} t^*=t^*$ and $\tilde{s}t^*
\tilde{s}=\tilde{s}$. Similarly, $\overline{ t^* \tilde{s}}= (t^*
\tilde{s})^* $ and $ \overline{\tilde{s}t^* }=( \tilde{s}t^*)^*$
are orthogonal projections onto $\overline{Ran(t^*)}$ and $
\overline{Ran(\tilde{s})}$, respectively.

We prove that $s$ is a regular operator and $s^*=\widetilde{s}$.
Consider the isometry $U \in B(E \oplus F, F \oplus E)$ by
$U(x,y)=(y,x)$, then by Proposition 9.3 of \cite{LAN} we have $ F
\oplus E=U G(t) \oplus G(-t^*)$ and so
\begin{eqnarray*}
F \oplus E &=& \{ (t(x_1),x_1): x_1 \in Dom(t)\cap
                \overline{Ran(t^*)} \} \oplus
                \{(0,y_3): y_3 \in Ker(t) \} \\
           & & \oplus \,\, \{ (y_1,-t^*(y_1)): y_1 \in Dom(t^*)\cap
                \overline{Ran(t)} \} \oplus
                \{(x_3,0): x_3 \in Ker(t^*) \} \\
           &=& \{( t(x_1)+x_3 \, ,x_1) \, : x_1 \in Dom(t)
                \cap \overline{Ran(t^*)}~,~ x_3 \in Ker(t^*) \}\\
           & & \oplus \,\, \{( y_1,-t^*(y_1)-y_3) : y_1 \in Dom(t^*)
                \cap \overline{Ran(t)}~,~ y_3 \in Ker(t) \}\\
           &=&  G(s) \oplus VG( \widetilde{s}) \, ,
\end{eqnarray*}

where $ V \in B(E \oplus F, F \oplus E)$ is an isometry defined
by $V(x,y)=(y,-x)$. The equality $ F \oplus E= G(s) \oplus V
G(\tilde{s})$ and Corollary 3.2 of \cite{F-S} imply that the
operator $s$ is adjointable, closed and the range of $1+s^*s$ is
dense in $F$. In particular $s^*= \tilde{s}$. Clearly, the regular
operators $s$ and $s^*$ with the properties of generalized
inverses of the operators $t$ and $t^*$, respectively, are unique.

(iii)$\Rightarrow$(ii) Let $s$ be generalized inverse of $t$, so
that $tst=t$, $sts=s$, $(ts)^*= \overline{ts}$ and $(st)^*=
\overline{st}$. Therefore $(ts) ^{2}=(ts)$ and $Ran(ts)=Ran(t)$,
what implies that $ \overline{ts}$ is an orthogonal projection on $
\overline{Ran(t)}$, i.e. $\overline{Ran(t)}$ is orthogonally
complemented. By the hypothesis $s^*$ is the generalized inverse
of $t^*$, therefore $\overline{t^*s^*}$ is an orthogonal
projection onto $\overline{Ran(t^*)}=\overline{Ran(|t|)}$, i.e.
$\overline{Ran(|t|)}$ is orthogonally complemented.

Note that $ \mathcal{V^*}\mathcal{V}$ is the orthogonal
projection onto $\overline{Ran(|t|)}$ so $
|t|=\mathcal{V^*}\mathcal{V}|t|$. This together with the polar
decomposition of $t$, gives $ \mathcal{V^*}t=|t|$ and $
t=\mathcal{V}\mathcal{V^*}t$.
\end{proof}

The previous theorem and its proof imply some interesting
results as follows:

\begin{corollary} \label{cor_transfer}
If $t \in R(E,F)$ and $F_{t}$ is its bounded transform, then t
has polar decomposition $t= \mathcal{V}|t|$ if and only if
$F_{t}$ has polar decomposition $F_{t}= \mathcal{V}|F_{t}|$, if
and only if  $F_{t}$ has polar decomposition $F_{t}= {\mathcal V}
F_{ |t| }$, for the partial isometry $ \mathcal{V}$ which was
introduced in Theorem \ref{thm_polar_decomp}.
\end{corollary}

For the proof, just recall that $t$ and $F_t$ have the same kernel
and the same range and that $F_{t^*}=F_t^*$. Note that $Q_{ |t| }
= Q_{t}$ and so $F_{ |t| } = | F_{t} |$.

\begin{corollary}
An operator $t \in R(E,F)$ has polar decomposition
$t= \mathcal{V}|t|$ if and only if its adjoint $t^*$ has
polar decomposition $t^*= \mathcal{V^*}|t^*|$,  for the partial
isometry $\mathcal{V}$ which was introduced in Theorem
\ref{thm_polar_decomp}.
\end{corollary}

\begin{proof}If $t= \mathcal{V}|t|$ then $F_{t}=
\mathcal{V}|F_{t}|$, so $F_{t}^*=|F_{t}|\mathcal{V^*}$. Define
$G:=\mathcal{V} |F_{t}| \mathcal{V^*}$, then $G$ is selfadjoint
and $G^{2}=G.G=\mathcal{V} |F_{t}| \mathcal{V^*} \,  \mathcal{V}
|F_{t}| \mathcal{V^*}=\mathcal{V} |F_{t}| \,
|F_{t}|\mathcal{V^*}=F_{t} F_{t}^*=|F_{t}^*|^{2}$, i.e.
$G=|F_{t}^*|=\mathcal{V} |F_{t}| \mathcal{V^*}$. Thus
$F_{t}^*=|F_{t}|\mathcal{V^*}=\mathcal{V^*} \mathcal{V} |F_{t}|
\mathcal{V^*}= \mathcal{V^*} |F_{t}^*|$. Following the proof of
Theorem \ref{thm_polar_decomp}, (ii) $ \Rightarrow$ (i), we get
$t^*=\mathcal{V^*}|t^*|$. The converse direction can be shown
taking into account that ${\mathcal V}^{**}=\mathcal V$ and
interchanging the roles of $t$ and $t^*$ in the first part of the
proof.
\end{proof}

\begin{corollary}
If $t \in R(E,F)$ has closed range, then $t$ has
polar decomposition. In this case the generalized inverse of $t$
is a bounded adjointable operator.
\end{corollary}

\begin{proof}If $t$ has closed range then Proposition 2.2 of
\cite{F-S} implies that $E=Ran(t^*) \oplus Ker(t)$ and $F=Ran(t)
\oplus Ker(t^*)$, so $t$ has polar decomposition by Theorem
\ref{thm_polar_decomp}.
The operators $s$ and $ \tilde{s}$ were defined in part ''(ii)
$\Rightarrow $ (iii)''. They are bounded because $Dom(s)=Ran(t)
\oplus Ker(t^*)=F$ and $Dom( \tilde{s} )=Ran(t^*) \oplus Ker(t)=E$,
i.e.~the generalized inverse of $t$ is a bounded adjointable
operator.
\end{proof}

Theorem \ref{thm_polar_decomp} and Corollary \ref{cor_transfer}
together with a recent result by Lun Chuan Zhang \cite{Zhang} and
by Qingxiang Xu and Lijuan Sheng \cite{Xu/Sheng} give
us the opportunity to derive a criterion for bounded $C^*$-linear
operators between Hilbert $C^*$-modules to admit a generalized
inverse in the sense of Banach algebra theory. These authors proved
independently that a bounded adjointable $C^*$-linear operator between
two Hilbert $C^*$-modules admits a bounded generalized inverse if
and only if the operator has closed range.

\begin{proposition}
Let $T \in R(E,F)$ be a bounded $\mathcal{A}$-linear operator
between two Hilbert  ${\mathcal{A}}$-modules $E,F$ over some
fixed $C^*$-algebra ${\mathcal{A}}$. Suppose, $T$ has polar
decomposition. Then $T$ admits a regular
operator $s$ as its generalized inverse. Moreover, $s$ is
bounded if and only if the range of $T$ is closed.
\end{proposition}

\begin{corollary}
For $t \in R(E,F)$ the bounded transform $F_t$ has a bounded
generalized inverse if and only if $F_t$ has closed range, if and
only if $t$ has closed range.
\end{corollary}

Magajna and Schweizer have shown, respectively, that
$C^*$-algebras of compact operators can be characterized by the
property that every norm closed (coinciding with its biorthogonal
complement, respectively) submodule of every Hilbert $C^*$-module
over them is automatically an orthogonal summand, cf. \cite{MAG},
\cite{SCH}. Recently further generic properties of the category
of Hilbert $C^*$-modules over $C^*$-algebras which characterize
precisely the $C^*$-algebras of compact operators have been found
by the authors in \cite{FR1} and \cite{F-S}. All in all,
$C^*$-algebras of compact operators turn out to be of unique
interest in Hilbert $C^*$-module theory.

\begin{theorem} \label{thm-3-7}
Let $\mathcal{A}$ be a $C^*$-algebra. The
following conditions are equivalent, among others:
\begin{list}{(\roman{cou001})}{\usecounter{cou001}}

\item $\mathcal{A}$ is an arbitrary $C^*$-algebra of compact operators.

\item For every (maximal) norm closed left ideal $I$
of $ \mathcal{A}$ the corresponding open projection $p \in
\mathcal{A}^{**}$ is an element of the multiplier $C^*$-algebra
$M(\mathcal{A})$ of $ \mathcal{A}$.

\item For every Hilbert $\mathcal{A}$-module $E$ every
Hilbert $\mathcal{A}$-submodule $F \subseteq E$ is automatically
orthogonally complemented, i.e. $F$ is an orthogonal summand.

\item For every Hilbert $\mathcal{A}$-module $E$ Hilbert
$\mathcal{A}$-submodule $F \subseteq E$ that coincides
with its biorthogonal complement $F^{\perp \perp}
\subseteq E$ is automatically orthogonally complemented in $E$.

\item For every Hilbert $ \mathcal{A}$-module $E$
every Hilbert $ \mathcal{A}$-submodule is automatically
topologi\-cally complemented there, i.e. it is a topological
direct summand.

\item For every pair of Hilbert $ \mathcal{A}$-modules
$E, F$, every densely defined closed operator $t:
Dom(t)\subseteq E \to F$ possesses a densely defined
adjoint operator $t^*: Dom(t^*)\subseteq  F \to E$.

\item For every pair of Hilbert $ \mathcal{A}$-modules
$E, F$, every densely defined closed operator $t:
Dom(t)\subseteq E \to F$ is regular.

\item The kernels of all densely defined closed
operators between arbitrary Hilbert $\mathcal{A}$-modules
are orthogonal summands.

\item The images of all densely defined closed operator
with norm closed range between ar\-bitrary Hilbert
$\mathcal{A}$-modules are orthogonal summands.
\end{list}
\end{theorem}

\begin{corollary}
Let $\mathcal{A}$ be a $C^*$-algebra. The
following conditions are equivalent:
\begin{list}{(\roman{cou001})}{\usecounter{cou001}}
\item $\mathcal{A}$ is an arbitrary $C^*$-algebra of
compact operators.
\end{list}
\begin{list}{(\roman{cou001})}{\usecounter{cou001}}
   \setcounter{cou001}{9}
\item For every pair of Hilbert $ \mathcal{A}$-modules
$E, F$, every densely defined closed operator $t: Dom(t)\subseteq
E \to F$ has polar decomposition, i.e. there exists a unique
partial isometry $ \mathcal{V}$ with initial set $
\overline{Ran(|t|)}$ and the final set $ \overline{Ran(t)}$ such
that $t=\mathcal{V}|t|$.

\item For every pair of Hilbert $ \mathcal{A}$-modules
$E, F$, every densely defined closed operator $t: Dom(t)\subseteq
E \to F$ and its adjoint have generalized inverses.
\end{list}
\end{corollary}

\begin{proof}The statements are deduced from
Theorem \ref{thm_polar_decomp} and from the conditions (vii), (viii)
of Theorem \ref{thm-3-7}.
\end{proof}

After searching in A.M.S.' MathSciNet data base we believe
that part of the results of Theorem 3.1, Proposition 3.5, and Corollaries
3.4, 3.6 and 3.8 are essentially new even in the case of Hilbert
spaces.

\medskip
{\bf Acknowledgement}: This research was done while the second
author stayed at the Ma\-the\-matisches Institut of the Westf\"alische
Wilhelms-Universit\"at in M\"unster, Germany, in 2007. He would like to
express his thanks to professor Michael Joachim and his
colleagues in the topology and operator algebras groups for
warm hospitality and great scientific atmosphere.
The authors would like to thank the referee for his/her very
useful comments which helped to improve parts of the proof
of Theorem 3.1 and to shorten the presentation of some facts.

config: arxiv


TITLE: Finding all real $a$ such that $16x^4-(a)x^3+(2a+17)x^2−(a)x+16=0$ has four distinct roots in geometric progression
QUESTION [1 upvotes]: Determine all real values of the parameter, $a$, for which the equation  $$16x^4−(a)x^3+(2a+17)x^2−(a)x+16=0$$ has exactly four distinct real roots that form a geometric progression?

I noticed that the coefficients are symmetric: namely, the first coefficient is the same as the fifth one, the second is the same as the fourth, and the third is the same as the third.
I don’t know how to proceed using Vieta's formula.

REPLY [0 votes]: Let's do like derivation of Vieta's formulas rather than the formulas themselves.
Let $b,\,bq,\,bq^2,\,bq^3$ be the roots of $$p(x)=(x-b)(x-bq)(x-bq^2)(x-bq^3)$$ $$\hbox{and }f(x)=x^4−\frac{a}{16}x^3+\frac{2a+17}{16}x^2−\frac{a}{16}x+1.$$
It can be verified that $$p(x)=b^4 q^6 - b^3 (q^3 + q^2 + q + 1) q^3 x + \\
b^2 (q^4 + q^3 + 2 q^2 + q + 1) q x^2 - b (q^3 + q^2 + q + 1) x^3 + x^4$$
But $p(x)$ and $f(x)$ are same degree polynomials with highest degree coefficient $1$ and the same roots $\Rightarrow$ they have the same coefficients.
$$\begin{cases}
b^4 q^6=1\\
16 b^3 (q^3 + q^2 + q + 1) q^3=a\\
16 b^2 (q^4 + q^3 + 2 q^2 + q + 1) q = 2a+17\\
16 b (q^3 + q^2 + q + 1) = a
\end{cases}$$
Hence $a=170$.


TITLE: How many permutations can there be?
QUESTION [0 upvotes]: Suppose I have a set $$\{1,...,1,2,...,2,3,...,3,...,n,...,n\}.$$ How many permutations can I have? I did try to solve this but I could only do it by brute force and for small $n$. I think the answer is $$\frac{N!}{k_1!k_2!\cdot\cdot\cdot k_n!}$$ Where $k_i$ is the amount of $i$ in the sum and $N$ is the total number of elements.
I have trouble excluding cases where I count a permutation multiple times.

REPLY [0 votes]: Ok, I thought about it and it's not that hard after all.
Suppose you have this set $\{x_1,x_2,...,x_n\}.$ For this set you will have $n!$ permutations. Every permutation will contain $x_1, x_2$ and $x_3$ somewhere in them. Suppose you have an arbitrary permutation, where you remove $x_1,x_2$ and $x_3$ and fill their spots with a blank. Given this one 'permutation' how many permutations can you generate by placing your three elements? $3!,$ because that is how many ways you have to arrange $x_1,x_2$ and $x_3$ onto those blanks. Now suppose $x_1=x_2=x_3.$ This means all the $3!$ permutations are the same. So for any arbitrary permutation you have $3!-1$ permutations that are the same. In other words the $3!$ permutations that were unique when all your elements where unique are now the same. So the total amount of permutations is $n!/3!.$ - The argument holds for any number of identical elements.
Now suppose $x_1=x_2=x_3\neq x_4=x_5.$ Assume all the elements are the same so the total number of permutations is $n!$ then say that $x_1=x_2=x_3$ and say that the total number of permutations is $n!/3!.$ Finally say that $x_4=x_5$ and say that now the total number of permutations is $n!/(3!2!).$


text:
\begin{document}

\title[Askey-Wilson algebra revisited and tridiagonalization]
{Little and Big $q-$Jacobi Polynomials and the Askey-Wilson algebra}

\author{Pascal Baseilhac$^{\dagger}$, Xavier Martin$^{\dagger}$}
\address{$^\dagger$ Institut Denis-Poisson CNRS/UMR 7013 - Universit\'e de Tours - Universit\'e d'Orl\'eans
Parc de Grammont, 37200 Tours,
FRANCE}
\email{pascal.baseilhac@idpoisson.fr; xavier.martin@idpoisson.fr}

\author{Luc Vinet$^{*}$}
\address{$^{*}$ Centre de recherches math\'ematiques Universit\'e de Montr\'eal, CNRS/UMI 3457, P.O. Box 6128, Centre-ville Station, Montr\'eal (Qu\'ebec), H3C 3J7 CANADA}
\email{vinet@CRM.UMontreal.CA}

\author{Alexei Zhedanov$^{\diamond,*}$}
\address{$^{\diamond}$ Department of Mathematics, Information School, Renmin University of China, Beijing 100872,
CHINA}
\email{zhedanov@yahoo.com}

\date{May 31, 2018}

\begin{abstract}
The little and big q-Jacobi polynomials are shown to arise as basis vectors for representations of the Askey-Wilson algebra. The operators that these polynomials respectively diagonalize are identified within the Askey-Wilson algebra generated by twisted primitive elements of $\mathfrak U_q(sl(2))$. The little q-Jacobi operator and a tridiagonalization of it are shown to realize the equitable embedding of the Askey-Wilson algebra into $\mathfrak U_q(sl(2))$.
\end{abstract}

\maketitle

\vskip -0.4cm

{\small MSC:\ 81R50;\ 81R10;\ 81U15;\ 39A70;\ 33D50;\ 39A13.}

{{\small  {\it \bf Keywords}:  Askey-Wilson algebra; Tridiagonalisation: Orthogonal polynomials}}

\vspace{7mm}

\section{Introduction}

This paper indicates how the little and big q-Jacobi polynomials \cite{KS} occur in the context of representations of the Askey-Wilson algebra. This structure was originally identified in \cite{7} as the algebra generated by the Askey-Wilson operator and the multiplication by the independent variable; its representations were shown to account for the bispectral properties of the Askey-Wilson polynomials. It has thereafter appeared in various contexts.

The little q-Jacobi polynomials \cite{KS} have received algebraic interpretations some time ago, either as matrix elements of co-representations of the quantum group $SL_q(2)$ \cite{VS, K1, MMNNU1, MMNNU2} or equivalently \cite{FV1}, as matrix elements of q-exponentials in the generators of the quantum algebra $\mathfrak U_q(sl(2))$ \cite {FV2}. The big q-Jacobi polynomials have been connected to quantum 2-spheres \cite {NM}. We shall provide here a rather different algebraic setting for these q-polynomials as basis vectors for modules of the Askey-Wilson algebra.

On the one hand, consider twisted primitive elements in $\mathfrak U_q(sl(2))$ \cite{K2,KJ}. It is known that such elements provide an embedding of the Askey-Wilson algebra in $\mathfrak U_q(sl(2))$ \cite{GZ}. Take the holomorphic realization of $\mathfrak U_q(sl(2))$ and a specialization of the generators, it will be seen that the operators of which the little and big q-Jacobi polynomials are eigenfunctions belong to this realization of the Askey-Wilson algebra.

On the other hand, the polynomials of the Askey scheme and especially those under consideration here, are solutions of  bispectral problems. It has been shown recently how algebraic Heun operators can be associated to any such problems \cite {GrVZ}. For some choices of the generic parameters, the resulting Heun operator can be one of those corresponding to the polynomials of the Askey tableau and this provides a procedure often referred to as tridiagonalization \cite{IK1, IK2} to construct and study higher special functions from simpler ones. Clearly, bilinear combinations of the bispectral operators of classical polynomials will be tridiagonal in the eigenbasis . The main point is that such Heun operators generate when combined with one of the operators entering in the bispectral problem, an algebra that encodes the properties of the higher functions/polynomials. This algebraic viewpoint on tridiagonalization has been developed recently. The 4-parameter Wilson polynomials have thus been recovered and analysed in that perspective through the tridiagonalization of the 2-parameter hypergeometric operator \cite{GIVZ}. Similarly the characterization of the 4-parameter Bannai-Ito polynomials could be retrieved by observing that the embedding of the Bannai-Ito algebra into the superalgebra $\mathfrak osp$(1,2) involved tridiagonalizing the operator of which the little -1 Jacobi polynomials are eigenfunctions \cite {BGVZ}. It is hence natural at this point, to complete the picture by determining how the Askey-Wilson polynomials lend themselves to a tridiagonalization analysis. This is one of the goals of this paper. Typically one would work with the differential or difference operator of which the polynomials are eigenfunctions. This will also be the main approach here. It should be said that tridiagonalization can as well be applied to recurrence operators. As a matter of fact, it has already been shown that the recurrence coefficients of the Askey-Wilson polynomials can be obtained from those of the big q-Jacobi polynomials from the tridiagonalization of the recurrence operator of the latter polynomials \cite{TVZ}.

It will be seen that the tridiagonalization of the q-difference operator, of which the little q-Jacobi polynomials are eigenfunctions, leads to the Askey-Wilson algebra in the fashion described above. It will be further observed that if the equitable presentation of $\mathfrak U_q(sl(2))$ is called upon \cite{ITW,T0}, the little  q-Jacobi operator and its tridiagonalized companion turn out to naturally take the form that generators of the Askey-Wilson algebra have when symmetrically embedded in $\mathfrak U_q(sl(2))$.

The paper will be organized as follows.  After having provided some relevant facts about  $\mathfrak U_q(sl(2))$, the Askey-Wilson algebra and its embedding in  $\mathfrak U_q(sl(2))$ are presented in Section 2. In Section 3, the little q-Jacobi polynomials are introduced via their q-difference equation, which is identified under special choices of the parameters as  the eigenvalue equation for one generator of the Askey-Wilson algebra through the realization inherited from that of  $\mathfrak U_q(sl(2))$. The eigenfunctions of a companion generator of the Askey-Wilson algebra are obtained and the expansion of the little q-Jacobi polynomials in this eigenbasis will be shown to yield an explicit expression of these polynomials. The tridiagonalization of the little q-Jacobi operator will be carried out in Section 4. This will provide an operator that is tridiagonal on the little q-Jacobi polynomials and that generates a new Askey- Wilson algebra together with the original little q-Jacobi operator. This will be obtained by noting that these operators take  the known form \cite{T}  of the Askey-Wilson generators under the embedding in  $\mathfrak U_q(sl(2))$ in terms the equitable generators of this quantum algebra \cite{T0}.
In Section 5 we shall turn to the big q-Jacobi polynomials and show that they are eigenfunctions of the same generator as for the little ones, in the initial embedding of the Askey-Wilson algebra in  $\mathfrak U_q(sl(2))$ albeit with  different and more general choices of the parameters. Their explicit expression will also be obtained as in section 3. As supplementary information, we shall show in Section 6 that a large class of big q-Jacobi polynomials can be obtained via tridiagonalization from special little q-Jacobi polynomials.
 The paper will end with a summary and outlook in Section 7.

\subsection{Notations}

In this paper, we fix a nonzero complex number $q$ which is not a root of unity.  We will use the standard $q$-shifted factorials (also called $q-$Pochhammer functions) \cite{KS}:

\beqa
(a;q)_n=\prod_{k=0}^{n-1}(1-aq^{k}).\label{poch}

\eeqa

\section{The Askey-Wilson algebra and its embedding in $\mathfrak U_q(sl(2))$}

We shall recap in this section the basic tools that shall be used in the remainder of the paper: first the standard Chevalley presentation of $\mathfrak U_q(sl(2))$  and second the equitable one, together with isomorphisms between the two. The realization of $\mathfrak U_q(sl(2))$  in terms of q-difference operators will then be recalled in subsection \ref{23} and  the Askey-Wilson algebra will be introduced via its embedding into $\mathfrak U_q(sl(2))$ in subsection \ref{24}.

\subsection{The Chevalley presentation of  $\mathfrak U_q(sl(2))$}

The Chevalley presentation of  $\mathfrak U_q(sl(2))$  consists of three generators denoted $S_\pm,s_3$. They  satisfy

\beqa
[s_3,S_\pm]=\pm S_\pm \qquad \mbox{and} \qquad
[S_+,S_-]=\frac{q^{2s_3}-q^{-2s_3}}{q-q^{-1}}\ .\label{Uqsl2}
\eeqa

The central element of  $\mathfrak U_q(sl(2))$ is the Casimir operator:

\beqa
\Omega = \frac{q^{-1}q^{2s_3}+ q q^{-2s_3}}{(q-q^{-1})^2} + S_+S_-.\label{Casimir}
\eeqa

\subsection{The equitable presentation of $\mathfrak U_q(sl(2))$}

The equitable presentation of  $\mathfrak U_q(sl(2))$  consists of three generators denoted $X,Y,Z$ \cite{ITW}.

\beqa
&& YY^{-1}=Y^{-1}Y=1,\quad
\frac{qXY-q^{-1}YX}{q-q^{-1}}=1, \quad \frac{qYZ-q^{-1}ZY}{q-q^{-1}}=1, \quad \frac{qZX-q^{-1}XZ}{q-q^{-1}}=1.\label{equit}
\eeqa

In \cite[Lemma 5.1]{T0},  an isomorphism with the presentation (\ref{Uqsl2}) is given. Here we will use a special case. In our notation, it reads:

\beqa
X &=& q^{-2s_3} -(q-q^{-1})q^{1/2}S_+q^{-s_3},\label{XYZ}\\
 Y&=&q^{2s_3},\nonumber\\
Z&=& q^{-2s_3} + (1-q^{-2})q^{1/2} S_-q^{-s_3}.\nonumber
\eeqa

\subsection{The $q-$difference operators realization of $\mathfrak U_q(sl(2))$}\label{23}

An irreducible infinite dimensional representation $V_\nu$ with $2\nu \notin {\mathbb Z}_+$ can be realized by $q-$difference operators acting on the space of formal power series $f(z) = \sum_{k\in {\mathbb Z}_+} \mu_k z^k$ in the variable $z$. The lowest weight vector corresponds to $1$. Define $T_\pm(f(z))=f(q^{\pm 1} z)$. There exists an homomorphism \cite{S1}:
\beqa
&& q^{s_3}\mapsto q^{-\nu}T_+, \qquad \qquad \qquad \qquad q^{-s_3}\mapsto q^{\nu}T_-, \label{holo}\\
&& S_+ \mapsto z\frac{(q^{2\nu}T_- - q^{-2\nu}T_+)}{(q-q^{-1})},\qquad S_-\mapsto -z^{-1}\frac{(T_- - T_+)}{(q-q^{-1})}.\nonumber
\eeqa

On $V_\nu$, the eigenvalue of $\Omega$ (\ref{Casimir}) is given by:

\beqa
\omega_\nu=\frac{(q^{2\nu+1} + q^{-2\nu-1})}{(q-q^{-1})^2}.\label{Casimir-val}
\eeqa

Note that if $2\nu \in {\mathbb Z}_+$, the representation becomes reducible as $(S_+)^{2\nu+1}=0$. The corresponding invariant subspace of polynomials of degree $2\nu$.

\subsection{The Askey-Wilson algebra}\label{24}
We now turn to the Askey-Wilson algebra and its embedding in $\mathfrak U_q(sl(2))$.
Let $c_0,\oc_0,c_1,\oc_1,\epsilon_0,\epsilon_1,\mu_0,\mu_1$ be arbitrary scalars. Define:

\beqa
W_0 &=& c_0 S_+q^{s_3} +  \oc_0 S_-q^{s_3} +  \epsilon_0 q^{2s_3} + \mu_0,\label{W}\\
W_1 &=& c_1 S_+q^{-s_3} +  \oc_1 S_-q^{-s_3} +  \epsilon_1 q^{-2s_3} + \mu_1.\nonumber
\eeqa

A third operator, namely

\beqa
G_1=qW_1W_0 -q^{-1}W_0W_1
\eeqa

 is constructed. By straightforward calculations, one finds:

\beqa
G_1=g_1 S_-^2 + g_2 S_-q^{-s_3} +  g_3 S_-q^{s_3} +   g'_2 S_+q^{-s_3} +  g'_3 S_+q^{s_3}  + g_4 q^{2s_3} + g_5q^{-2s_3} + g_6,\label{G1}
\eeqa

where $g_i,i=1,..,6$ and $g'_2,g'_3$ are expressed in terms of $c_j,\oc_j,\epsilon_j,j=0,1$:

\beqa
&&g_1=\oc_0\oc_1(q^2-q^{-2}),\nonumber\\
&& g_2=\oc_0\epsilon_1(q^2-q^{-2})q + \mu_0 \oc_1(q-q^{-1}),\quad g_3=\oc_1\epsilon_0(q^2-q^{-2})q^{-1} +  \mu_1\oc_0(q-q^{-1}),\nonumber\\
&& g'_2= \mu_0c_1(q-q^{-1}), \quad g'_3= \mu_1c_0(q-q^{-1}), \nonumber\\
&& g_4= -\frac{c_0\oc_1q^{-1}(q+q^{-1})}{(q-q^{-1})} + \mu_1\epsilon_0(q-q^{-1}),\quad g_5= -\frac{c_1\oc_0q(q+q^{-1})}{(q-q^{-1})} + \mu_0\epsilon_1(q-q^{-1}),\nonumber\\
&& g_6= (c_1\oc_0q +c_0\oc_1q^{-1})(q-q^{-1})\Omega + (\epsilon_0\epsilon_1 + \mu_0\mu_1)(q-q^{-1}).\nonumber
\eeqa

\begin{prop} $W_0,W_1,G_1$ satisfy the Askey-Wilson algebra:

\beqa
\big[W_1,W_0\big]_q&=&G_1,\nonumber\\
\quad \big[W_0,G_1\big]_q &=& \rho_0 W_1 + \omega W_0 + \gamma_0 (W_0W_1+W_1W_0) + \gamma_1 W_0^2 + \eta_0,  \nonumber\\
\big[G_1,W_1\big]_q &=& \rho_1 W_0 + \omega W_1 + \gamma_1 (W_0W_1+W_1W_0) + \gamma_0 W_1^2 + \eta_1,  \nonumber
\eeqa

where

\beqa
\rho_0&=&-c_0\oc_0 (q+q^{-1})^2 - \mu_0^2(q-q^{-1})^2,\quad
\rho_1= -c_1\oc_1 (q+q^{-1})^2 - \mu_1^2(q-q^{-1})^2,\nonumber\\
\omega &=& (q-q^{-1})g_6 - 2\mu_0\mu_1 (q-q^{-1})^2,\nonumber\\
\gamma_0 &=& \mu_0(q-q^{-1})^2,\quad \gamma_1 = \mu_1(q-q^{-1})^2,\nonumber\\
\eta_0 &=& (q+q^{-1})\left( c_0\oc_0 \epsilon_1(q-q^{-1})^2 \Omega - \epsilon_0 (qc_1\oc_0 + q^{-1}c_0\oc_1)\right) - \mu_0(q-q^{-1})g_6 + \mu_1c_0\oc_0(q+q^{-1})^2 + \mu_0^2\mu_1(q-q^{-1})^2,\nonumber\\
\eta_1 &=& (q+q^{-1})\left( c_1\oc_1 \epsilon_0(q-q^{-1})^2 \Omega - \epsilon_1 (qc_0\oc_1 + q^{-1}c_1\oc_0)\right) - \mu_1(q-q^{-1})g_6 + \mu_0c_1\oc_1(q+q^{-1})^2 + \mu_1^2\mu_0(q-q^{-1})^2,\nonumber
\eeqa

\end{prop}
\begin{proof} Straightforward, using (\ref{Uqsl2}) and (\ref{Casimir}).\end{proof}

In the text below, we consider successively special values of the structure constants of the AW algebra. In each case, a realization of $W_0,W_1,G_1$ in terms of $\mathfrak U_q(sl(2))$ Chevalley generators is given.  For a first choice of $c_i,\oc_i,\epsilon_i,\mu_i$ namely $\mu_0=c_0=\oc_1=0$ the operator $G_1$ will be denoted by $\tG_1$. It is diagonalized by the little q-Jacobi polynomials.
For a second choice of $c_i,\oc_i,\epsilon_i,\mu_i$, namely $\mu_0=c_0=0$ the operator $G_1$ is diagonalized by the big q-Jacobi polynomials.  In both cases, the elements $W_0,W_1,G_1$ satisfy the `reduced' relations:

\beqa
\big[W_1,W_0\big]_q&=&G_1,\nonumber\\
\quad \big[W_0,G_1\big]_q &=&  \omega W_0 + \gamma_1 W_0^2 + \eta_0,  \nonumber\\
\big[G_1,W_1\big]_q &=& \rho_1 W_0 + \omega W_1 + \gamma_1 (W_0W_1+W_1W_0)  + \eta_1,  \nonumber
\eeqa

where the structure constants are specializations of $\rho_i,\omega,\gamma_i,\eta_i$ as given above.

\section{The little $q-$Jacobi polynomials}

We shall now make precise the circumstances under which the little q-Jacobi polynomials are eigenfunctions of $\tG_1$ and form a representation basis of a specialized Askey-Wilson algebra. Given the ensuing specific choice of parameters, the eigenfunctions of $W_0$ will be determined to be q-Pochammer symbols. It will be observed that $\tG_1$ acts bidiagonally on these functions, as well as on monomials, and this will entail explicit expressions for the little q-Jacobi polynomials.

\subsection{The second order q-difference operator $\tG_1$}

If we now compare the operator $G_1$ of (\ref{G1}) using (\ref{holo}) with (3.12.4) of \cite{KS}, one has:

\beqa
\tG_1\equiv G_1|_{g'_2=g'_3=g_1=0} &\mapsto& \underbrace{\left( \frac{g_3q^{-\nu}}{(q-q^{-1})}\frac{1}{z} + g_4q^{-2\nu} \right)}_{\tA_0} T_+^2 + \underbrace{\left( -\frac{g_2q^{\nu}}{(q-q^{-1})}\frac{1}{z} + g_5q^{2\nu} \right)}_{\tB_0} T_-^2  \label{tG1exp}\\
&& + \underbrace{\left(  -\frac{(g_3q^{-\nu} - g_2q^{\nu} ) }{(q-q^{-1})}\frac{1}{z} + g_6\right)}_{\tC_0}.\nonumber
\eeqa

\begin{prop}\label{p41} For the specialization\footnote{We choose $c_0=\oc_1=\mu_0=0$ in (\ref{W}). } $g'_2=g'_3=g_1=0$, the `reduced' operator $ \tG_1$ is diagonalized by the little $q-$Jacobi polynomials.
\end{prop}

\begin{proof} Denote the little q-Jacobi polynomials as:

\beqa
y(z) \equiv p_n(z;\ma,\mb;q^2)\nonumber
\eeqa

They satisfy the second-order q-difference equation:

\beqa
q^{-2n}(1-q^{2n})(1-\ma\mb q^{2n+2})y(z) = B(z) y(q^2z) + \overline{B}(z)y(q^{-2}z) - (B(z)+\overline{B}(z))y(z),\nonumber
\eeqa

where

\beqa
B(z) &=& -\ma\frac{1}{z}+   \ma\mb q^2 ,\nonumber\\
\overline{B}(z)&=&-\frac{1}{z}+ 1,\nonumber
\eeqa

The exact relation between the parameters $g_i$ and the parameters entering in the little q-Jacobi polynomials is as follows:

\beqa
\tG_1 y(z) =\left( g_5q^{2\nu}(q^{-2n}+\ma\mb q^{2n+2}) +g_6\right)y(z)\nonumber
\eeqa

and

\beqa
&&\ma = -\frac{g_3}{g_5}\frac{q^{-3\nu}}{(q-q^{-1})},\quad \mb= - \frac{g_4}{g_3}(q-q^{-1})q^{-\nu-2} \quad \mbox{and}\quad \frac{g_2}{g_5}=(q-q^{-1})q^\nu.\label{lab}
\eeqa
\end{proof}

\subsection{Eigenfunctions and the explicit expression of the little $q-$Jacobi polynomials}

First, we construct the eigenfunctions of $W_0$ in (\ref{W}) for $g'_2=g'_3=g_1=0$ .

\begin{lem}\label{lem2}\label{lem1}  For $c_0=\mu_0=0$, one has:

\beqa
W_0f_n(z) =\lambda_n f_n (z) \quad \mbox{with} \quad \lambda_n=\epsilon_0 q^{2(n-\nu)} \quad \mbox{and}\quad f_n(z) = \left(\frac{\epsilon_0}{\oc_0}(1-q^2)q^{-\nu-1}z;q^2\right)_n.\label{f}
\eeqa
\end{lem}

\begin{proof} Considering (\ref{W}) for $c_0=\mu_0=0$ and using (\ref{holo}), the action on the monomials $z^n$ reads:

\beqa
W_0 z^n= \lambda_n  z^n + \nu_n z^{n-1}\quad \mbox{with}\quad \lambda_n=\epsilon_0 q^{2(n-\nu)}\quad  \mbox{and} \quad \nu_n= \oc_0 q^{-\nu+1}\frac{1-q^{2n}}{1-q^2}.\nonumber
\eeqa

Let $f_n(z)$ be such that $W_0f_n(z)=\lambda_nf_n(z)$.  Define $f_n(z)=\sum_{s=0}^\infty \alpha_{n,s} z^s$. The action of $W_0$ on $f_n(z)$ gives the recurrence relation:

\beqa
\lambda_n \alpha_{n,s} = \lambda_s\alpha_{n,s} + \nu_{s+1}\alpha_{n,s+1}. \nonumber
\eeqa

The solution reads

\beqa
\alpha_{n,s}&=&\alpha_{n,0} \frac{(\lambda_n-\lambda_0)(\lambda_n-\lambda_1)\cdots (\lambda_n-\lambda_{s-1})}{\nu_1\nu_2 \cdots \nu_{s}}\\
&=& \alpha_{n,0}  \left(  (1-q^2)\frac{\epsilon_0}{\oc_0}q^{2n-\nu-1}\right)^s \frac{(q^{-2n};q^2)_s}{(q^2;q^2)_s},\nonumber
\eeqa

from which, setting $\alpha_{n,0}=1$  and using the $q-$binomial theorem \cite{KS}, we finally obtain (\ref{f}).
\end{proof}

On the eigenfunctions of $W_0$, we now consider the action of $\tG_1$.

\begin{lem} One has:

\beqa
\tG_1 f_n(z) = a_n f_n(z) + b_n f_{n-1}(z)\label{acttG1}
\eeqa

with

\beqa
a_n&=& g_5q^{2\nu}(q^{-2n}+\ma\mb q^{2n+2}) +g_6,\nonumber\\
b_n&=&  -g_5 q^{2\nu}q^{-2n}(1-q^{2n})(1-q^{2n}\mb).\nonumber
\eeqa
\end{lem}

\begin{proof} Recall (\ref{tG1exp}). The l.h.s. of (\ref{acttG1}) reads:

\beqa
\tG_1 f_n(z) &=& \left(\tA_0 (1-\gamma_0 q^{2n-2}z)(1-\gamma_0 q^{2n}z) + \tB_0 (1-\gamma_0 q^{-2}z)(1-\gamma_0 z) \right.\nonumber\\
&& \left.+ \tC_0 (1-\gamma_0 z)(1-\gamma_0 q^{2n-2}z)\right)(1-\gamma_0 q^{2}z)\cdots (1-\gamma_0 q^{2n-4}z),\nonumber
\eeqa

where we have denoted  $\gamma_0=\frac{\epsilon_0}{\oc_0}(1-q^2)q^{-\nu-1}$. The r.h.s reads:

\beqa
\left(a_n (1-\gamma_0 z)(1-\gamma_0 q^{2n-2}z) + b_n(1-\gamma_0 z)\right)(1-\gamma_0 q^{2}z)\cdots (1-\gamma_0 q^{2n-4}z).\nonumber
\eeqa

Equating both sides of the equation (\ref{actG1}), the coefficients $a_n$ and $b_n$ are determined uniquely.
\end{proof}

\begin{rem} Observe that the coefficient $a_n$ coincides with the spectrum of $\tG_1$, as expected ($\tG_1$ is lower triangular in the basis $\{f_n\}$).
\end{rem}

Next, we are interested in the overlap coefficients between the eigenfunctions of $W_0$ (the $f_n(z)$) and the eigenfunctions of $\tG_1$ (the little q-Jacobi polynomials).

\begin{prop}\label{prop2} One has:

\beqa
p_n(z;\ma,\mb;q^2)= \sum_{s=0}^{\infty} \gamma_{n,s}f_s(z) \label{exppn}
\eeqa

where

\beqa
\gamma_{n,s}=   (-q^2 \mb)^{-n} q^{-n(n-1)} \frac{(q^2\mb;q^2)_n}{(q^2\ma;q^2)_n} \frac{ (q^{-2n};q^2)_s (\ma\mb q^{2n+2};q^2)_s}{(q^2\mb;q^2)_s (q^2;q^2)_s}q^{2s} .\label{gam}
\eeqa

\end{prop}

\begin{proof} By Prop. \ref{p41} recall that $p_n(z;\ma,\mb;q^2)$ are eigenfunctions of $\tG_1$ with the identification (\ref{lab}).  Consider the expansion (\ref{exppn}).
The action of $\tG_1$ on $p_n(z;\ma,\mb;q^2)$ gives the recurrence relation:

\beqa
a_n \gamma_{n,s} = a_s\gamma_{n,s} + b_{s+1}\gamma_{n,s+1}. \nonumber
\eeqa

The solution reads

\beqa
\gamma_{n,s}&=&\gamma_{n,0} \frac{(a_n-a_0)(a_n-a_1)\cdots (a_n-a_{s-1})}{b_1b_2 \cdots b_{s}}.\nonumber
\eeqa

By straightforward calculations, one finds:

\beqa
(a_n-a_0)(a_n-a_1)\cdots (a_n-a_{s-1}) &=& (-g_5q^{2\nu})^s q^{-s(s-1)} (q^{-2n};q^2)_s (\ma\mb q^{2n+2};q^2)_s,\nonumber\\
b_1b_2\cdots b_s &=&  (-g_5q^{2\nu})^sq^{-s(s+1)}(q^2\mb;q^2)_s (q^2;q^2)_s.\nonumber
\eeqa

Setting the normalization such that:

\beqa
 \gamma_{n,0}= (-q^2 \mb)^{-n} q^{-n(n-1)} \frac{(q^2\mb;q^2)_n}{(q^2\ma;q^2)_n},
\eeqa
we get (\ref{gam}).
\end{proof}

\begin{rem} The expansion formula (\ref{exppn}) coincides with (2.46) of \cite{Kor} with $q \rightarrow q^2$:

\beqa
p_n(z;\ma,\mb;q^2) =  (-q^2 \mb)^{-n} q^{-n(n-1)} \frac{(q^2\mb;q^2)_n}{(q^2\ma;q^2)_n} \    _3\phi_2\left[\begin{matrix} q^{-n} , q^{n+1}\ma\mb, q^2\mb z \\
q^2\mb, 0 \end{matrix}\,; q^2,q^2\right].
\eeqa

\end{rem}

\begin{rem} Note that the little q-Jabobi operator $\tG_1$ is also bidiagonal in the monomial basis $z^n$.  By analogy with the proof of Lemma \ref{lem2} one can retrieve the following familar explicit expression for the little q-Jacobi polynomial \cite[eq. (3.12.1)]{KS} (see also \cite[eq. (2.45)]{Kor}):

\beqa
p_n(z;\ma,\mb;q^2) = \   _2\phi_1(q^{-n} , q^{n+1}\ma\mb, q^2\ma ; q^2;q^2 z).
\eeqa

\end{rem}

\vspace{2mm}

\section{Tridiagonalization of the little $q-$Jacobi operator and the equitable presentation of $\mathfrak U_q(sl(2))$}

Calling upon the equitable embedding of the Askey-Wilson algebra \cite{T} which will be recalled next, we shall now make the observation that the little q-Jacobi operator and a tridiagonalization of this operator realizes also the Askey-Wilson algebra.

\begin{rem} Define

\beqa
g_2=(1-q^{-2})q^{1/2}b^{-1},\quad g_3=-q^{-3/2}(q-q^{-1})ca^{-1},\quad g_4=g_5^{-1}=b, \quad g_6=0.
\eeqa

Then

\beqa
\tG_1&=& (1-q^{-2})q^{1/2}b^{-1}  S_-q^{-s_3} -  q^{-3/2}(q-q^{-1})ca^{-1}  S_-q^{s_3} + b q^{2s_3} + b^{-1}q^{-2s_3}\label{B}\\
 &=& bY + b^{-1}Z +qca^{-1}(1-YZ)\ ,\nonumber
\eeqa

where $Y$ and $Z$ are the equitable $\mathfrak U_q(sl(2))$ generators introduced in (\ref{XYZ}).

\end{rem}

Using the equitable presentation of $\mathfrak U_q(sl(2))$  (\ref{XYZ}), by Proposition 1.1 and Lemma 3.4 in \cite{T}, one gets:

\begin{prop} Define

\beqa
A&=& aX + a^{-1}Y + qbc^{-1}(1-XY),\label{ABC}\\
B&=&  bY + b^{-1}Z +qca^{-1}(1-YZ),\nonumber\\
C&=& cZ + c^{-1}X + qab^{-1}(1-ZX).\nonumber
\eeqa

The generators $A,B,C$ satisfy the Askey-Wilson algebra:

\beqa
A + \frac{qBC-q^{-1}CB}{q^2-q^{-2}} &=& \frac{\Lambda(a+a^{-1}) + (b+b^{-1})(c+c^{-1})}{q+q^{-1}}\ ,\nonumber\\
B + \frac{qCA-q^{-1}AC}{q^2-q^{-2}} &=& \frac{\Lambda(b+b^{-1}) + (c+c^{-1})(a+a^{-1})}{q+q^{-1}}\ ,\nonumber\\
C + \frac{qAB-q^{-1}BA}{q^2-q^{-2}} &=& \frac{\Lambda(c+c^{-1}) + (a+a^{-1})(b+b^{-1})}{q+q^{-1}}\ ,\nonumber
\eeqa

where $\Lambda$ denotes the `normalized' Casimir element

\beqa
\Lambda=(q-q^{-1})^2\Omega.\nonumber
\eeqa

\end{prop}

\begin{rem} In terms of the Chevalley generators of $\mathfrak U_q(sl(2))$, one has $B=\tG_1$ with (\ref{B}) and

\beqa
A&=&  -a(q-q^{-1})q^{1/2}S_+q^{-s_3} + bc^{-1}(q^2-1)q^{1/2} S_+q^{s_3} +   a^{-1}q^{2s_3} +  aq^{-2s_3} ,\label{A}\\
C&=& (c+c^{-1})q^{-2s_3}  - ab^{-1}(q+q^{-1})q^{-4s_3} + ab^{-1}(q-q^{-1})^2\Omega q^{-2s_3} \nonumber\\
&& + (1-q^{-2}) \left(S_-(cq^{1/2} q^{-s_3} -q^{3/2}ab^{-1}q^{-3s_3})  - S_+(c^{-1}q^{3/2}q^{-s_3} -ab^{-1}q^{1/2}q^{-3s_3})\right).\nonumber
\eeqa
\end{rem}

For the analysis to follow, let us introduce the operators $O$ and $\bO$:

 \beqa
&& O = (q-q^{-1}) q^{-\nu} S_+ q^{s_3}(1-q^{-2\nu}q^{2s_3})^{-1}, \quad \bO = (q-q^{-1}) q^{-\nu-1} q^{-s_3}S_-(1-q^{-2\nu}q^{-2s_3})^{-1},\label{ObO}
\eeqa

where $(1-q^{-2\nu}q^{\pm 2s_3})^{-1}$ is understood as a power series in the Cartan element of $\mathfrak U_q(sl(2))$. Note that
$O\bO=\bO O=1$. Also,  the action of each operator  on the space of monomials in $z$, according to (\ref{holo}), is such that:

\beqa
O \mapsto z, \quad \bO \mapsto z^{-1}.
\eeqa

\begin{lem}\label{ABz} Consider $A,B$ as the $q-$difference operators obtained from (\ref{ABC}) with (\ref{XYZ}). One has:

\beqa
A \equiv \alpha O B + \beta B O + \gamma O  + \delta ,\label{Texp}
\eeqa

with

\beqa
\alpha&=& -\frac{q(q^{3/2+\nu}ab-q^{-3/2-\nu}c^{-1})}{q^2-q^{-2}} ,\quad \beta= \frac{(q^{1/2+\nu}ab-q^{3/2-\nu}c^{-1})}{q^2-q^{-2}} ,\label{sol1}\\
 \gamma &=&  bc^{-1}q^{3/2+\nu} +aq^{1/2-\nu},\nonumber\\
 \delta&=&  \frac{(q^{2\nu+1}a+ q^{-2\nu-1}a^{-1}  +bc + b^{-1}c^{-1})}{q+q^{-1}}. \nonumber
\eeqa

\end{lem}

\begin{proof}
Recall that $B=\tG_1$ is given by (\ref{B}). Consider the r.h.s. of (\ref{Texp}). The action of $O$ on the monomials $S_-q^{\pm s_3},q^{\pm 2s_3}$ is such that:

\beqa
OS_-q^{-s_3} &=& \frac{q^{\nu} -q^{-\nu}q^{-2s_3}}{q-q^{-1}}, \quad  OS_-q^{s_3} = \frac{-q^{-\nu} +q^{\nu}q^{2s_3}}{q-q^{-1}},\nonumber\\
S_-q^{-s_3}O &=& \frac{q^{\nu} -q^{-\nu-2}q^{-2s_3}}{q-q^{-1}}, \quad  S_-q^{s_3}O = \frac{-q^{-\nu} +q^{\nu+2}q^{2s_3}}{q-q^{-1}},\nonumber\\
q^{2s_3}O &=& -(q-q^{-1})q^{\nu+2}S_+q^{s_3} +q^{2\nu+2}O,\quad q^{-2s_3}O = (q-q^{-1})q^{-\nu-2}S_+q^{-s_3} +q^{-2\nu-2}O.\nonumber
\eeqa

 By straightforward calculations, we obtain:

\beqa
\alpha OB + \beta BO + \gamma O  + \delta &=& -b(q-q^{-1})q^\nu(\alpha+\beta q^2)S_+q^{s_3} + b^{-1}(q-q^{-1})q^{-\nu}(\alpha+\beta q^{-2})S_+q^{-s_3} \nonumber\\
&& -  ca^{-1}q^{-3/2}q^\nu(\alpha+\beta q^2)q^{2s_3} -   b^{-1}q^{-1/2}q^{-\nu}(\alpha+\beta q^{-2})q^{-2s_3} \nonumber\\
&&   + \left( \alpha(bq^{2\nu} + b^{-1}q^{-2\nu}) + \beta (bq^{2\nu+2} + b^{-1}q^{-2\nu-2}) + \gamma \right)O\nonumber\\
&&  + \left(b^{-1}q^{\nu-1/2} + ca^{-1}q^{-\nu-3/2}\right)(\alpha+\beta) + \delta.\nonumber
\eeqa

Compare the expression above with (\ref{A}), we obtain the constraints:

\beqa
&& a=-b^{-1}q^{-1/2-\nu}(\alpha+\beta q^{-2}), \quad c = -\frac{q^{-\nu+3/2}}{(\alpha+\beta q^{2})}, \nonumber\\
&&  \alpha(bq^{2\nu} + b^{-1}q^{-2\nu}) + \beta (bq^{2\nu+2} + b^{-1}q^{-2\nu-2}) + \gamma = 0,\nonumber\\
&& \left(b^{-1}q^{\nu-1/2} + ca^{-1}q^{-\nu-3/2}\right)(\alpha+\beta) + \delta =0\nonumber
\eeqa

from which we get (\ref{sol1}).

\end{proof}

This lemma thus states that the generator $A$ of the equitable presentation of the Askey-Wilson algebra is obtained from $B$ by tridiagonalization. Stated differently, the upshot is that $B$ i.e. $\tG_1$ together with its tridiagonalization (\ref{Texp})
generate the Askey-Wilson algebra.

The tridiagonalized form of $A$ can be inverted to express $B$ in terms of $A$ as follows.

\begin{lem}  Consider $A,B$ as the $q-$difference operators obtained from (\ref{ABC}) with (\ref{XYZ}). One has:

\beqa
B \equiv \alpha \bO A + \beta A \bO + \gamma \bO  + \delta ,\label{TBexp}
\eeqa

with

\beqa
\alpha&=& -\frac{q^{-1}(q^{3/2+\nu}c-q^{-3/2-\nu}a^{-1}b^{-1})}{q^2-q^{-2}} ,\quad \beta= \frac{(q^{-3/2+\nu}c-q^{-1/2-\nu}a^{-1}b^{-1})}{q^2-q^{-2}} ,\label{sol2}\\
 \gamma &=&  b^{-1}q^{-1/2+\nu} +ca^{-1}q^{-3/2-\nu},\nonumber\\
 \delta&=&  \frac{(q^{2\nu+1}b+ q^{-2\nu-1}b^{-1}  + ac + a^{-1}c^{-1})}{q+q^{-1}}.\nonumber
\eeqa
\end{lem}

\begin{proof}
Recall $A$ is given by (\ref{A}). Consider the r.h.s. of (\ref{TBexp}). The action of $\bO$ on the monomials $S_+q^{\pm s_3},q^{\pm 2s_3}$ is such that:

\beqa
\bO S_+q^{s_3} &=& \frac{q^{\nu} -q^{-\nu}q^{2s_3}}{q-q^{-1}}, \quad  \bO S_+q^{-s_3} = \frac{-q^{-\nu} +q^{\nu}q^{-2s_3}}{q-q^{-1}},\nonumber\\
S_+q^{s_3}\bO &=& \frac{q^{\nu} -q^{-\nu-2}q^{2s_3}}{q-q^{-1}}, \quad  S_+q^{-s_3}\bO = \frac{-q^{-\nu} +q^{\nu+2}q^{-2s_3}}{q-q^{-1}},\nonumber\\
q^{-2s_3}\bO &=& -(q-q^{-1})q^{\nu+2}S_-q^{-s_3} +q^{2\nu+2}\bO,\quad q^{2s_3}\bO = (q-q^{-1})q^{-\nu-2}S_-q^{s_3} +q^{-2\nu-2}\bO.\nonumber
\eeqa

 By straightforward calculations, we obtain:

\beqa
\alpha \bO A + \beta A\bO + \gamma \bO  + \delta &=& a^{-1}(q-q^{-1})q^{-\nu}(\alpha+\beta q^{-2})S_-q^{s_3} - a(q-q^{-1})q^{\nu}(\alpha+\beta q^{2})S_-q^{-s_3} \nonumber\\
&& -  bc^{-1}q^{3/2-\nu}(\alpha+\beta q^{-2})q^{2s_3} -   aq^{\nu+1/2}(\alpha+\beta q^{2})q^{-2s_3} \nonumber\\
&&   + \left( \alpha(aq^{2\nu} + a^{-1}q^{-2\nu}) + \beta (aq^{2\nu+2} + a^{-1}q^{-2\nu-2}) + \gamma \right)\bO\nonumber\\
&&  + \left(aq^{1/2-\nu} + bc^{-1}q^{3/2+\nu}\right)(\alpha+\beta) + \delta.\nonumber
\eeqa

Compare the expression above with (\ref{B}), we obtain the constraints:

\beqa
&& c= -(\alpha + q^{-2}\beta)q^{3/2-\nu},\quad b^{-1}= - (\alpha + q^2\beta)aq^{\nu+1/2}, \nonumber\\
&&  \alpha(aq^{2\nu} + a^{-1}q^{-2\nu}) + \beta (aq^{2\nu+2} + a^{-1}q^{-2\nu-2}) + \gamma = 0,\nonumber\\
&& \left(aq^{1/2-\nu} + bc^{-1}q^{3/2+\nu}\right)(\alpha+\beta) + \delta =0\nonumber
\eeqa

from which we get (\ref{sol2}).
\end{proof}

\section{The big $q-$Jacobi polynomials}

We shall now carry for the big q-Jacobi polynomials an analysis similar to the one that was given in Section 3 for the little q-Jacobi polynomials; namely, identify the specialization of the parameters in $G_1$ that will lead to the big q-Jacobi operator, identify the eigenfunctions of $W_0$ in that case, determine the action of the restricted $G_1$ on these functions and arrive at the explicit expression of the big q-Jacobi polynomials by expanding one eigenbasis over the other.

\subsection{The second order q-difference operator $G_1$}

Recall (\ref{holo}). By straightforward replacements, from (\ref{G1}) we get for the specialization\footnote{We choose $\mu_0=c_0=0$ in (\ref{W}).} $g'_2=g'_3=0$:

\beqa
G_1|_{g'_2=g'_3=0} &\mapsto& \underbrace{ \left( \frac{g_1q^{-1}}{(q-q^{-1})^2}\frac{1}{z^2}  +\frac{g_3q^{-\nu}}{(q-q^{-1})}\frac{1}{z} + g_4q^{-2\nu} \right) }_{A_0}T_+^2 \label{G1exp}\\
&& + \underbrace{\left( \frac{g_1q}{(q-q^{-1})^2}\frac{1}{z^2}  -\frac{g_2q^{\nu}}{(q-q^{-1})}\frac{1}{z} + g_5q^{2\nu} \right) }_{B_0}T_-^2  \nonumber\\
&& + \underbrace{\left(  -\frac{g_1(q+q^{-1})}{(q-q^{-1})^2}\frac{1}{z^2}  -\frac{(g_3q^{-\nu} - g_2q^{\nu} ) }{(q-q^{-1})}\frac{1}{z} + g_6\right)}_{C_0}.\nonumber
\eeqa

\begin{prop} For  $g'_2=g'_3=0$, the operator $G_1$ is diagonalized by the big q-Jacobi polynomials.
\end{prop}
\begin{proof} Compare the operator $G_1$ written as (\ref{G1exp}) with the second-order q-difference operator\footnote{Here $q^2$ is $q$ of \cite{KS}.} (3.5.4) in \cite{KS}. Denote the big q-Jacobi polynomials as:

\beqa
y(z) \equiv P_n(z;\ma,\mb,\mc;q^2)\nonumber
\eeqa

They satisfy the second-order q-difference equation:

\beqa
q^{-2n}(1-q^{2n})(1-\ma\mb q^{2n+2})y(z) = B(z) y(q^2z) + \overline{B}(z)y(q^{-2}z) - (B(z)+\overline{B}(z))y(z),\label{qdiff}
\eeqa

where

\beqa
B(z) &=& \ma\mc q^2 \frac{1}{z^2} - \ma(\mb+\mc)q^2 \frac{1}{z} + \ma\mb q^2,\nonumber\\
\overline{B}(z)&=&\ma\mc q^4 \frac{1}{z^2} - (\ma+\mc)q^2 \frac{1}{z} + 1,\nonumber
\eeqa

The exact relation between the parameters $g_i$ and the parameters entering in the big q-Jacobi polynomials is as follows:

\beqa
G_1 y(z) =\left( g_5q^{2\nu}(q^{-2n}+\ma\mb q^{2n+2}) +g_6\right)y(z)\nonumber
\eeqa

and

\beqa
&&\ma\mc = \frac{g_1}{g_5}\frac{q^{-2\nu-3}}{(q-q^{-1})^2},\quad \ma(\mb+\mc)= -\frac{g_3}{g_5}\frac{q^{-3\nu-2}}{(q-q^{-1})},\quad  \ma\mb = \frac{g_4}{g_5}q^{-4\nu-2},\quad \ma+\mc =\frac{g_2}{g_5}\frac{q^{-\nu-2}}{(q-q^{-1})}.
\eeqa

\vspace{2mm}

\begin{rem}Note that if instead we would like to consider the spectral problem for the big q-Jacobi polynomials

\beqa
y(q^2\mb z) \equiv P_n(q^2\mb z;\ma,\mb,\mc;q^2),
\eeqa

the substitution $z \rightarrow q^2\mb z$ into (\ref{qdiff}) gives the following identification:

\beqa
&&\frac{\ma\mc}{\mb^2} = \frac{g_1}{g_5}\frac{q^{-2\nu+1}}{(q-q^{-1})^2},\quad \frac{\ma(\mb+\mc)}{\mb}= -\frac{g_3}{g_5}\frac{q^{-3\nu}}{(q-q^{-1})},\quad  \ma\mb = \frac{g_4}{g_5}q^{-4\nu-2},\quad \frac{\ma+\mc}{\mb} =\frac{g_2}{g_5}\frac{q^{-\nu}}{(q-q^{-1})}.
\eeqa

For the special choice $\mc=0$, note that the big q-Jacobi polynomial $P_n(q^2\mb z;\mb,\ma,0;q^2)$ can be expressed in terms of the little q-Jacobi polynomial $p_n( z;\ma,\mb;q^2)$. See \cite{KS} for details.
\end{rem}

\end{proof}

\subsection{Eigenfunctions and the explicit expression of the big $q-$Jacobi polynomials}

Consider the eigenfunctions of $W_0$ in (\ref{W}) for $g'_2=g'_3=0$. They coincide with the ones given in Lemma \ref{lem1}.
On these eigenfunctions of $W_0$, we now consider the action of $G_1$.

\begin{lem} One has:

\beqa
G_1 f_n(z) = a_n f_n(z) + b_n f_{n-1}(z)\label{actG1}
\eeqa

with

\beqa
a_n&=& g_5q^{2\nu}(q^{-2n}+\ma\mb q^{2n+2}) +g_6,\nonumber\\
b_n&=&  - g_5 q^{2\nu}q^{-2n}(1-q^{2n})(1-q^{2n}\ma)(1-q^{2n}\mc).\nonumber
\eeqa
\end{lem}

\begin{proof} Recall (\ref{G1exp}). The l.h.s. of (\ref{actG1}) reads:

\beqa
G_1 f_n(z) &=& \left(A_0 (1-\gamma_0 q^{2n-2}z)(1-\gamma_0 q^{2n}z) + B_0 (1-\gamma_0 q^{-2}z)(1-\gamma_0 z) \right.\nonumber\\
&& \left.+ C_0 (1-\gamma_0 z)(1-\gamma_0 q^{2n-2}z)\right)(1-\gamma_0 q^{2}z)\cdots (1-\gamma_0 q^{2n-4}z),\nonumber
\eeqa

where  $\gamma_0=\frac{\epsilon_0}{\oc_0}(1-q^2)q^{-\nu-1}$. The r.h.s reads:

\beqa
\left(a_n (1-\gamma_0 z)(1-\gamma_0 q^{2n-2}z) + b_n(1-\gamma_0 z)\right)(1-\gamma_0 q^{2}z)\cdots (1-\gamma_0 q^{2n-4}z).\nonumber
\eeqa

Equating both sides of the equation (\ref{acttG1}), the coefficients $a_n$ and $b_n$ are determined uniquely.
\end{proof}

\begin{rem} Observe that the coefficient $a_n$ coincides with the spectrum of $G_1$, as expected ($G_1$ is lower triangular in the basis $\{f_n\}$).
\end{rem}

Next, we are interested in the overlap coefficients between the eigenfunctions of $W_0$ (the $f_n(z)$) and the eigenfunctions of $G_1$ (the big q-Jacobi polynomials). The proof of the following proposition is straightforward, by analogy with the proof of Prop. \ref{prop2}

\begin{prop} One has:

\beqa
P_n(q^2 \mb z;\ma,\mb,\mc;q^2)= \sum_{s=0}^{\infty} \gamma'_{n,s}f_s(z) \label{expPn}
\eeqa

where

\beqa
\gamma'_{n,s}=  \frac{ (q^{-2n};q^2)_s (\ma\mb q^{2n+2};q^2)_s}{(q^2\ma;q^2)_s (q^2\mc;q^2)_s (q^2;q^2)_s}q^{2s} .\label{gam2}
\eeqa

\end{prop}

\begin{rem} In particular,  note that:

\beqa
 p_n(z;\ma,\mb;q^2) =  (-q^2 \mb)^{-n} q^{-n(n-1)} \frac{(q^2\mb;q^2)_n}{(q^2\ma;q^2)_n}  P_n(q^2 \mb z;\mb,\ma,0;q^2).
\eeqa
\end{rem}

\section{A connection between some big $q-$Jacobi polynomials and special little $q-$Jacobi polynomials by tridiagonalization}

Much as in \cite{IK1} and \cite{IK2} we shall  supplement the observations of the preceding section by noting that a large class of big q-Jacobi polynomials can be obtained from a particular tridiagonalization of the special little q-Jacobi operator with (\ref{lab}) for $\ma = q^2$.

\begin{lem}  Consider the little q-Jacobi operator $\tG_1$ and the big q-Jacobi operator $G_1$ of the form:

\beqa
G_1&=&g_1 S_-^2 + g_2 S_-q^{-s_3} +  g_3 S_-q^{s_3} + g_4 q^{2s_3} + g_5q^{-2s_3} + g_6,\\
\tG_1&=& \tg_2 S_-q^{-s_3} +  \tg_3 S_-q^{s_3}  + \tg_4 q^{2s_3} + \tg_5q^{-2s_3} + \tg_6.\label{bigqJ}
\eeqa

For $\tg_3=-q^{2\nu+2}\tg_2$, one has:

\beqa
G_1 \equiv  \beta \tG_1 \bO +  \gamma \tG_1 + \gamma' \bO + \delta \label{TGGexp}
\eeqa

with (\ref{ObO}) and

\beqa

\qquad  \beta&=&  \frac{(q^{\nu+1}g_2 + q^{-\nu-1}g_3)}{(q-q^{-1})(q^{-2\nu-3}\tg_4 - q^{2\nu+3}\tg_5)},\label{sol3}\\
 \gamma &=&   \frac{q^{-\nu-1}(q^{\nu+2}g_3\tg_5 + q^{-\nu-2}\tg_4 g_2)}{\tg_2(q^{-2\nu-3}\tg_4 - q^{2\nu+3}\tg_5)}  ,\quad \gamma'=
-\frac{(q^{-2\nu-2}\tg_4 + q^{2\nu+2}\tg_5 + \tg_6)(q^{\nu+1}g_2 +q^{-\nu-1}g_3  )}{(q-q^{-1})(q^{-2\nu-3}\tg_4 - q^{2\nu+3}\tg_5)},
\nonumber\\
 \delta&=& g_6-\gamma \tg_6.\  \nonumber
\eeqa

\end{lem}

\begin{proof}
Consider the l.h.s. of (\ref{TGGexp}). The monomial $S_-^2$ is easily reduced in terms of $\bO S_-q^{\pm s_3}$, namely:

\beqa
S_-^2 = -\frac{q^{1-\nu}}{q-q^{-1}}\bO S_- q^{-s_3}  +  \frac{q^{\nu-1}}{q-q^{-1}} \bO S_-q^{s_3}.
\eeqa

Consider the r.h.s.  of (\ref{TGGexp}). Using:

\beqa
S_-q^{-s_3}\bO = q^2 \bO S_- q^{-s_3} - \bO^2 q^{\nu+1},\qquad S_-q^{s_3}\bO = q^{-2} \bO S_- q^{s_3} - \bO^2 q^{-\nu-1},
\eeqa

by straightforward calculations, for a slightly more general combination we obtain:

\beqa
\alpha \bO \tG_1 + \beta \tG_1\bO  +\gamma \tG_1 + \gamma' \bO   + \delta &=&
-  (\alpha + \beta q^{2})(q-q^{-1})q^{\nu-1}\tg_2  S_-^2 \nonumber\\
&& +  \left((\alpha + \beta q^{-2})(q-q^{-1})q^{-\nu}\tg_4 + \gamma \tg_3\right) S_-q^{s_3} \nonumber\\
&& +  \left(-(\alpha + \beta q^{2})(q-q^{-1})q^{\nu}\tg_5 + \gamma \tg_2\right) S_-q^{-s_3} \nonumber\\
&& + \gamma \tg_4 q^{2s_3} + \gamma \tg_5 q^{-2s_3} + \gamma \tg_6 + \delta\nonumber\\
&& + \left( (\alpha + \beta q^{-2})\tg_3 + (\alpha + \beta q^{2})\tg_2 q^{2\nu-2}\right)\bO S_-q^{s_3} \nonumber\\
&& + \left( (\alpha + \beta q^{-2})\tg_4 q^{-2\nu}  +  (\alpha + \beta q^{2})\tg_5 q^{2\nu}+ (\alpha+\beta)\tg_6+\gamma'\right) \bO \nonumber\\
&& - \beta (\tg_2 q^{\nu+1} + \tg_3 q^{-\nu-1} ) \bO^2. \nonumber
\eeqa

Comparing the r.h.s and l.h.s of (\ref{TGGexp}), we identify:

\beqa
&&   g_1 = -(\alpha + \beta q^2)(q-q^{-1}) q^{\nu-1}\tg_2\ ,\label{coefg}\\
&& g_2=-(\alpha + \beta q^{2})(q-q^{-1})q^{\nu}\tg_5 + \gamma \tg_2\ ,\nonumber\\
&&  g_3 = (\alpha + \beta q^{-2})(q-q^{-1})q^{-\nu}\tg_4 + \gamma \tg_3 \ ,\nonumber\\
&& g_4= \gamma \tg_4,\quad  g_5 =  \gamma \tg_5 \ \nonumber, \quad g_6 = \gamma \tg_6 + \delta \nonumber
\eeqa

together with the constraints:

\beqa
&&\tg_2 q^{\nu+1} + \tg_3 q^{-\nu-1}=0 \ ,\\
&&   (\alpha + \beta q^{-2})\tg_3 + (\alpha + \beta q^{2})\tg_2 q^{2\nu-2} =0\ ,\nonumber\\
&& (\alpha + \beta q^{-2})\tg_4 q^{-2\nu}  +  (\alpha + \beta q^{2})\tg_5 q^{2\nu} + (\alpha+\beta)\tg_6+\gamma'=0\ .\nonumber
\eeqa

The first constraint gives $\tg_3=-q^{2\nu+2}\tg_2$. Inserting in the second constraint, one gets $\alpha=0$. The other equations imply
 (\ref{sol3}).
\end{proof}

Note from eq. (\ref{lab}) that the condition $\tg_3 =  - q^{2\nu +2} \tg_2$ implies that $\ma=q^2$. Thus, for a large parameter set, big q-Jacobi polynomials can be constructed from little q- Jacobi polynomials with that special value of the parameter $\ma$ by using the one-sided tridiagonalization introduced in \cite{IK1} and \cite{IK2}.

\section{Conclusion}

This paper has shown how the little and big q-Jacobi polynomials form bases  for representations of the (specialized) Askey-Wilson algebra. The starting point has been the embedding of the Askey-Wilson algebra in $\mathfrak U_q(sl(2))$ realized in terms of q-difference operators. It then proved possible to identify within the model the defining operators of the polynomials; once this was done the algebraic structure could be used to obtain, for instance, the explicit expressions of these polynomials in a simple fashion. The realization was also instrumental in the observation that the little q-Jacobi operator and a tridiagonalization of this operator realized together the Askey-Wilson algebra since they could readily be identified as generators in the equitable embedding of this algebra in $\mathfrak U_q(sl(2))$.\vspace{1mm}

It should similarly prove interesting to adopt the somewhat opposite viewpoint, that is to start from the little and big q-Jacobi operators and their tridiagonalizations and to look at the conditions for these operators to form Askey-Wilson algebras. Recall that the most general tridiagonalization of the ordinary Jacobi operator led to the standard Heun operator \cite{GrVZ}. The program thus suggested would therefore involve certain q-analogs of the Heun equation. An interesting question that is complementary to those pursued in the present paper is: under what specializations will these q-Heun operators form Askey-Wilson algebras with the original operators. We plan to pursue this matter in a future publication.

\vspace{4mm}

\noindent{\bf Acknowledgements:} We thank Paul Terwilliger for comments. L.V. would like to express his gratitude for the hospitality extended to him
by the Institut Denis-Poisson of the Universit\'e Fran\c cois-Rabelais de Tours as Chercheur
Invit\'e where most of this research was carried out. The research of L.V. is funded in part by
a discovery grant from the Natural Sciences and Engineering Research Council (NSERC) of Canada.
 P.B. is supported by C.N.R.S.  Work of A.Z. is supported by the National Science Foundation of China (Grant No.11771015).

\vspace{0.2cm}

\begin{appendix}

\end{appendix}

\vspace{0.5cm}


TITLE: Why do particles in the molten wax near a burning candle wick get pushed away?
QUESTION [32 upvotes]: Have a look at this clip I made of my candle burning:

I can accept that the molten wax gets sucked into the wick (though not quite sure why: capilary action?).
But why does it also seem to get thrown outwards when it reaches the wick? I would expect a build-up of residue (particles, dust) near the wick, if anything.

REPLY [1 votes]: @rodrigo answer is partially correct. One thing he forgot to mention was the effect of the surface tension. The motion of the particles in the molten wax pool is a combination of cellular convection currents coupled with diffusion due to the so-called Marangoni effect. The viscosity of wax is dependent on its temperature. Wax close to the flame surface is less viscous, so the particles can move faster in it, with less friction,when close to the flame. The particles begin to slow down as they approach the cooler wax further away from the flame which has higher viscosity. As the particle slows down, a diffusion dominated force due to the surface tension difference allows the particle to move back towards the flame.
For additional information: see
Fluid Mechanics of a Shallow Fuel Layer Near a Burning Wick


TITLE: + (plus) notation in regular expression
QUESTION [1 upvotes]: I'm a student learning regular expression. today while studying I saw the question
Regular expression (A*B*)* = (A+B)* proof.
for me + notation means one or more, while * means zero or more.
So $(A^*B^*)^*$ accepts $A$ but $(A+B)^*$ doesn't because the will be always a $B$ after an $A$. in the other words for me, $(A+B)$ accepts $A...AB$ strings. so the question is trivially wrong and doesn't need to be asked.
But as I can see, the questioner and answerers don't think like that. So the definition of + notation must be different for us.
I'm here to ask what does it mean here if I'm wrong? thanks in advance.

REPLY [1 votes]: Here the operator $+$ means alternation. It does not mean repeating the previous symbol one or more times. So, $(A+B)^{*}$ means: zero or more occurrences of $A+B$, which means zero or more occurrences of either $A$ or $B$.
Valid strings are: $\varepsilon,A,AA,AAA,B,BB,BBB,ABBA,BABA,\ldots$
We observe $(A+B)^*$ produces the same words as $(A^{*}B^{*})^*$.