TITLE: Why degree of a reducible projective variety is the sum of the degree of its irreducible components
QUESTION [6 upvotes]: Could anyone show me how to prove that

The degree of a reducible projective variety is the sum of the degree of
  its irreducible components?

The definition of the degree I know is quite vague, saying that the degree of a projective variety $X$ is the maximal number of possible finite intersections of $X$ with a linear hyperspace in general position.
I'm not comfortable with the concept in general position. The note I'm reading explains it by an example. I guess that in general position just means there isn't multiple intersection.

REPLY [12 votes]: As mentioned in the comments, the best way to prove statements like this is to use the Hilbert polynomial definition of degree. For this point of view, see section I.7 of Algebraic Geometry by Hartshorne.
Let $R = k[x_0,\ldots, x_n]$ where $k$ is an algebraically closed field viewed as a graded ring. For any graded $R$-module $M$, the Hilbert function is the function
$$
h(l) = \dim_k M_l
$$
giving the dimension as a $k$-vector space of the graded pieces of $M$. The idea is that for large enough $l$ this agrees with a polynomial $P_M(l)$ which is the Hilbert polynomial of $M$. Then for a projective variety $X \hookrightarrow \mathbb{P}^n$, the Hilbert polynomial $P_X(l)$ is just the Hilbert polynomial of the homogeneous coordinate ring of $X$ as a graded module over $R$. You can show that the degree of $P_X(l)$ is $d = \dim X$ and then we define the degree of $X$ to be $d!$ times the leading coefficient of $P_X(l)$
The nice thing about the Hilbert polynomial is that it behaves well with exact sequences and this gives it the geometric properties we want and expect. This is because $\dim_k$ is additive on exact sequences.
In particular, if we have that $X = Y_1 \cup Y_2$ with $Y_1$ and $Y_2$ the same dimension and intersecting in a lower dimension, then we can write the exact sequence
$$
0 \to R/I \to R/I_1 \oplus R/I_2 \to R/(I_1 + I_2) \to 0
$$
where $I_i$ is the homogeneous ideal of $Y_i$ and $I$ is the homogeneous ideal of $X$. Then by additivity of the Hilbert polynomial,
$$
P_{R/I_1 \oplus R/I_2} = P_{R/I} + P_{R/(I_1 + I_2)}.
$$
Applying additivity of the Hilbert polynomial again, we see that the left hand side of this equation is in fact $P_{R/I_1} + P_{R/I_2}$. Rephrasing this geometrically, we see that
$$
P_{Y_1} + P_{Y_2} = P_{X} + P_{Y_1 \cap Y_2}.
$$
Since $Y_i$ were assumed to be the same dimension, the leading coefficient of the left hand side is the sum $\deg{Y_1}/d! + \deg{Y_2}/d!$. Similarly, on the right hand side, since we assumed $Y_1 \cap Y_2$ is lower dimensional than all of $X$, we have that the leading coefficient of the right hand side is just that of $P_{X}$, that is, $\deg{X}/d!$, giving us the equality $\deg{X} = \deg{Y_1} + \deg{Y_2}$.
Now you can deduce the more general case with a little more work by applying this to the irreducible components.
The proof that this gives the same definition of degree as the one you gave is a little involved but it uses exactly the same technique. Write down an exact sequence whose terms correspond to the varieties we are intersecting and compare the two sides of the equation we get for the Hilbert polynomials. However, it requires some commutative algebra.
Edit: I wanted to add a bit about the dimension considerations since you brought that up as something you had issue with. If you notice, my argument above implies something a little different from your statement. It says that the degree of a variety is the sum of degrees of the highest dimensional irreducible components. This is because the contribution of the lower dimensional components to the Hilbert polynomial will not affect the leading coefficient which is the same degree as the dimension.
How does this reconcile with the classical notion of degree? The idea is that the lower dimensional components won't affect the intersection with your general plane. The reason for this is that if we have a $k$ dimensional subvariety $Y$ of $\mathbb{P}^n$, the classical degree is the number of points in the intersection with an $n - k$ plane. "Most" $n-k$ planes will certainly miss any components of dimension less than $k$ (think for example a point and a line in $\mathbb{P}^3$) and thus shouldn't contribute to the classical notion of degree, and indeed with the Hilbert polynomial argument, we see that they don't. Hopefully this fixes part of your confusion about how the dimension affects things.


TITLE: Prove using weak law of large numbers
QUESTION [0 upvotes]: Can someone help me out?

Let $X_1,X_2,...,X_n\sim N(0,1)$ with
  $$
  \mathbb{C}\textrm{ov}(X_{i},X_{j})=\begin{cases}
  \begin{array}{ccc}
\rho & \textrm{if} & \left|i-j\right|=1\\
0 & \textrm{if} & \left|i-j\right|>1
\end{array}\end{cases}
$$
Prove that $\dfrac{S_{n}}{n}\xrightarrow{\text{P}}0$.

REPLY [0 votes]: I have the solution.
We have to prove:
$\dfrac{S_{n}}{n}\overset{\mathrm{P}}{\longrightarrow}0\;\Leftrightarrow\;\mathbb{P}\left(\left|\dfrac{S_{n}}{n}-0\right|>\epsilon\right)\underset{n\rightarrow\infty}{\longrightarrow}0$.
Since $\mathbb{E}\left(\dfrac{S_{n}}{n}\right)=\dfrac{1}{n}\cdot\mathbb{E}\left(S_{n}\right)=\dfrac{1}{n}\cdot\mathbb{E}\left(\sum_{i=1}^{n}X_{i}\right)=\dfrac{1}{n}\cdot\sum_{i=1}^{n}\mathbb{E}\left(X_{i}\right)=0$ and using Chevichev's inequality we have:
$\mathbb{P}\left(\left|\dfrac{S_{n}}{n}-0\right|>\epsilon\right)\leq\dfrac{\mathbb{V}\textrm{ar}\left(\dfrac{S_{n}}{n}\right)}{\epsilon^{2}}=\dfrac{\dfrac{1}{n^{2}}\cdot\mathbb{V}\textrm{ar}\left(S_{n}\right)}{\epsilon^{2}}=\dfrac{\mathbb{V}\textrm{ar}\left(S_{n}\right)}{n^{2}\epsilon^{2}}=\dfrac{\mathbb{V}\textrm{ar}\left(\sum_{i=1}^{n}X_{i}\right)}{n^{2}\epsilon^{2}}=\dfrac{\sum_{i=1}^{n}\mathbb{V}\textrm{ar}\left(X_{i}\right)+2\sum_{i<j}\mathbb{C}\textrm{ov}\left(X_{i},X_{j}\right)}{n^{2}\epsilon^{2}}=\dfrac{n+2(n-1)\rho}{n^{2}\epsilon^{2}}\underset{n\rightarrow\infty}{\longrightarrow}0$


TITLE: Finding roots of equation
QUESTION [5 upvotes]: $$r\left(\frac 1p -1\right)- \left(\frac 1p\right)^\frac89 + \left(\frac 1p\right)^\frac19 =0$$
where $ r = 2^\frac89 - 2^\frac19$
How do I solve this without a computer?

REPLY [3 votes]: First of all, there no need to express as 1/p so
$(2^{\frac 8 9} - 2^\frac 1 9)(m - 1) - m^{\frac 8 9} + m^{\frac 1 9} = 0$
$(2^{\frac 8 9} - 2^\frac 1 9)(m - 1) - (m^{\frac 8 9} - m^{\frac 1 9}) = 0$
Oh look! The $(m^{\frac 8 9} - m^{\frac 1 9})$ looks like  $(2^{\frac 8 9} - 2^\frac 1 9)$!  What if $m = 2$?  Then $m - 1$ = 1 and we'd get an equality.  So $p = \frac 1 2$ is a solution.
But that was us seeing something.  Ho do we solve in general?
$(2^{\frac 8 9} - 2^\frac 1 9) = \frac{ (m^{\frac 8 9} - m^{\frac 1 9})}{m-1} $
$2^{\frac 1 9}(2^8 - 2) = \dfrac {m'^8 - m} {m'^9 - 1}$ for $m' = m^{\frac 1 9}$....
You know what? let's just say $p = \frac 1 2$ and call it a day.


TITLE: Circle bisecting the circumference of another circle
QUESTION [2 upvotes]: If the circle $x^2+y^2+4x+22y+l=0$ bisects the circumference of the circle $x^2+y^2-2x+8y-m=0$,then $l+m$ is equal to
(A)$\ 60$
(B)$\ 50$
(C)$\ 46$
(D)$\ 40$
I don't know the condition when one circle intersects the circumference of other circle. So could not solve this question. Can someone help me in this question?

REPLY [1 votes]: The condition for a circle (Circle A) to bisect the circumference of another circle (Circle B) is:
The common chord of A and B should pass through the center of B


TITLE: Pointwise multiplication by unbounded function throws us out of $L^2(\mu)$
QUESTION [5 upvotes]: Let $(X,\mathcal{A},\mu)$ be a $\sigma$-finite measure space and let $\phi$ be a measurable function that is not an element of $L^\infty(\mu)$, i.e. $\phi\not\in L^\infty(\mu)$. I am trying to construct a function $g\in L^2(\mu)$ such that $\phi\cdot g\not\in L^2(\mu)$.
I tried partitioning $X$ in the sets $A_k=\{k\leq|\phi|<k+1\}$ so i could somehow control the behavior of $g$ on those sets relative to $\phi$, but I needed something more. I considered a partition of $X$ in disjoint sets of finite measure say $(X_n)$ and then I tried to take their co-partition. The problem arising is that I cannot determine which of the sets $A_k\cap X_n$ are of non-zero measure. I am really stuck here. Any ideas?

REPLY [0 votes]: Another way of seeing this, is the following:
If $T:g\mapsto\phi g$ acted as $L^2(\mu)\to L^2(\mu)$, then by the closed graph theorem it would be $T\in\mathcal{B}(L^2(\mu))$. Indeed, if $g_n\to0$ in $L^2(\mu)$-norm and $\phi g_n\to h$, then $\|h\|_2^2=\langle h,h\rangle=\displaystyle{\lim_{n\to\infty}\langle \phi g_n, h\rangle=\lim_{n\to\infty}\langle g_n, \overline{\phi} h\rangle=0}$.
But $T$ is the multiplication operator by $\phi$, hence $\|T\|=\|\phi\|_\infty$, a contradiction.
(To see that $\|T\|\geq\|\phi\|_\infty,$ let $a<\|\phi\|_\infty$. Choose a measurable set $E$ of positive and finite measure such that $E\subset\{|\phi|>a\}$, consider the characteristic function of $E$ and apply $T$ to it).


TITLE: Kurtz randomness and supermartingales with infinite *limit*
QUESTION [0 upvotes]: Suppose you replace the usual success conditions for a supermartingale (lim sup is infinite) with the requirement that the actual limit is infinite, e.g. a supermartingale $B$ succeeds on $X \in 2^\omega$ just if
$$\lim_{n \to \infty} B(X\restriction_n) = \infty$$
I'm 90% sure that you don't get a 'valid' randomness notion because any sufficiently generic real should qualify.  However, if I had to guess I'd think this notion turns out to be equivalent to Kurtz 'randomness' (avoids all $\Pi^0_1$ null sets).  Since I'm guess this is a known result and my brain is feeling super foggy I thought I would ask

REPLY [1 votes]: Assuming you mean a c.e. (super)martingale, this is just Martin-Löf randomness.  What needs to be shown is that, for every non-random real, there is a martingale which limit-succeeds on the real.  You can see this from the proof of the equivalence of the martingale definition and the Kolmogorov complexity definition:
Let $U$ be a universal prefix-free machine.  For $\sigma \in 2^{<\omega}$, define $M_\sigma$ to be the (computable) martingale that begins with capital 1 and bets it all on $\sigma$, and bets evenly afterwards.  So $M(\tau) = 2^{|\sigma|}$ for any $\tau$ extending $\sigma$.  Define $M = \sum_{U(\rho)=\sigma} 2^{-|\rho|}M_\sigma$.  This is a c.e. martingale (with starting capital $\Omega$).  For any non-random $X$, for any $d$, there is an $n$ with $K(X\upharpoonright n) < n - d$.  So there is some $\rho$ with $|\rho| < n - d$ and $U(\rho) = X\upharpoonright n$.  So $2^{-|\rho|} M_{X\upharpoonright n}$ is a summand in the definition of $M$, and $2^{-|\rho|}M_{X\upharpoonright n}(\tau) > 2^d$ for any $\tau$ extending $X\upharpoonright n$.  So $\lim_{n \to \infty} M(X\upharpoonright n) = \infty$.
Note that this construction shows why your generic intuition is wrong: for every $d$, there is a neighborhood of $X$ on which the martingale never again dips below $2^d$.  So generics can't get out of it.


TITLE: Orientability of surface r(u,v)=(ucos(v),usin(v),v)
QUESTION [0 upvotes]: Say one wanted to take the surface integral $\iint_S F\cdot dS$ of the vector field $F(x,y,z)=\langle z,y,x \rangle$ over the surface $S$ parametrized by $r(u,v)=\langle u\cos(v),u\sin(v),v\rangle$, $0 \leq u \leq 1$, $0 \leq v \leq \pi$ with $\textit{upward orientation}$. This is an example from Stewart 7th ed.
We first see if $r_u \times r_v=\langle \sin v, -\cos v, u \rangle$ agrees with the requested upwards orientation. It does, since the $k$ component is nonnegative as $u$ ranges from $0$ to $1$. So we proceed to set up the integral and evaluate without any sign-change.
However, my question is, if we were asked to take $\iint_S F\cdot dS$ all the same, but the surface was slightly altered by changing the bounds on $u$ to $\mathbf{-1 \leq u \leq 1}$, what happens in terms of orientation? I have plotted $S$, and see it gives both "sides" of the helicoid rather than just one (when $0 \leq u \leq 1$).
Is this surface orientable? As $u$ varies, the normal vector field to our surface will have both positive and negative $k$ components. I'm not sure which orientation this would induce (if any) on $S$. Let me know if I can help clarify anything.

REPLY [1 votes]: The surface is orientable. It's important not to get too hung up on (the sign of) the components of the normal vector along any particular coordinate axis; a sphere, for instance, is orientable but the 'oriented' normal vector field will inevitably have vectors with both positive and negative values of any coordinate.
What orientable means in this context is that if we start at some point $p$ on the surface with a normal $\hat{n}_p$ and trace a loop along the surface continuously varying the normal, we can never get back to $p$ with the normal $-\hat{n}_p$. (For instance, imagine taking a Möbius strip and going once around it; you'll see that the normal continuously varies, but when you return to your original position it will be the inverse of the original normal.) In the case of the helicoid, it's true that the normal won't always be pointing upwards, but that doesn't mean that any path will ever flip it to the 'other side' of the surface.


text:
\begin{document}

\title[Mock Jacobi forms in basic hypergeometric series] {Mock Jacobi forms in \\basic hypergeometric series }
\author{Soon-Yi Kang}

\date{\today}

\address{Korea Advanced Institute for Science and Technology, Daejeon 305-701, KOREA}
\email{s2kang@kaist.ac.kr}

\subjclass[2000]{11F37, 11F50, 05A17, 33D15}
\keywords{basic hypergeometric series, mock Jacobi forms, mock modular forms, weakly holomorphic modular forms, ranks and cranks of partitions}

\begin{abstract}
We show that some $q$-series such as universal mock theta functions are linear sums of theta quotients and mock Jacobi forms of weight $1/2$, which become holomorphic parts of real analytic modular forms when they are restricted to torsion points and multiplied by suitable powers of $q$.  And we prove that certain linear sums of
$q$-series are weakly holomorphic modular forms of weight $1/2$ due to annihilation of mock Jacobi forms or completion by mock Jacobi forms.  As an application, we obtain a relation between the rank and crank of a partition.
\end{abstract}
\maketitle

\section{Introduction and summary of results}\label{intro}
A Jacobi form is a holomorphic function of two variables $z\in \mathbb{C}$ and $\t\in \mathbb{H},$ the upper half of the complex plane, which satisfies elliptic transformation properties with respect to $z$, modular transformation properties with respect to $\t$, and a certain growth condition. It is well known \cite[Theorem 1.3]{EZ} that a Jacobi form at torsion points, $z=r\t+s$, with $r$, $s\in \mathbb{Q}$, multiplied by a suitable rational power of $q=e^{2\pi i \t}$ is a modular form with respect to $\t$.  For example, the classical theta series
\begin{equation}\label{theta}\vartheta(z):=\vartheta(z;\t):=\sum_{n=-\i}^\i(-1)^{n} e^{\pi i (2n+1)z} q^{\frac{n(n+1)}{2}+\frac{1}{8}}\end{equation}
is the most famous Jacobi form and $q^{r^2/2}\vartheta(r\t+s;\t)$ is an ordinary modular form of weight $1/2$.

In his ground-breaking work on mock theta functions \cite{Zwegers}, Zwegers constructed a non-holomorphic but real analytic Jacobi form whose holomorphic part is a normalized Lerch sum and whose non-holomorphic part can be written in terms of a period integral of a weight $3/2$ unary theta series.  The holomorphic part of Zwegers' real analytic Jacobi form is defined by
\begin{equation}\label{mu}\mu(u,v):=\mu(u,v;\t):=\frac{a^{1/2}}{\vartheta(v)}\sum_{n=-\i}^\i\frac{(-b)^nq^{n(n+1)/2}}{1-aq^n},\end{equation}
where $a=e^{2\pi i u}$ and $b=e^{2\pi i v}$ for complex values $u$ and $v$.
Following Zagier, we call $\mu$ a \textit{mock Jacobi form}, because it behaves almost like a Jacobi form with two elliptic variables and when the elliptic variables are restricted to torsion points, it becomes a \textit{a mock modular form} up to a factor of a rational power of $q$.  Each mock modular form of weight $k$ is the holomorphic part of a real analytic modular form of weight $k$, of which the non-holomorphic part is determined by an ordinary modular form of weight $2-k$, called the \textit{shadow} of the mock modular form.  The real analytic modular form that has a mock modular form as its holomorphic projection is now called a harmonic weak Maass form.  And a mock theta function is a $q$-series that is a mock modular form of weight $1/2$ multiplied by a rational power of $q$ and the shadow is a unary theta series of weight $3/2$.  For a full description of the notions of mock modular forms and their shadows, the reader is referred to \cite{Zagier}.

We recall that the \textit{rank} of a partition is defined as its largest part minus its number of parts.  Then
the generating function for the number of partitions of given size and rank is given by
 $$\mathcal{R}(w;q):=\sum_{n=0}^\i \frac{q^{n^2}}{\prod_{k=1}^n(1-wq^k)(1-q^k/w)}.$$   When $w = -1$, this reduces to one of
Ramanujan's 3rd order mock theta functions $f(q)$, and Zwegers \cite{Zwegers1} proved that $q^{-1/24}f(q)$ is a component of a vector-valued real analytic modular form of weight $1/2$. With the aid of this work, Bringmann and Ono \cite{BO1,BO2} showed that $q^{-1/24}f(q)$ is the holomorphic part of a weak Maass form of weight $1/2$.  Furthermore, they \cite[Theorem 1.2]{BO2} generalized this to other roots of unity. By using Zwegers' mock Jacobi form $\mu$, Zagier \cite[Theorem 7.1]{Zagier} then found a result that subsumes and is also much simpler in formulation and proof than that by Bringmann and Ono,  as remarked by Zagier himself.  Once we represent $\mathcal{R}(w;q)$ in terms of a generalized Lambert series, we can easily deduce that
 \begin{equation}\label{CR}
\frac{q^{-1/24}\mathcal{R}(e^{2\pi i\a};q)}{e^{-\pi i \a}-e^{\pi i \a}}=\frac{\eta^3(3\t)}{\eta(\t)\vartheta(3\a;3\t)}-q^{-1/6}e^{-2\pi i \a}\mu(3\a,-\t;3\t)+q^{-1/6}e^{2\pi i \a}\mu(3\a,\t;3\t),\end{equation}
where $\eta(\t)$ is the Dedekind eta-function satisfying $q^{-1/24}\eta(\t)=\prod_{n=1}^\i(1-q^n)$.  ($q^{-1/6}$ in the last two terms of the right hand side are missing in \cite{Zagier}.) Applying Zwegers' result to (\ref{CR}), Zagier \cite[Theorem 7.1]{Zagier} proved that, for any root of unity $\zeta=e^{2\pi i\a}\neq 1$, $q^{-1/24}\mathcal{R}(\zeta;q)$ is a mock modular form of weight $1/2$ with shadow proportional to $(\zeta^{-1/2}-\zeta^{1/2})\sum_{n\in \mathbb{Z}}\lt(\frac{12}{n}\rt)n\sin(\pi n\a)q^{n^2/24}$.

In this note, we find more $q$-series which are composed of sums of theta quotients and mock Jacobi forms like the rank generating function $\mathcal{R}(w;q)$.  They become mock modular forms when the elliptic variables are restricted to torsion points. In particular, both of two \textit{universal mock-theta functions} are such series:
\begin{equation}\label{mockeven}g_2(w;q):=\sum_{n=0}^\i\frac{
(-q)_n q^{n(n+1)/2}}{(w;q)_{n+1}(q/w;q)_{n+1}},\end{equation}

\begin{equation}\label{mockodd}g_3(w;q):=\sum_{n=0}^\i\frac{
q^{n(n+1)}}{(w;q)_{n+1}(q/w;q)_{n+1}},\end{equation}
 where $(a)_0:=(a;q)_0:=1,\ \ (a)_n:=(a;q)_n:=\prod_{j=0}^{n-1}(1-aq^j),
\ \ n\geq1,$ and $(a)_\i:=(a;q)_\i:=\prod_{j=0}^{\i}(1-aq^j).$
 We call these functions universal mock-theta functions as all mock-theta functions of even (resp. odd) order are expressed in terms of $g_2$ (resp. $g_3$). This was first noticed by D. Hickerson \cite{H, H7} when he proved that mock theta conjectures on mock theta functions of orders 5 and 7 can be stated in terms of $g_3$.  Recently Gordon and McIntosh \cite{GM} observed that this is a uniform phenomenon for mock theta functions of all orders.  They noticed that $g_3$ can be written in terms of $g_2$, and hence named $g_2$ a universal mock theta function \cite{GM}.  In addition to those reviewed above, there are many results on these two functions.  However, we show that Zwegers' mock Jacobi form $\mu$ naturally reveals the properties of $g_2$ and $g_3$.

\begin{thm}\label{umt} Let $\a$ be any complex number such that $\a\neq 2n\t+m$ for $m$, $n\in Z$. If $w=e^{2\pi i \a}$, then
\begin{equation}\label{memjf}wg_2(w;q)=\frac{\eta^4(2\t)}{\eta^2(\t)\vartheta(2\a;2\t)}+wq^{-1/4}{\mu}(2\a,\t;2\t),\end{equation}
which is a mock modular form of weight $1/2$ times a rational power of $q$ at torsion points.  In particular, if $\zeta\neq 1$ is a root of unity, then $\zeta g_2(\zeta;q)$ is a mock modular form of weight $1/2$ with shadow proportional to $\sum_{n=-\i}^\i(-1)^n n \zeta^{-2n}q^{n^2}.$
\end{thm}

When $w=q^r$ for a rational number $r$,  Bringmann, Ono, and Rhoades \cite[Theorem 1.3]{BOR} proved that $\displaystyle{q^{r(1-r)}(g_2(w;q)+g_2(-w;q))}$ is a weight $1/2$ weakly holomorphic modular form on a subgroup of the full modular group $SL_2(\mathbb{Z})$ depending on $r$ by showing that the summands
are distinct Maass forms with equal non-holomorphic parts of opposite signs.  If we use elliptic transformation properties of $\mu$ and $\vartheta$, we can easily deduce a more general result from (\ref{memjf}):

 \begin{thm}\label{Anng2} Let $\a$ be any complex number such that $\a\neq 2n\t+m$ for $m$, $n\in Z$. If $w=e^{2\pi i \a}$, then we have a meromorphic Jacobi form
  \begin{equation}\label{ann}w(g_2(w;q)+g_2(-w;q))=\frac{2\eta^4(2\t)}{\eta^2(\t)\vartheta(2\a;2\t)}.\end{equation}
If $\a=r\t+s$ with $r$, $s\in \mathbb{Q}$, then it is a weight $1/2$ weakly holomorphic modular form on a subgroup of $SL_2(\mathbb{Z})$ depending on $r$ when it is multiplied by $q^{-r^2}$. \end{thm}

By writing $g_3(w;q)$ as a sum of a theta quotient and mock Jacobi forms, we can derive an analogous result for $g_3(w;q)$ as well:
\begin{thm}\label{Anng3} Let $\a$ be any complex number such that $\a\neq 3n\t+m$ for $m$, $n\in Z$. If $w=e^{2\pi i \a}$ and $\zeta_3=e^{2\pi i/3}$, then we have a meromorphic Jacobi form
  \begin{equation}\label{anng3}w^{3/2}q^{-1/24}(g_3(w;q)+g_3(\zeta_3w;q)+g_3(\zeta_3^2w;q))=\frac{3\eta^3(3\t)}{\eta(\t)\vartheta(3\a;3\t)}.\end{equation}
  If $\a=r\t+s$ with $r$, $s\in \mathbb{Q}$, then it is a weight $1/2$ weakly holomorphic modular form on a subgroup of $SL_2(\mathbb{Z})$ depending on $r$ when it is multiplied by  $q^{-3r^2/2}$.\end{thm}

We can apply Theorem \ref{Anng3} to the theory of partitions.  Let $p(n)$ denote the number of partitions of $n$.  The rank of a partition gives combinatorial interpretations for Ramanujan's famous congruences $p(5n+4)\equiv 0\pmod 5$ and
$p(7n+5)\equiv 0\pmod 7$ by splitting the partitions of $5n+4$
(resp. $7n+5$) into $5$ (resp. $7$) equal classes when sorted
according to the rank modulo $5$ (resp. $7$).  Besides the rank, there is another partition statistics, called the \textit{crank}, which gives combinatorial interpretations not only for the two congruences above but also for Ramanujan's
third partition congruence $p(11n+6)\equiv 0\pmod {11}$.   From Theorem \ref{Anng3}, we derive a relation between ranks and cranks which yields that

\begin{thm}\label{Rank-Crank} Let $N(m,n)$ and $M(m,n)$ denote the numbers of partitions of a nonnegative integer $n$ with rank $m$ and crank $m$, respectively, and $p_3(n)$ be the number of partitions of $n$ into parts congruent to $\pm 1$ modulo $3$. Then for each integer $m$, we have
\begin{equation}N(3m-1,n)+N(3m,n)+N(3m+1,n)=\sum_{0\leq 3k\leq n} M(m,k)p_3(n-3k).\end{equation}
\end{thm}

By the way, Theorem \ref{Anng2} is an immediate consequence of Ramanujan's $_1\psi_1$ summation formula \cite[p.~502]{AAR},
\begin{equation}\label{1psi1}
_1\psi_1(\a;\b;z):=\sum_{n=-\i}^\i\frac{(\a)_n}{(\b)_n}z^n
=\frac{(\b/\a)_\i(\a z)_\i(q/(\a z))_\i(q)_\i}{(q/\a)_\i(\b/(\a
z))_\i(\b)_\i(z)_\i},  \ \mathrm{for}\ |\b/\a|<|z|<1.\end{equation}
After some manipulations, we can observe that $g_2(w;q)$ arises from a positive indexed summation while $g_2(-w;q)$ arises from a negative indexed summation of the the bilateral series in (\ref{1psi1}).  This tells us the beauty of $_1\psi_1$ summation formula comes from annihilation of mock Jacobi forms in some specific cases.  The following theorem, however, shows another specific case of (\ref{1psi1}) holds due to a different reason:

\begin{thm}\label{q2cq} Define
$$K'(w;\t):=\sum_{n=0}^\i\frac{(-1)^nq^{n^2}(q;q^2)_n}{(wq^{2};q^2)_{n}(q^{2}/w;q^2)_{n}},$$
$$K''(w;\t):=\sum_{n=1}^\i\frac{(-1)^n q^{n^2}(q;q^2)_{n-1}}{(wq;q^2)_{n}(q/w;q^2)_{n}}.$$
Let $\a$ be any complex number such that $\a\neq m\t+n$ for $m$, $n\in Z$. If $w=e^{2\pi i \a}$, then we have a meromorphic Jacobi form
\begin{equation}\label{q2cqe}\frac{q^{-1/8}}{(w)^{-1/2}-(w)^{1/2}}K'(w;\t)+q^{-1/8}(w^{1/2}-w^{-1/2})K''(w;\t)=\frac{\eta^4(\t)}{\eta^2(2\t)\vartheta(\a;\t)}.\end{equation}
If $\a=r\t+s$ with $r$, $s\in \mathbb{Q}$, then it is a weight $1/2$ weakly holomorphic modular form on a subgroup of $SL_2(\mathbb{Z})$ depending on $r$ when it is multiplied by  $q^{-r^2/2}$.\end{thm}
The two terms in the left hand side of (\ref{q2cqe}) are mock Jacobi forms and each of them is about the half of a meromorphic Jacobi form in the right hand side. That is, there occurs a completion to a meromorphic Jacobi form by mock Jacobi forms not annihilation of mock Jacobi forms as in Theorem \ref{umt}.

This paper is presented as follows.  In Section 2, we review Zwegers' results on mock Jacobi forms and Zagier's notion on shadows of mock modular forms very briefly.  In Section 3, we decompose universal mock theta functions as sums of theta quotients and mock Jacobi forms and show they are mock modular forms at torsion points when they are multiplied by suitable powers of $q$. And using the results, we prove Theorems \ref{Anng2} and \ref{Anng3}.  In Section 4, using Theorem \ref{Anng3}, we establish a relation between rank and crank generating functions and discuss further relations between ranks and cranks, including Theorem \ref{Rank-Crank}.  Finally, in Section 5, we reprove some of the results in Section 3 using the theory of basic hypergeometric series and prove Theorem \ref{q2cq}.

\section{Properties of Mock Jacobi Forms and Mock Modular Forms}
We first recall important properties of a mock Jacobi form in (\ref{mu}) stated in Zwegers' Ph.D Thesis \cite{Zwegers}.
Note that our $\vartheta$ is $-i$ times of Zwegers'  $\vartheta$ function. (See \cite[Proposition 1.3]{Zwegers}.) It is well known that
the theta series in (\ref{theta}) is an odd function of $z$ having a product representation
\begin{equation}\label{jtp}\vartheta(z;\t)=q^{1/8}x^{-1/2}\prod_{n=1}^\i(1-q^n)(1-xq^{n-1})(1-x^{-1}q^{n}),\end{equation}
where $x=e^{2\pi i z}$ by the Jacobi triple product identity.

\begin{prop}\cite[Proposition 1.4 and 1.5]{Zwegers}\label{mup} The mock Jacobi form $\mu(u,v;\t)$ in (\ref{mu}) is a meromorphic function of $u$ with simple poles in the points $u=n\t+m$ for $n,$ $m\in \mathbb{Z}$, symmetric in $u$ and $v$, and satisfies elliptic transformation properties:
\begin{equation}\mu(u+1,v)=-\mu(u,v)=e^{-2\pi i(u-v)-\pi i \t}\mu(u+\t,v)+e^{-\pi i (u-v)-\pi i \t/4},\end{equation}
and modular transformation properties:
\begin{equation}e^{\pi i /4}{\mu}(u,v;\t+1)=-(-i\t)^{-1/2}e^{\pi i (u-v)^2/\t}{\mu}(\frac{u}{\t},\frac{v}{\t};\frac{-1}{\t})+\frac{1}{2}h(u-v;\t)={\mu}(u,v;\t),\end{equation}
where $\displaystyle{h(z):=h(z;\t):=\int_\mathbb{R}\frac{e^{\pi i \t x^2-2\pi z x}}{\cosh{\pi x}}\, dx,}$ a Mordell integral.
\end{prop}

Zwegers then constructed a non-holomorphic function $R(z;\t)$ for $z\in \mathbb{C}$ so that \begin{equation}\label{zw}\widehat{\mu}(u,v;\t)=\mu(u,v;\t)-\frac{1}{2}R(u-v;\t)\end{equation} is a real analytic Jacobi form.
For rational numbers $a$ and $b$, $q^\lambda\mu(a\t,b;\t)$ is a mock modular form of weight $1/2$ for a certain rational number $\lambda$ and its shadow  is a weight $3/2$ unary theta series.

In general, the shadow $g(\t)$ of a mock modular form $f(\t)$ of weight $k$ is proportional to $\displaystyle{y^{k}\frac{\overline{\partial}}{\partial \bar{\t}}}(f+E)(\t)$, where $y=\mathrm{Im}(\t)$ and $E(\t)$ is the associated non-holomorphic function to $f(\t)$ such that $f(\t)+E(\t)$ is a real analytic modular form of weight $k$.  Under this differential operator, the holomorphic part is mapped to zero, and thus the shadow is determined by the correction function $E(\t)$ as follows:
\begin{prop}\cite[Section 5]{Zagier}\label{za} If $E(\t)$ is the correction function of a mock modular form of weight $k$ with the shadow $g(\t)$ and $y=\mathrm{Im}(\t)$, then $\displaystyle{y^{k}\frac{{\partial}}{\partial \bar{\t}}}E(\t)$ is proportional to $\overline{g(\t)}$ and
\begin{equation}\label{Err} E(\t)=(i/2)^{k-1}\int_{-\bar{\t}}^{i\i}\frac{\overline{g(-\bar{z})}}{\sqrt{-i(z+\t)}}\, dz.\end{equation}
\end{prop}
\noindent The differential operator $\displaystyle{y^{k}\frac{\overline{\partial}}{\partial \bar{\t}}}$ is also discussed as a map from the space of harmonic weak Maass forms of weight $k$ to the space of weakly holomorphic modular forms of weight $2-k$ in \cite[Proposition 3.2]{BF}.
In finding the shadow of $q^\lambda\mu(a\t,b;\t)$ then, the following result of Zwegers would be useful.
\begin{prop}\cite[(1.5)]{Zwegers}\label{dif} If $a$, $b\in \mathbb{R}$, then
 \begin{equation}\label{difr}\frac{\partial}{\partial \bar{\t}}R(a\t-b;\t)=-\frac{i}{\sqrt{2y}}e^{-2\pi a^2 y}\sum_{\nu\in \frac{1}{2}+\mathbb{Z}}(-1)^{\nu-1/2}(\nu+a)e^{-\pi i \nu^2 \bar{\t}-2\pi i \nu (a\bar{\t}-b)}.\end{equation}
\end{prop}
But if $a\in (-\frac{1}{2}, \frac{1}{2})$ and $b\in \mathbb{R}$, we may use another formula for  $R(a\t-b)$ that writes $R(a\t-b)$ in terms of the period integral of a unary theta function of weight $3/2$.
\begin{prop}\cite[Theorem 1.16]{Zwegers}\label{Rcor}  For $a$, $b\in \mathbb{R}$, define a unary theta series of weight $3/2$
\begin{equation}\label{theta32}g_{a,b}(\t):=\sum_{\nu\in a+\mathbb{Z}}\nu e^{\pi i \nu^2\t+2\pi i \nu b}.\end{equation}
Then, for $a\in (-\frac{1}{2}, \frac{1}{2})$ and $b\in \mathbb{R}$,
\begin{equation}\label{cor}\int_{-\bar{\t}}^{i\i}\frac{g_{a+\frac{1}{2},b+\frac{1}{2}}(z)}{\sqrt{-i(z+\t)}}\, dz=-e^{-\pi i a^2\t+2\pi i a(b+\frac{1}{2})}R(a\t-b).
\end{equation}\end{prop}

Although (\ref{cor}) holds for $a\in (-\frac{1}{2},\frac{1}{2})$ only, we can extend the result to all real numbers $a$ except $a\in -1/2+\mathbb{Z}$ thanks to the periodicity of $g$ as a function of $a$:

\begin{prop}\cite[Proposition 1.15]{Zwegers}\label{gp}
$g_{a,b}$ satisfies
\begin{equation}g_{a+1,b}(\t)=g_{a,b}(\t)=e^{-2\pi i a}g_{a,b+1}(\t)=-g_{-a,-b}(\t).\end{equation}
\end{prop}

\section{Universal mock theta functions in terms of mock Jacobi forms}

We first recall generalized Lambert series expansions of universal mock theta functions.  As the standard method of deriving a generalized Lambert series expansion of a $q$-series is using a limiting case of the Watson-Whipple transformation formula, we state the formula here.  For convenience, we use the notation
$(a_1,a_2,\cdots,a_k)_n:=(a_1)_n(a_2)_n\cdots(a_k)_n.$
Letting $N\to \i$ in \cite[eq.~(2.5.1),~p.~43]{GR}, we have
\begin{eqnarray}\label{watson} \sum_{n=0}^\i\frac{(\a,\b,\r,\delta,\epsilon)_n(1-\a q^{2n})q^{n(n+3)/2}}
{(\a q/\b,\a q/\r,\a q/\delta ,\a q/\epsilon,q)_n(1-\a)} \lt(-\df{\a^2}{\b\r\delta\epsilon}\rt)^n \cr
=\frac{(\a q ,\a q/(\delta\epsilon))_\i}{(\a q/\delta, \a
q/\epsilon)_\i} \sum_{n=0}^\i\frac{(\delta,\epsilon, \a
q/(\b\r))_n}{(\a q/\b,\a q/\r,q)_n}\lt(\frac{\a q}{\delta\epsilon}\rt)^n.\end{eqnarray}
Setting $\a=q$, $\b\r=q^2$, $\delta=x$, and $\epsilon=q/x$ in (\ref{watson}) and then multiplying both sides of the resulting equation by $(1-q)/((1-w)(1-q/w))$ yields the following well known Lambert series expansion of a theta quotient.

\begin{lem}\label{thetaLam} Let $\a$ be any complex number such that $\a\neq n\t$ for $n\in Z$. If $w=e^{2\pi i \a}$, then
\begin{equation}\sum_{n=-\i}^\i\frac{(-1)^nq^{n(n+1)/2}}{1-wq^{n}}=\frac{(q;q)^2_\i}{(w;q)_\i(q/w;q)_\i}.\end{equation}\end{lem}

The generalized Lambert series expansions for both of $g_2$ and $g_3$ are already known in the literature as well (for example, \cite{GM}, \cite{H}).  Utilizing (\ref{watson}) again, we have

\begin{lem}\label{uniLam} If $g_2(w;q)$ and $g_3(w;q)$ are defined as in (\ref{mockeven}) and (\ref{mockodd}), respectively, then
\begin{equation}\label{g2}g_2(w;q)=\frac{(-q)_\i}{(q)_\i}\sum_{n=-\i}^\i\frac{(-1)^nq^{n(n+1)}}{1-wq^{n}},\end{equation} and
\begin{equation}\label{g3}g_3(w;q)=\frac{1}{(q)_\i}\sum_{n=-\i}^\i\frac{(-1)^nq^{3n(n+1)/2}}{1-wq^{n}}.\end{equation}
\end{lem}

Now, we can decompose both $g_2(w;q)$ and $g_3(w;q)$ as sums of theta quotients and mock Jacobi forms. First let us prove Theorem \ref{umt}.

\begin{proof}[proof of Theorem \ref{umt}]
Using the fact $\displaystyle{\frac{1}{1-x}=\frac{1+x}{1-x^2}}$, we deduce  from (\ref{g2}) that
$$g_2(w;q)=\frac{(-q)_\i}{(q)_\i}\lt(\sum_{n=-\i}^\i\frac{(-1)^nq^{n^2+n}}{1-w^2q^{2n}}+w\sum_{n=-\i}^\i\frac{(-q)^nq^{n^2+n}}{1-w^2q^{2n}}\rt).$$

Using Lemma \ref{thetaLam} and the definitions of $\vartheta(z)$  and $\mu(u,v)$  in  (\ref{theta})  and (\ref{mu}), respectively, we establish (\ref{memjf}).
Hence by the results in Section 2, we can see that $w g_2(w;q)$ is a mock modular form of weight $1/2$ at torsion points when it is multiplied by a rational power of $q$. In particular, if we restrict $\a$ to a rational number, then in (\ref{memjf}), the first term is a weakly holomorphic modular form of weight $1/2$ and the second term is a mock modular form of weight $1/2$ with the correction term $-e^{2\pi i\a}q^{-1/4}R(2\a-\t;2\t)/2$.  By setting $a=-1/2$ in (\ref{difr}), we find that  \begin{equation}\label{difr2}\frac{\partial}{\partial \bar{\t}}R(-\frac{\t}{2}-b)=-\frac{i}{\sqrt{2y}}e^{\pi i \t/4+\pi i b}\sum_{n\in \mathbb{Z}}(-1)^{n}ne^{-\pi i n^2\bar{\t}+2\pi i n b},\end{equation}
and hence it follows from Proposition \ref{za} that the shadow of $e^{-\pi i b}q^{-1/4}\mu(-\t/2,b;\t)$ is
\begin{equation}\label{sh}{\frac{\sqrt{i}}{2}\sum_{n\in \mathbb{Z}}(-1)^{n}ne^{\pi i n^2{\t}+2\pi i n b}}.\end{equation}  Therefore, substituting $b=-2\a$ and replacing $\t$ by $2\t$ in (\ref{sh}) completes the proof for $\zeta g_2(\zeta;q)$, where $\zeta=e^{2\pi i\a}\neq 1$.
\end{proof}

We have similar results for $g_3$.
\begin{thm}\label{ag} Let $\a$ be any complex number such that $\a\neq 3m\t+n$ for $m$, $n\in Z$, then
\begin{equation}\label{momjf}q^{-1/24}e^{3\pi i\a}g_3(e^{2\pi i\a};q)=\frac{\eta^3(3\t)}{\eta(\t)\vartheta(3\a;3\t)}+q^{-1/6}e^{2\pi i \a}\mu(3\a,\t;3\t)+q^{-2/3}e^{4\pi i \a}\mu(3\a,2\t;3\t),\end{equation}
which is a mock modular form of weight $1/2$ times a rational power of $q$ at torsion points.  In particular, if $\zeta\neq 1$ is a root of unity, then
 $q^{-1/24}\zeta^{3/2} g_3(\zeta;q)$ is a mock modular form of weight $1/2$ with shadow proportional to $\sum_{n=-\i}^\i(\frac{12}{n}) n \sin(\pi n \a)q^{n^2/24}.$\end{thm}
\textit{Remark.} As stated in \cite[Theorem 7.1]{Zagier},
$$\mathcal{R}(w;q)=\frac{1-w}{(q)_\i}\sum_{n=-\i}^\i\frac{(-1)^nq^{(3n^2+n)/2}}{1-wq^n}.$$
By comparing this with the Lambert series expansion of $g_3(w;q)$ in (\ref{g3}), we can easily see that $\displaystyle{g_3(w;q)=-\frac{1}{w}+\frac{1}{w(1-w)}R(w;q)}$.  Hence we may obtain the result for $g_3(\zeta;q)$ above from the fact that $q^{-1/24}\mathcal{ R}(\zeta;q)$ is a mock modular form stated in the introduction as well.
\begin{proof}
The proof is similar to that of Theorem \ref{umt}.  It follows from  (\ref{g3}) and ${\frac{1}{1-x}=\frac{1+x+x^2}{1-x^n}}$ that
$$g_3(w;q)=\frac{1}{(q)_\i}\lt(\sum_{n=-\i}^\i\frac{(-1)^nq^{3n(n+1)/2}}{1-w^3q^{3n}}+\sum_{n=-\i}^\i\frac{(-q)^nwq^{3n(n+1)/2}}{1-w^3q^{3n}}
+\sum_{n=-\i}^\i\frac{(-q^2)^nw^2q^{3n(n+1)/2}}{1-w^3q^{3n}}\rt).$$  From Lemma \ref{thetaLam} and the definitions of $\vartheta(z)$  and $\mu(u,v)$  in  (\ref{theta})  and (\ref{mu}), respectively, we deduce (\ref{momjf}), where $w=e^{2\pi i \a}$.  Hence by the results in Section 2, we can see that $q^{-1/24}e^{3\pi i\a}g_3(e^{2\pi i\a};q)$ is a mock modular form of weight $1/2$ at torsion points when it is multiplied by a rational power of $q$. In particular, if we restrict $\a$ to a rational number, the first term in (\ref{momjf}) is a weakly holomorphic modular form of weight $1/2$ and the last two terms are mock modular forms of weight $1/2$ with correction terms $-\frac{e^{2\pi i\a} q^{-1/6}}{2}R(3\a-\t;3\t)$ and $-\frac{e^{4\pi i\a} q^{-2/3}}{2}R(3\a-2\t;3\t)$ by (\ref{zw}). Using Propositions \ref{Rcor} and \ref{gp}, we can find that
\begin{eqnarray}&&\frac{e^{2\pi i\a} q^{-1/6}}{2}R(3\a-\t;3\t)+\frac{e^{4\pi i\a} q^{-2/3}}{2}R(3\a-2\t;3\t)\cr

&=&-\frac{1}{2}\int_{-3\bar{\t}}^{i\i}\frac{e^{\pi i/3}g_{\frac{1}{6},-3\a+\frac{1}{2}}(z)-e^{2\pi i/3}g_{\frac{1}{6},3\a-\frac{1}{2}}(z)}{\sqrt{-i(z+3\t)}}\, dz,\end{eqnarray} and the shadow is $$\frac{1}{2}\lt(e^{\pi i/3}g_{\frac{1}{6},-3\a+\frac{1}{2}}(3\t)-e^{2\pi i/3}g_{\frac{1}{6},3\a-\frac{1}{2}}(3\t)\rt)=\frac{1}{6}\sum_{n=-\i}^\i (-1)^n (6n+1)q^{(6n+1)^2/24}\sin(\pi(6n+1)\a).$$
\end{proof}

Next, let us prove Theorem \ref{Anng2} and Theorem \ref{Anng3}.

\begin{proof}[Proof of Theorem \ref{Anng2}]
The result follows immediately from (\ref{memjf}), as we have
$$g_2(-e^{2\pi i\a};q)=g_2(e^{2\pi i(\a+\frac{1}{2})};q)=\frac{\eta^4(2\t)e^{-2\pi i\a}}{\eta^2(\t)\vartheta(2\a;2\t)}-q^{-1/4}\mu(2\a,\t;2\t)$$
by elliptic properties of $\vartheta$ and $\mu$ such as $\vartheta(z+1)=-\vartheta(z)$ and $\mu(u+1,v)=-\mu(u,v)$.
\end{proof}

\begin{proof}[Proof of Theorem \ref{Anng3}] Applying (\ref{momjf}) three times with the elliptic properties of $\vartheta$ and $\mu$, we obtain that
\begin{eqnarray}&&q^{-1/24}(g_3(w;q)+g_3(\zeta_3w;q)+g_3(\zeta_3^2w;q))=\frac{3\eta^3(3\t)e^{-3\pi i\a}}{\eta(\t)\vartheta(3\a;3\t)}\cr
&&\ +e^{-\pi i\a}q^{-1/6}\mu(3\a,\t;3\t)(1-e^{-\pi i/3}+e^{-2\pi i/3})+e^{\pi i\a}q^{-2/3}\mu(3\a,2\t;3\t)(1-e^{\pi i/3}+e^{2\pi i/3}),\nonumber\end{eqnarray}
where the last two terms are zero.
\end{proof}

\section{Applications to partitions}

Using (\ref{CR}) or using Theorem \ref{Anng3}, we also have a similar result to Theorem \ref{Anng3} for the rank generating function $\mathcal{R}(w;q)$: If $w=e^{2\pi i\a}$ for any complex number $\a\neq 3n\t+m$ with $m$, $n\in \mathbb{Z}$ and $\zeta=e^{2\pi i/3}$, then
\begin{equation}\label{AnnCR}
w^{3/2}q^{-1/24}\lt(\frac{\mathcal{R}(w;q)}{w(1-w)}+\frac{\mathcal{R}(w\zeta;q)}{w\zeta(1-w\zeta)}+\frac{\mathcal{R}(w\zeta^2;q)}{w\zeta^2(1-w\zeta^2)}\rt)
=\frac{3\eta^3(3\t)}{\eta(\t)\vartheta(3\a;3\t)}.\end{equation}

We recall the \textit{crank} of a partition, which is defined by
$$\mathrm{crank}=\left\{
\begin{array}{ll}
    \mathrm{the\ largest\ part}, & \hbox{{if} part $1$ does not appear;} \\
    e(k)-k, & \hbox{{if} part $1$ appears $k$ times,} \\
\end{array}
\right.$$ where $e(k)$ denotes the number of parts in the partition that are strictly
larger than $k$. Let $M(m,n)$ denote the number of partitions of a nonnegative integer $n$ with crank $m$.  If we define $M(0,1)=-1$, $M(-1,1)=M(1,1)=1$ and $M(m,1)=0$ otherwise, the generating function for $M(m,n)$ is given by, \cite{AG},
\begin{equation}\label{gencrank}
\mathcal{C}(w;q):=1+\sum_{m=-\i}^\i\sum_{n=1}^\i M(m,n)w^mq^n=\frac{(q;q)_\i}{(wq;q)_\i(q/w;q)_\i}.\end{equation}
Then from \eqref{AnnCR}, we deduce the following relation between $\mathcal{R}(w;q)$ and $\mathcal{C}(w;q)$:
\begin{equation}\label{rank-crank}
\frac{\mathcal{R}(w;q)}{w(1-w)}+\frac{\mathcal{R}(w\zeta;q)}{w\zeta(1-w\zeta)}+\frac{\mathcal{R}(w\zeta^2;q)}{w\zeta^2(1-w\zeta^2)}
=\frac{3(q^3;q^3)_\i\mathcal{C}(w^3;q^3)}{(q;q)_\i(1-w^3)}.\end{equation}

Furthermore, we note that the left hand side of equation \eqref{rank-crank} can be written as
\begin{equation}\label{left}
L:=\frac{1}{1-w^3}\lt(\mathcal{R}(w;q)(\frac{1}{w}+1+w)+\mathcal{R}(w\zeta;q)(\frac{1}{w\zeta}+1+w\zeta)
+\mathcal{R}(w\zeta^2;q)(\frac{1}{w\zeta^2}+1+w\zeta^2)\rt).\end{equation}

If we apply
$$\mathcal{R}(w;q)=1+\sum_{m=-\i}^\i\sum_{n=1}^\i N(m,n)w^mq^n$$  into \eqref{left}, we find that
\begin{equation}
L=\frac{3}{1-w^3}\sum_{m=-\i}^\i\sum_{n=0}^\i\lt(N(3m,n)+N(3m+1,n)+N(3m-1,n)\rt)w^{3m}q^n,\end{equation}
from the fact $1+\zeta+\zeta^2=0$.

Hence, from \eqref{rank-crank}, we derive the theorem below after replacing $w^3$ by $w$:
\begin{thm}\label{rc} Let $N(m,n)$ and $M(m,n)$ denote the numbers of partitions of a nonnegative integer $n$ with rank $m$ and crank $m$, respectively. Then we have
\begin{eqnarray}\label{rcc}&&\sum_{m=-\i}^\i\sum_{n=0}^\i\lt(N(3m,n)+N(3m+1,n)+N(3m-1,n)\rt)w^{m}q^n\cr
&&\hspace{1in}=\frac{(q^3;q^3)_\i}{(q;q)_\i}\sum_{m=-\i}^\i\sum_{n=0}^\i M(m,n)w^{m}q^{3n}.\end{eqnarray}
\end{thm}

Theorem \ref{Rank-Crank} follows from Theorem \ref{rc} immediately.  For example, when $n=4$, there are 5 different partitions, and
$N(-1,4)+N(0,4)+N(1,4)=3=M(0,0)p_3(4)+M(0,1)p_3(1)$.

For another consequence of Theorem \ref{rc}, we substitute $w=-1$ in \eqref{rcc} and obtain:
\begin{corol} Let $N(t,m,n)$ and $M(t,m,n)$ denote the number of partitions of $n$ with rank and crank, respectively, congruent to $t$ modulo $m$.  Then
 \begin{equation}\label{rcc-1}\sum_{n=0}^\i N_{(6)}(n)q^n=\frac{(q^3;q^3)_\i}{(q;q)_\i}\sum_{n=0}^\i M_{(2)}(n)q^{3n}=\frac{(q^3;q^3)^2_\i}{(q;q)_\i(-q^3;q^3)^2_\i},\end{equation}
where $\displaystyle{N_{(6)}(n):=\sum_{t=0,\pm 1}N(t,6,n)-\sum_{t=3,\pm 2}N(t,6,n)}$ and $\displaystyle{M_{(2)}(n):=M(0,2,n)-M(1,2,n)}.$

And hence, \begin{equation}\label{rcc-2}N_{(6)}(n)=\sum_{0\leq 3k\leq n} M_{(2)}(k)p_3(n-3k).\end{equation}
\end{corol}
From the work of Treneer \cite{T}, we know that $N_{(6)}(n)$ has infinitely many congruences in arithmetic progressions modulo any prime coprime to $6$. Although $M_{(2)}(n)$ satisfies a Ramanujan type congruence modulo $5$, namely, $M_{(2)}(5n+4)\equiv 0\pmod 5$ which is proved in \cite{CKL}, there seems no such simple congruences for $N_{(6)}(n)$.  However, it seems that $N_{(6)}(2n)>0$ and $N_{(6)}(2n+1)<0$ for any non-negative integers $n$ such as $M_{(2)}(n)$. But note that $N_{(6)}(1)>0$ while $M_{(2)}(1)<0$.  We will leave this observation as a conjecture.

In closing this section, we should mention that Theorem \ref{Anng3} is a special case of a more general identity due to Atkin and Swinnerton-Dyer \cite[(5.1)]{AtS} who used the general identity to prove Dyson's results for the rank modulo 5 and 7.  It also leads to the rank-crank PDE, \cite[Theorem 1.1]{AtG}, which is another relation between the rank and crank.

\section{Mock Jacobi forms that arise in $_1\psi_1$ summation formula}

 It is much more convenient to discuss our subject in terms of the following equivalent form of Ramanujan's $_1\psi_1$ summation formula (\ref{1psi1}):
\begin{prop}[{Three variable Reciprocity
Theorem} \cite{K07}]\label{3var} If $c\neq -aq^{-n}$, $-bq^{-m}$ for non-negative integers $n$ and $m$, then
\begin{equation}\label{3rec}
\rho(a,b,c)-\rho(b,a,c)=\left(\frac{1}{b}-\frac{1}{a}\right)
\frac{(c)_\i(aq/b)_\i(bq/a)_\i(q)_\i}{(-c/a)_\i(-c/b)_\i(-aq)_\i(-bq)_\i},
\end{equation}
 where
\begin{equation}\label{rho3}\rho(a,b,c):=(1+\frac{1}{b})\sum_{n=0}^\i\frac{(c)_n(-1)^nq^{n(n+1)/2}(a/b)^n}{(-aq)_n(-c/b)_{n+1}}.\end{equation}\end{prop}

If we set $c=0$ in (\ref{3rec}), we obtain Ramanujan's reciprocity theorem \cite[Theorem 1.1]{K07} and identity (\ref{ann}) is
a special case of (\ref{3rec}) when $c=-q$.

\begin{proof}[Second proof of Theorem \ref{Anng2}]\label{c=-q}
Letting $b=-a=-w$ and $c=-q$ in the three variable reciprocity theorem (\ref{3rec}), we find that
$$\sum_{n=0}^\i
\frac{(-q)_n q^{n(n+1)/2}}{(-w)_{n+1}(-q/w)_{n+1}}+\sum_{n=0}^\i
\frac{(-q)_n
q^{n(n+1)/2}}{(w)_{n+1}(q/w)_{n+1}}=\frac{2(-q;q)^2_\i(q^2;q^2)_\i}{(w^2;q^2)_\i(q^2/w^2;q^2)_\i}.$$
Applying Jacobi triple product identity (\ref{jtp}), we obtain (\ref{ann}).\end{proof}

As commented in the introduction, Bringmann, Ono, and Rhoades proved Theorem \ref{Anng2} when $w=q^r$ for a rational number $r$ in \cite[Theorem 1.3]{BOR}. The theorem has one more result, and we state it here again with their notations:
\begin{prop}\cite[Theorem 1.3]{BOR}\label{k} Let $K'$ and $K''$ be defined as in Theorem \ref{q2cq}.
If $\zeta_\r:=e^{2\pi i/\r}$, $f_\r:=2\r/\gcd(\r,4)$, and
$$\tilde{K}(\a,\r;\t):=\frac{1}{4}\csc(\a\pi/\r)q^{-\frac{1}{8}}K'(\zeta_\r^\a;\t)+\sin(\a\pi/\r)q^{-\frac{1}{8}}K''(\zeta_\r^\a;\t),$$
then $\tilde{K}(\a,\r;2f_\r^2\t)$ is a weight 1/2 weakly holomorphic
modular form on $\Gamma_1(64f_\r^4)$.
\end{prop}

Theorem \ref{q2cq} is more general than Proposition \ref{k} and it is an immediate consequence of the reciprocity theorem above.

\begin{proof}[1st proof of Theorem \ref{q2cq}]
Replacing $q$ by $q^2$, and letting $a=-w$, $b=-wq$ and $c=q$ in (\ref{3rec}), and simplifying the resulting equation, we deduce that
\begin{equation}\label{ktheta}\frac{1}{1-w}K'(w;\t)+(1-\frac{1}{w})K''(w;\t)=\frac{(q;q^2)^3_\i(q^2;q^2)_\i}{(w;q)_\i(q/w;q)_\i}.\end{equation}
Using Jacobi triple product identity (\ref{jtp}), we finish this proof.\end{proof}

\begin{proof}[2nd proof of Theorem \ref{q2cq}]
We first find generalized Lambert series expansions for $K'(w;\t)$ and $K''(w;\t)$ so that we can observe what happens in terms of mock Jacobi forms.
If we let $\epsilon\to\i$ and then replace $q$ by $q^2$ in (\ref{watson}), we have
\begin{eqnarray}\label{einfty} 1+\sum_{n=1}^\i\frac{(\a q^2;q^2)_{n-1}(\b,\r,\delta ;q^2)_n(1-\a q^{4n})q^{2n(n+1)}}
{(\a q^2/\b,\a q^2/\r,\a q^2/\delta ,q^2;q^2)_n} \lt(\df{\a^2}{\b\r\delta }\rt)^n \cr
=\frac{(\a q^2;q^2)_\i}{(\a q^2/\delta;q^2 )_\i} \sum_{n=0}^\i\frac{(\delta , \a
q^2/(\b\r);q^2)_nq^{n(n+1)}}{(\a q^2/\b,\a q^2/\r,q^2;q^2)_n}\lt(-\frac{\a }{\delta }\rt)^n,\end{eqnarray}
since $\lim(\epsilon)_n\epsilon^{-n}=(-1)^nq^{n(n-1)/2}.$
Setting $\a=1$, $\delta=q$, $\b=w$, and $\r=1/w$  in (\ref{einfty}), we deduce that
\begin{equation}\label{k'}K'(w;\t)=\frac{(q;q^2)_\i}{(q^2;q^2)_\i}\lt\{1+(1-w)(1-\frac{1}{w})\sum_{n=1}^\i\frac{(1+q^{2n})q^{2n^2+n}}{(1-q^{2n}/w)(1-wq^{2n})}\rt\}.\end{equation}
Hence \begin{equation}\label{k'1}\frac{1}{1-w}K'(w;\t)=\frac{(q;q^2)_\i}{(q^2;q^2)_\i}\sum_{n=-\i}^\i\frac{q^{2n^2+n}}{1-wq^{2n}}.\end{equation}
Next, after letting $\epsilon\to\i$, setting $q$ by $q^2$, $\a=q^2$, $\delta=q$, $\b=wq$, and $\r=q/w$  in (\ref{watson}), we find that
\begin{equation}\label{k''}K''(w;\t)=-\frac{(q;q^2)_\i}{(q^2;q^2)_\i}\sum_{n=0}^\i\frac{(1+q^{2n+1})q^{2n^2+3n+1}}{(1-q^{2n+1}/w)(1-wq^{2n+1})}.\end{equation}
Thus \begin{equation}\label{k''1}(1-\frac{1}{w})K''(w;\t)=-\frac{(q;q^2)_\i}{(q^2;q^2)_\i}\sum_{n=-\i}^\i\frac{q^{2n^2+3n+1}}{1-wq^{2n+1}}.\end{equation}
Adding (\ref{k'1}) and (\ref{k''1}), we derive that
\begin{equation}\label{k'addk''}\frac{1}{1-w}K'(w;\t)+(1-\frac{1}{w})K''(w;\t)=\frac{(q;q^2)_\i}{(q^2;q^2)_\i}\sum_{n=-\i}^\i\frac{(-1)^n q^{n(n+1)/2}}{1-wq^{n}}.\end{equation}
Now by using Lemma \ref{thetaLam} and Jacobi triple product identity (\ref{jtp}), we complete the proof.\end{proof}

According to (\ref{k'1}) and (\ref{k''1}), each of $\displaystyle{\frac{1}{1-w}K'(w;\t)}$ and $\displaystyle{(1-\frac{1}{w})K''(w;\t)}$ belongs to a more general type of mock Jacobi forms than $\mu$ in (\ref{mu}), called a level 2 Appell function by Zagier and Zwegers \cite{Zwegers2}.  The proof above shows that when the index $n$ in the series in the right hand side of (\ref{k'addk''}) is even, the sum equals $\displaystyle{\frac{1}{1-w}K'(w;\t)}$ and when it is odd, the sum is $\displaystyle{(1-\frac{1}{w})K''(w;\t)}$.  Hence the theta quotient in (\ref{q2cqe}) is obtained by adding two mock Jacobi forms not by canceling two mock jacobi forms such as in Theorems \ref{Anng2} and \ref{Anng3}.

\section{Concluding Remarks}

We have observed that once we write $q$-series as generalized Lambert series

$\displaystyle{\frac{1}{\theta(\t;k\t)}\sum_{n=-\i}^\i\frac{(-1)^nq^{kn(n+1)/2}}{1-wq^{n}}}$
($k<3$), Zwegers' work on mock Jacobi forms provide short but beautiful and complete information on the $q$-series. When $k=1$ and  $\theta(\t;k\t)=1$, the series is a theta quotient and when $k=2$ or $3$ and $\theta=\vartheta$, it is a sum of a theta quotient and $k-1$ mock Jacobi forms.  For larger values of $k$, Zagier and Zwegers \cite{Zwegers2} observed that it is still a holomorphic part of a mock modular form which is a sum of a theta quotient and $k-1$ correction terms.
Further study in this direction may help us to see which $q$-series is a mock modular form.

config: arxiv


TITLE: Can the natural boundary be part of the unit circle?
QUESTION [2 upvotes]: It is well-known that the function $f(z)=\sum_{n=0}^\infty z^{n!}$ is analytic in the open unit disk and it can not be extended analytically to any proper open superset of the unit disk, i.e., the unit circle is the natural boundary of $f$.
Is there an example of a function, analytical in the unit disk, with "natural boundary" a part (of positive 1-dimensional measure) of the unit circle, which extends analytically to rest of $\mathbb C$.
I am not asking for a variation of $g(z)=\sqrt{z^2-1}$, which is defined in $\mathbb C\smallsetminus [-1,1]$, since the segment $[-1,1]$ is not a "natural boundary" of $g$, as $g$ can be analytically continued through this segment.

REPLY [8 votes]: Every closed set $F$ on the unit circle is the set of singularities of some analytic function.
Take a countable dense subset $z_k$ of $F$ and then choose positive $a_k$ so small
that the series
$$f(z)=\sum_k\frac{a_k}{z-z_k}$$
converges uniformly on compact subsets of $C\backslash F$.


TITLE: How to find the value of $e^{-x} x^n$ at x = $\infty$?
QUESTION [1 upvotes]: How to find the value of $e^{-x} x^n$ at x = $\infty$ ? Actually while proving that $\Gamma(n+1) = n\Gamma(n)$ there is a step where I need to evaluate  ($e^{-x} x^n$)$\big|_0^\infty$ . Now, since this is not an improper integral but the expression after calculating an improper integral(by parts) for which I need to find the value by substituting upper and lower limit, I donot know how to calculate the value at infinity since infinity is not a finite value. Thanks for any help.

REPLY [1 votes]: There is a result that $\lim_{x \rightarrow \infty}\frac{p(x)}{e^x} = 0$ for any polynomial $p(x)$. You can see this intuitively looking at Taylor's Series expansion of $\exp(x)$.
Then, $$\frac{x^n}{e^x} |_0^\infty = \lim_{\alpha \rightarrow \infty} \frac{\alpha^n}{e^\alpha} - \frac{0}{1} = 0. $$
Otherwise, you could use L'Hopital since numerator and denominator are going to $\infty$ and are continuous and differentiable in all $\mathbb R$.


TITLE: methods for finding the limit of $n$-step probability
QUESTION [1 upvotes]: I have the following transition matrix of a markov chain:
$$\mathbb{P}=
\begin{bmatrix}
1 & 0 & 0 & 0 & 0\\
q & \delta & p & 0 & 0\\
0 & q & \delta & p & 0\\
0 & 0 & 0 & \delta+q & p\\
0 & 0 & 0 & q & \delta+p
\end{bmatrix}$$
I am trying to calculate the limit of $\mathbb{P}^{n}$.
There are two closed sets $\{1\}$ and $\{4,5\}$. In addition, these two sets are irreducible. States $2$ and $3$ communicate those two closed sets, they will get lost in those two sets eventually. So, $p_{2,2}=p_{2,3}=p_{3,2}=p_{3,3}=0$ in the long run. State $1$ is a closed state, it stays there forever. For the rest of the entries, I applied the kolmogorov equation.
There are a lot of computations when using the kolmogorov equation. I am wondering if there are better ways.

REPLY [1 votes]: With a heavy calculation (that's the crux of this problem), you can verify that $\mathbb{P}$ has the following Jordan canonical form (a more secured and expedient way is to use Wolfram Alpha Jordan Form Calculator):
\begin{equation}
J = \text{diag}(1, \delta, \delta - \sqrt{pq}, \delta + \sqrt{pq}, 1),
\end{equation}
i.e., there exists an invertible matrix $Q$, such that $\mathbb{P} = QJQ^{-1}$. One choice of $Q$ is as follows:
\begin{equation}
Q = \begin{bmatrix}
q^{-2}\alpha & 0 & 0 & 0 & 0 \\
q^{-1}(1 - \delta) & \rho^2 & -\sqrt{\rho} & \sqrt{\rho} & \alpha^{-1}p^2   \\
1 & 0 & 1 & 1 & \alpha^{-1}p(p + q)\\
0 & -\rho & 0 & 0 & 1 \\
0 & 1 & 0 & 0 & 1
\end{bmatrix},
\end{equation}
where
$\alpha = p^2 + pq + q^2, \rho = \frac{p}{q}$.
Therefore,
$$\mathbb{P}^n = QJ^nQ^{-1}.$$
Under the constraint $p \in (0, 1), q \in (0, 1), \delta \in (0, 1), p + q + \delta = 1$, it follows that
$$ |\delta \pm \sqrt{pq}| \leq \delta + \sqrt{pq} \leq \delta + \frac{1}{2}(p + q) < \delta + p + q = 1.$$
Hence
$$\lim_{n \to \infty} J^n = \text{diag}(1, 0, 0, 0, 1).$$
Consequently,
$$\lim_{n \to \infty} \mathbb{P}^n = Q\text{diag}(1, 0, 0, 0, 1)Q^{-1} = \begin{bmatrix}
1 & 0 & 0 & 0 & 0 \\
\alpha^{-1}q(p + q) & 0 & 0 & \alpha^{-1}p^2(1 - \beta) & \alpha^{-1}p^2\beta \\
\alpha^{-1}q^2 & 0 & 0 & \alpha^{-1}pq & \alpha^{-1}p^2 \\
0 & 0 & 0 & 1 - \beta & \beta \\
0 & 0 & 0 & 1 - \beta & \beta
\end{bmatrix},$$
where $\beta = \frac{p}{p + q}$.


TITLE: How is the definition of work motivated?
QUESTION [3 upvotes]: For most dynamical variables in classical physics, I can understand how one may have decided to introduce them as a result of some "incompleteness" in Newton's laws of motion. For example:

One would like to say that applying a force to an object over time "gives" it something, so momentum must be introduced as
\begin{align} \mathbf p = \int _0 ^t \mathbf F(t') dt', \text{ or as some quantity such that }\mathbf F = \frac{d\mathbf p}{dt}. \end{align}
It is useful to describe forces that result in changes in angular velocity $\mathbf{ \omega = v \times r}$, so that leads to the definition of angular momentum and torque.
For some forces it is convenient to say that they serve to make a system approach some stable state, no matter what their initial condition is. Take a spinning top, for example. No matter how you spin it, it will always approach higher "stability" by falling onto the table. The definition of potential energy in the equation $\mathbf F = -\nabla U$ emphasizes the idea of increasing "stability", or decreasing potential, when a force acts on something.

These examples all involved some kind of physically obvious observation (exerting a force gives an object something, forces sometimes rotate things, some forces make things more stable), to which the response was to model the idea using Newton's laws. One can even do something similar with kinetic energy, by using the definition of work, by saying that a change in kinetic energy is just an alternate form for the work done on some object.
But there's something about work which just does not feel motivated at all to me. The work done on an object moving along a path $\gamma$ is defined as
\begin{align} W = \int _\gamma \mathbf F \cdot d\mathbf r, \end{align}
but I simply do not see how this is a quantity which "needed" to exist. Of course it is a very useful quantity, but I'm still having some trouble understanding what it really means intuitively.
Internet definitions (and even textbook ones) don't help much; often either kinetic energy is randomly defined from which the definition of work follows, work is randomly defined from which the definition of kinetic energy follows, or (my least favourite) a circular definition is employed, where energy is defined to be the ability to do work, and work is defined as the energy transferred to an object.
So what is the real meaning of the term "work", before even discussing anything about energy?
(I should clarify that I do not necessarily want to know the historical motivation for the definition of work. I am aware that sometimes things just "turn up" in physics because of experimentation with numbers, etc. I suppose I'm wondering what the "constructivist" approach to the definition of work is.)

REPLY [1 votes]: The fundamental forces in newtonian mechanics, namely gravity and columb forces depend on distance instead of time, thus it becomes sensible to write $F=mv\frac {dv}{dx}$ instead of $F= m\frac {dv}{dt}$
The following images are from my personal notes, I couldn't copy and paste the text due to some issues Apologies for that.


TITLE: What do I have to do different when taking derivative of $2^{-x}$ opposed to $2^x$
QUESTION [4 upvotes]: So for $2^x$ I know the derivative would be
$$2^x \ln(2)$$
What would be the thing or step I'd do different for $2^{-x}$ ?
I've trying to take the derivative of
$$40\over1+2^{-t}$$
& I keep getting the wrong asnwer, I think it's because I'm not sure about how to handle the $2^{-t}$

REPLY [0 votes]: First of all, the derivative of $2^x = e^{(\ln 2) x}$ is $(\ln 2) e^{(\ln 2) x} = (\ln 2)2^x$, not $1 / ((\ln 2) 2^x)$ (which admittedly was given in an earlier revision of the question).  This follows from the general formula
$\dfrac{de^{u(x)}}{dx} = e^{u(x)}\dfrac{du(x)}{dx} \tag{1}$
by taking $u(x) = (\ln 2) x$; we also used the identity $a = e^{\ln a}$, $a > 0$ to write $2^x= e^{(\ln 2) x}$.  The derivative of $2^{-x}$ can be computed in essentially the same manner; we simply note that $2^{-1} = e^{\ln 2^{-1}} = e^{-\ln 2}$ and set $u(x) = -(\ln 2)x$ to obtain
$\dfrac{d2^{-x}}{dx} = \dfrac{de^{(-\ln 2)x}}{dx} = (-\ln 2)e^{(-\ln 2)x} = (-\ln 2) 2^{-x}. \tag{2}$
As for
$\dfrac{40}{1 + 2^{-t}} = 40(1 + 2^{-t})^{-1}, \tag{3}$
we simply apply the chain rule (which has already occured in the formula (1)) a couple of times, first to obtain
$\dfrac{d}{dt}(40(1 + 2^{-t})^{-1}) = -40(1 + 2^{-t})^{-2}\dfrac{d2^{-t}}{dt}, \tag{4}$
then once again to $d2^{-t}/dt$, which we basically have already done in (2):
$\dfrac{d}{dt}(40(1 + 2^{-t})^{-1}) = -40(1 + 2^{-t})^{-2}(-\ln 2)2^{-t} = \dfrac{40 (\ln 2)2^{-t}}{(1 + 2^{-t})^2} = \dfrac{40 (\ln 2)}{(2^{2t} + 2^t)^2}.  \tag{5}$
And that's as far as I'll take it.  I can't see the advantage to fiddling around too much with powers of $2$ and so forth, or why Steve Wolfram's engine gives a better formula!
Hope this helps.  Cheers!
and as ever,
Fiat Lux!!!


TITLE: Hyperbolic PDE that is Non-Homogenous
QUESTION [1 upvotes]: Say I have a PDE of form $$Au_{xx} + Bu_{xy} + Cu_{yy} = xy$$
and that A,B, and C are constants chosen so that the PDE is hyperbolic. How do I go about solving this system? I solved the homogeneous case first, but I am not sure if this is the first step and if so how to extend it to solve the actual PDE.

REPLY [0 votes]: You need to find a particular solution of the complete equation. Since $x\,y$ is a polynomial of degree $2$, it seems reasonable to look for a solution that is a polynimial of degree $4$. We see that
$$
(x^2\,y^2)_{xy}=4\,x\,y.
$$
This suggets a solution of the form
$$
u=a\,x^4+b\,x^2\,y^2+c\,y^4.
$$


TITLE: Geometry of a d10
QUESTION [2 upvotes]: A d10, in roleplayers' lingo, is a 10-sided dice.
Several different shapes for this item exist, but the most common is that of two 5-sided pyramids with their equilateral bases lying on the same plane and offset by 36° (1/10, or half a side of a pentagon), as shown in the lower part of the image:

Both pyramids have their apex on the vertical of A, and the upper part of the drawing (shown at a different zoom below, so that all letters can be read) is built upon points B and C for the upper pyramid (projected: H and J) and I and F for the lower one (projectet: K and G).

It is trivial to see that the x component of AC is ABcos(36°)
Now, the planes defined by the 10 side faces of the pyramids meet on a jagged line that goes up and down the plane where the original bases lie. I suppose the meeting point between a face and the edge of an opposite pyramid are N and P but I'm not sure I'm right.
Suppose that ∠HAJ (and therefore also ∠KLG) is 90° (then, because of simmetry, ∠LPM and ∠LNM are 90° too), what's the LN/LP ratio?
I'm sure I can calculate ∠JHL and ∠HJL from the ratio between the sides (which is the same as the ratio between the x component of AC and and ABcos(36°)) and that I could get JL and HL from there but I can't think about a way to find NJ and PH.

Images created with geogebra

REPLY [2 votes]: Not a very geometric solution but it gets the job done. Take three consecutive points on the jagged equator:
$$(\cos\frac{\pi}5, -\sin\frac{\pi}5, x), (1,0,-x), (\cos\frac{\pi}5, \sin\frac{\pi}5, x)$$
where $x$ is some still to be determined deviation. The plane through these points determines the apex at $(0,0,(5+2\sqrt 5) x)$. Now $x$ needs to be taken such that the three points $$(-1, 0, x), (0,0,(5+2\sqrt 5) x), (1, 0, -x)$$  in the $XZ$ plane (vertex apex vertex) form a right angle. This gives $$(4+2\sqrt 5)(6+2\sqrt 5)x^2 = (44 + 20 \sqrt 5)x^2=1.$$ For this value of $x$ the length of the long edge is $\sqrt{1+(4+2\sqrt 5)^2x^2}=\tfrac12\sqrt{5+\sqrt5}$ and of the long diagonal $\sqrt{1+(6+2\sqrt 5)^2x^2} = \sqrt{\sqrt 5}$. The ratio long diagonal to long edge is therefore $\sqrt{\sqrt 5 -1}$.


TITLE: How do I know there are only 5 different groups of order 8?
QUESTION [1 upvotes]: How many different groups are there in order 8?
And how do I know which groups they are?
I mean, is there anyone can teach me to calculate them?
I want a proof, thank you!
They are $C_8$, $D_4$, $Q_8$, $C_{4h}=C_4 \times V_2$, $D_{2h}=D_2 \times V_2$

REPLY [3 votes]: If the group has an element of order $8$ then it is cyclic. If all non-identity elements have order $2$ it is abelian ($1=(ab)^2=abab$ so that $ab=a(abab)b=a^2bab^2=ba$) - and there is only one option.
So any other group must have at least one element $a$ of order $4$. Note that the subgroup $<a>$ generated by $a$ has index $2$ and is therefore normal. Consider the elements $b \notin <a>$.
If there is such a $b$ of order $2$ which commutes with $a$, then the group is abelian $\mathbb Z_4 \times \mathbb Z_2$.
If there is an element $b$ of order $2$ which does not commute with $a$ then $b^{-1}ab$ must be an element of $<a>$ (normal subgroup) of order $4$ which is not equal to $a$ so must be $a^{-1}=a^3$.
The only other possibility is that all the elements outside $<a>$ have order $4$, making six elements of order $4$, one of order $2$ and the identity. The square of any element of order $4$ must be the element of order $2$. So we take $b\notin <a>$ of order $4$ - which cannot commute with $a$ as this would make the whole group commutative, and this doesn't work with all those elements of order 4. As before $<a>$ is normal, so we must have $b^{-1}ab=a^{-1}$, and together with $a^4=b^4=1$ and $a^2=b^2$
There are five groups, which are readily seen to be distinct by considering the orders of the elements.


TITLE: Finding all positive real roots of $e^x - x -2 = 0$ using Newton Raphson's method.
QUESTION [0 upvotes]: How should I go approach this problem if I need to solve this problem by hand? Is there a general formula that I can use?

REPLY [0 votes]: Without Lambert function.
Consider that you look for the zero's of function
$$f(x)=e^x - x -2$$ for which
$$f'(x)=e^x-1 \qquad \text{and} \qquad f''(x)=e^x \,\, >0 \,\,\forall x$$  To start Newton method, you need a guess for each possible root.
In this problem, the first derivative cancels at $x_*=0$ where $f(x_*)=-1$ and the second derivative test confirms that this point is a minimum; so two roots.
Expand $f(x)$ as a Taylor series around $x=0$; this gives
$$f(x)=-1+\frac{x^2}{2}+O\left(x^3\right)$$ and as estimates $\pm \sqrt 2$.
Now, just run.


TITLE: Prove that $G$ is a vector space over $\mathbb Z_2$
QUESTION [1 upvotes]: Suppose $G$ is an abelian group such that all non-identity elements in $G$ has order $2$.
Prove that $G$ is a vector space over $\mathbb Z_2$
Since $G$ is an abelian group only thing to show is to find an external composition .define $\circ :G\times \mathbb Z_2\rightarrow G$ by $g\circ 0=0$ and $g \circ 1=g$
Is this the solution
?Where $g^2=e$ forall g is used I cant identify?

REPLY [4 votes]: "Is this the solution?" - well if I were your instructor I would want lots more detail.  But I'm not, so you should ask your instructor.
"Where is $g^2=e$ used?" - first of all, the vector "addition" will be the group operation.  So it would be much less confusing to write the group operation as addition: write $g+h$ instead of $gh$, and $g+g$ instead of $g^2$, and $0$ instead of $e$.  Then it is not too hard to see that you need the property $g+g=0$ to prove the scalar distributive law
$$(\alpha+\beta)g=(\alpha g)+(\beta g)\ .$$
Proof: for the case $\alpha=\beta=1$ we have
$$\eqalign{
  LHS&=(1+1)g=0g=0\cr
  RHS&=(1g)+(1g)=g+g=0\cr}$$
so $LHS=RHS$; the other three cases are easy.

REPLY [1 votes]: Perhaps using $+$ as the group operation might help see things clearly. Need also that $(a+b) \circ x = a\circ x+ b \circ x$ and the distributivity for vector addition across scalar multiplication. I think if not for order 2, the equation $x+x=2x=0$ would give problems as generally $x+x$ is not zero in most abelian groups, but, the interlinking of $\mathbb{Z}_2$ through distributive axioms demands this.


text:
{\bf Problem.} Andy, Beth, Charlie, and Daniel take a test with thirty questions. Andy and Beth together get the same number of questions wrong as Charlie and Daniel together. Andy and Daniel together get four more questions wrong than Beth and Charlie do together. If Charlie gets five questions wrong, how many questions does Andy get wrong?
{\bf Level.} Level 4
{\bf Type.} Algebra
{\bf Solution.} Let $a$ denote the number of questions Andy got wrong, $b$ the number of questions Beth missed, $c$ the number of questions Charlie answered incorrectly, and $d$ the number of questions Daniel got wrong. Using the information given in the problem, we can form the following system of linear equations:  \begin{align*}
a + b &= c + d \\
a + d &= b+c +4 \\
c = 5
\end{align*} Adding the first two equations together yields $2a + b + d = 2c + b + d + 4$, which simplifies to $a = c + 2$. Since $c = 5$, Andy must have answered $\boxed{7}$ questions incorrectly.

set_name: MATH


TITLE: Is $k[x,y]_{(x,y)}/(x^2-y^3)$ regular local?
QUESTION [1 upvotes]: Let $k$ be field,  and $A:=k[x,y]_{(x,y)}/(x^2-y^3). $
Now, $A$ is 1-dimension Noetherian local normal domain.
In general, 1-dimension Noetherian local domain is DVR iff normal.
So I think $A$ is DVR, in particular regular local.
What is my mistake?

REPLY [4 votes]: The element $T = x/y$ in the fraction field of $A$ satisfies
$$
T^3 - x = 0,
$$
but $T$ is not in $A$. Therefore $A$ is not integrally closed in its fraction field and is not normal.


TITLE: $\lim_{x \to 0}\frac{x- \sin{x}}{x^2}$
QUESTION [2 upvotes]: Calculate:
$$\lim_{x \to 0}\frac{x- \sin{x}}{x^2}$$
I would like to try but i don't find any idea i don't know how to use Hopital rule i tried to return $\cos$ to $\sin$
But it doesn't work a help

REPLY [1 votes]: L'Hopital is the easy way to do this, but here's another approach, using only $(\sin x)'=\cos x$ and $\sin x\lt x$ (for $x$ positive):
Since ${x-\sin x\over x^2}$ is an odd function, it suffices to consider the one-sided limit with $x\gt0$.  We have
$$0\le{x-\sin x\over x^2}={1\over x^2}\int_0^x(1-\cos u)du={1\over x^2}\int_0^x{\sin^2u\over1+\cos u}du\le{1\over x^2}\int_0^x\sin^2udu\\={1\over x^2}\int_0^xu^2{\sin^2u\over u^2}du\le{1\over x^2}\int_0^xu^2du={1\over x^2}\cdot{1\over3}x^3={1\over3}x\to0\text{ as }x\to0^+$$


TITLE: Interpolation inequalities
QUESTION [3 upvotes]: Let $\Omega$ be a regular domain of $\mathbb{R}^d$, $d=2,3$. Let $\mathcal{T}_h$ be a triangulation of $\Omega$ of size $h>0$.
Assume we can prove
\begin{equation*}
\begin{aligned}
\|v\|_{L^2(\Omega)}&\le C h\|v\|_{H^1(\Omega)}\\
\|v\|_{L^2( \Omega)}&\le C \|v\|_{L^2(\Omega)},
\end{aligned}
\end{equation*}
then by interpolation we get
\begin{equation}
\|v\|_{L^2(\Omega)}\le C h^s\|v\|_{H^s_i(\Omega)}\qquad(1),
\end{equation}
where $H^s_i(\Omega)=H^s(\Omega)$ if $0\le s<\frac{1}{2}$, $H^{\frac{1}{2}}_i(\Omega)=H^{\frac{1}{2}}_{00}(\Omega)$ and $H^s_i(\Omega)=H^s_{0}(\Omega)$ if $\frac{1}{2}< s\le 1$.
How to prove the inequality $(1)$?

REPLY [0 votes]: If you have two couple of Banach spaces $(X_{0},X_{1})$ and $(Y_{0},Y_{1})$
and $T:X_{0}+X_{1}\rightarrow Y_{0}+Y_{1}$ a linear operator such that $T$ is
bounded from $X_{i}$ to $Y_{i}$, $i=0,1$, then if you consider the
interpolation spaces $X_{\theta}:=(X_{0},X_{1})_{q,\theta}$ and $Y_{\theta
}:=(Y_{0},Y_{1})_{q,\theta}$ you have the interpolation inequality$$
\Vert T\Vert_{L(X_{\theta};Y_{\theta})}\leq C\Vert T\Vert_{L(X_{0};Y_{0}%
)}^{1-\theta}\Vert T\Vert_{L(X_{1};Y_{1})}^{\theta}.
$$
See for example
wikipedia-interpolation
From this inequality you get
\begin{align*}
\Vert T(x)\Vert_{Y_{\theta}}  & \leq\Vert T\Vert_{L(X_{\theta};Y_{\theta}%
)}\Vert x\Vert_{X_{\theta}}\\
& \leq C\Vert T\Vert_{L(X_{0};Y_{0})}^{1-\theta}\Vert T\Vert_{L(X_{1};Y_{1}%
)}^{\theta}\Vert x\Vert_{X_{\theta}}%
\end{align*}
for all $x\in X_{\theta}$. Now you have to use the facts that (see the book of
Lions and Magenes Non-homogeneous boundary problems and applications, volume
I)
\begin{align*}
L^{2}(\Omega)  & =(L^{2}(\Omega),L^{2}(\Omega))_{2,s},\\
H^{s}(\Omega)  & =(L^{2}(\Omega),H_{0}^{1}(\Omega))_{2,s},\quad0\leq
s<\frac{1}{2},\\
H_{00}^{1/2}(\Omega)  & =(L^{2}(\Omega),H_{0}^{1}(\Omega))_{2,1/2},\\
H_{0}^{s}(\Omega)  & =(L^{2}(\Omega),H_{0}^{1}(\Omega))_{2,s},\quad s>\frac
{1}{2}.
\end{align*}
So you will take $X_{0}=L^{2}(\Omega)$ and $X_{1}=H_{0}^{1}(\Omega)$ and
$Y_{0}=L^{2}(\Omega)$ and $Y_{1}=L^{2}(\Omega)$ and consider the operator
$T:L^{2}(\Omega)+H_{0}^{1}(\Omega)\rightarrow L^{2}(\Omega)+L^{2}(\Omega)$ to
be the identity $T(v):=v$. Then by your hypothesis $T:H_{0}^{1}(\Omega
)\rightarrow L^{2}(\Omega)$ is bounded with$$\Vert T(v)\Vert_{L^{2}}\leq Ch\Vert v\Vert_{H^{1}}%
$$
for all $v\in H_{0}^{1}(\Omega)$, which gives that $\Vert T\Vert_{L(H_{0}%
^{1}(\Omega);L^{2}(\Omega))}\leq Ch$. Similarly $\Vert T\Vert_{L(L^{2}%
(\Omega);L^{2}(\Omega))}=1$. Since $H_{s}=H_{i}^{s}(\Omega)$ and $Y_{s}%
=L^{2}(\Omega)$ it follows from the inequality%
\begin{align*}
\Vert v\Vert_{L^{2}}  & =\Vert T(v)\Vert_{L^{2}(\Omega)}\leq C\Vert
T\Vert_{L(L^{2}(\Omega);L^{2}(\Omega))}^{1-s}\Vert T\Vert_{L(H_{0}^{1}%
(\Omega);L^{2}(\Omega))}^{s}\Vert v\Vert_{H_{i}^{s}(\Omega)}\\
& \leq C1^{1-\theta}(Ch)^{s}\Vert v\Vert_{H_{i}^{s}(\Omega)}=C(Ch)^{s}\Vert
v\Vert_{H_{i}^{s}(\Omega)}%
\end{align*}
for all $v\in H_{i}^{s}(\Omega)$.


TITLE: Does this "Index condition" always implies that subgroup is normal?
QUESTION [1 upvotes]: Let $G$ be a finite group and $p$ be the smallest prime dividing $o(G)$,let $q(>p)$ be the next prime that divides $o(G)$.Suppose $G$ does not have any subgroup of index $p$ and has a subgroup say $H$ of index $q$ then $H$ is normal in $G$ ?

Edit: A simple example $A_5$ works as counterexample,So now additionally I assume that there is no prime between $p$ and $q$.
The above question is motivated from the example of  $A_4$ the Alternating group on 4 symbols,since $A_4$ does not have any subgroup of index $2$ and Klein-4 group is subgroup of $G$ of index $3$ which is clearly normal.

REPLY [4 votes]: Let $P$ be the image of the permutation action of $G$ on the cosets of the subgroup of index $q$. If the subgroup is not normal, then the action is not regular, so $|P| > q$. But since $|P|$ divides $q!$, $P$ must have order  $p^kq$ for some $k>0$.
So $P$ is solvable by Burnside's Theorem, and hence it has a normal subgroup $N$ of order $q$. But then the inverse image of $N$ in $G$ has index $p^k$ in $G$, so $G$ has a subgroup of index $p$, contrary to assumption.


TITLE: Algorithm for solving line line intersection in 3d
QUESTION [1 upvotes]: I am trying to find an algorithm that a computer can execute that finds the intersection point between two lines each defined by a point on the line and a direction vector. Does anyone know of one? It is preferable that I can do any solving ahead of time to make the code a matter of plugging numbers into a formula.

REPLY [0 votes]: Let's call your points $A, B$, with corresponding vectors $u,v$.
Then your intersection $M$, if it exists, satisfies
$$\overrightarrow{AM}=\lambda \vec{u}$$
$$\overrightarrow{BM}=\mu \vec{v}$$
That's two unknowns, which is not pleasing. But you can rewrite this as: $\overrightarrow{AM}=\lambda \vec{u}$, and $\overrightarrow{BM}$ is parallel to $\vec v$. The latter is expressed with cross product, so you have to find $\lambda$ such that $\overrightarrow{BM} \wedge \vec v=0$, with $\overrightarrow{BM}=\overrightarrow{BA}+\overrightarrow{AM}$, or:
$$\left(\overrightarrow{BA}+\lambda\vec u\right)\wedge \vec v=0$$
This will give you three linear equations (one per axis), but one is enough, the other are just safe checks, since $M$ may not exist at all.
By the way, for such a point $M$ to exist, you need that $\vec u,\vec v$ and $\overrightarrow{AB}$ be coplanar, that is
$$\det(\vec u,\vec v,\overrightarrow{AB})=0$$
It's not necessarily enough, since you could have parallel coplanar lines.


TITLE: The Michelson-Morley experiment (fringe shift)
QUESTION [1 upvotes]: What i don't understand about the experiment is why wasn't the first calculated path drift $ds$ enough but they had to rotate the expiriment by 90 degrees and calculate $dx= ds- ds'$? (Where the $ds'$ is the second calculated path drift). If there as a difference between the fringes wasn't the first calculation enought?

REPLY [1 votes]: Yeah, if you know the precise positions of the mirrors (down to <1 micrometer), and you know that the arms are aligned to the ether, then you could measure the effect of the ether in a single measurement.
Unfortunately, Michelson and Morley didn’t have the tech to know the first, and they couldn’t know the second in principle. So they did what’s known as a relative measurement. By subtracting the fringe measurements for the orthogonal orientations, the mirror position error is eliminated, and the effect of the ether (should it exist) is doubled.


TITLE: Generalizing Cantor's Nested Intervals Theorem
QUESTION [3 upvotes]: I am trying to prove the following: If $\{A_n\}_{n=1}^{\infty}$ is a nested sequence of sets in $\mathbb{R}$ each of which is closed, bounded, and non-empty, then $\cap_{n=1}^{\infty} A_n$ is non-empty.
I know the nested interval theorem, so I tried to make progress as follows. Let $a_n = \inf A_n$ and $b_n = \sup A_n$, which are well-defined since each $A_n$ is non-empty and bounded. The intervals $[a_n, b_n]$ are a nested sequence of non-empty, closed intervals and as such $\cap_{n=1} [a_n, b_n]$ is not empty. Also, $\cap_{n=1}^{\infty} A_n \subset \cap_{n=1} [a_n, b_n]$.
However, it is unclear if I can make further progress in this direction. Simple counterexamples show that $\cap_{n=1}^{\infty} A_n \neq \cap_{n=1} [a_n, b_n]$, e.g., $A_n = [0, 1/n] \cup [1, (n+1)/n]$ whence $\cap_{n=1}^{\infty} A_n = \{0, 1\}$, while $\cap_{n=1} [a_n, b_n] = [0, 1]$.
I have noticed some other things while trying to prove this statement. For example, if $A_n$ is a finite set for any $n$, say $A_n = \{a_1, ..., a_r\}$, then there must be some $i = 1,...,r$ such that $a_i \in A_m$ for all $m \geq n$ (because the sequence is nested and each set in the sequence is non-empty). Thus $\cap_{n=1}^{\infty} A_n$ contains $a_i$ and is not empty as desired.
Also, if $a$ is a limit point of $A_n$, then for $m \geq n$ either $a$ is in $A_m$ or $a$ is not a limit point of $A_m$ (since $A_m$ is closed).

REPLY [2 votes]: You can make progress. You don't need to consider endpoints. What I would do is choose $x_n\in A_n$ for all $n$. Since the $A_1$ is bounded and $x_n\in A_n\subseteq A_1$ for all $n$ it follows that $\{x_n\}$ is bounded. It now follows from the Bolzano-Weierstrass property that $\{x_n\}$ has at least one convergent subsequence, let $x$
be a limit of one such subsequence. Since $x$ is a limit of a subsequence of$\{x_n:n\in\mathbb{N}\}$ it follows that it is a limit of a subsequence of
$$\{x_m:m\in\mathbb{N} \ \text{and } m\geq n\}\subseteq A_n$$
for all $n$. Well then as $A_n$ is closed and $x$ is a limit of a sequence in $A_n$ we have that $x\in A_n$. Since $n$ was arbitrary $x\in \bigcap_{n\in\mathbb{N}} A_n$. Thus the set is not-empty.


TITLE: Suppose that $f (x)$ is continuous at $x = 0$. Prove that $\lim x f (x) = 0$ as $x→0$
QUESTION [0 upvotes]: (a) Suppose that $f (x)$ is continuous at $x = 0$. Prove that $\lim xf(x) = 0$ as $x→0$.
(b) Prove that the converse of (a) is not true. That is, find a function $f(x)$ such that $\lim xf(x) = 0$ as $x→0$, but $f$ is not continuous at $x = 0$.
It's obvious that any large number multiplied by $0$ is $0$ but how would I go about proving this? I don't understand.

REPLY [1 votes]: HINT: For part (a), since $f$ is continuous at 0, $\lim_{x \to 0} f(x) = f(0)$.  Now $\lim_{x \to 0} x f(x) = ?$ (does the limit of a product equal the product of the limits?)
For part (b), think about the function $f(x) = \frac{1}{\sqrt{x}}$.


TITLE: upper bound of outer measure on compact support of continuous function
QUESTION [4 upvotes]: Given $(\mathbb R^n,\mathcal B,\mu)$, where $\mu$ is a positive finite  Radon measure, define the function
$$
\mu^*(A) = \inf \left\{
\sum_{i=1}^n\mu(B_i) \,\,|\,\, A\subseteq \cup_i B_i, \,\, n \in \mathbb N
\right\}
$$
where $B_i$ are balls (both closed and open).
Take now $G:\mathbb R^n \to \mathbb R_+$ a continuous function with compact support $E$ and call
$$
E_r = \{x\in \mathbb R^n | G(x)>r \}.
$$
Given any $\varepsilon>0$, is it true that
$$
\mu^*(E_\epsilon) \le \mu(E_0)?
$$

REPLY [1 votes]: Yes, it is true, and to prove it I will use the following proposition.
Proposition. Let $\mu$ be any positive outer-regular Borel measure on $\mathbb R^n$, and let $K$ be any compact subset of $\mathbb R^n$. Then $\mu(K)= \inf \sum_{i=1}^k \mu(B_i)$,
where the infimum is taken over all finite open covers $\lbrace B_1,\ldots,B_k\rbrace$ of $K$ by open balls $B_i$.
Before presenting the (somewhat lengthy) proof of the proposition, I will first show how it resolves the question. Consider the set $K=G^{-1}\bigl([\epsilon,\infty)\bigr)$, which is closed (by continuity of $G$) and compact (since it is a subset of the support of $G$, which was assumed to be compact). Since Radon measures are outer-regular and Borel, the proposition applies, yielding that $\mu^\star(E_\epsilon)\leq \mu(K)$. Since $K\subseteq E_0$, the bound $\mu^\star(E_\epsilon)\leq \mu(E_0)$ follows.
Proof of the proposition.
Obviously $\mu(K)\leq \mu(B_1\cup\cdots\cup B_k)\leq \sum_{i=1}^k \mu(B_i)$ by subbaditivity, so $\mu(K)\leq \inf\sum_{i=1}^k \mu(B_i)$.
To establish the other inequality, we apply the following result:

Besicovitch Covering Theorem. Let $A$ be any subset of $\mathbb R^d$ and let $r\colon A\to\mathbb R_+$ be any function. Then
  there exists an integer $m\leq 5^d$ and disjoint subsets
  $A_1,\ldots,A_m$ of $A$ with the following properties:

For all $1\leq i\leq m$, the collection of open balls centered at $x\in A_i$ with radius $f(x)$ is mutually disjoint.
The union of all balls appearing in (1) contains $A$.

Fix $\epsilon>0$. If $\mu(K)=\infty$ there is nothing to prove, so suppose that $\mu(K)<\infty$. We will iteratively construct a cover of $K$ using open balls $B_i$ that are nearly disjoint. To start the iteration, set $K_0=K$. By outer regularity of $\mu$, there exists an open set $U_0$ containing $K_0$ such that $\mu(U_0)\leq(1+\epsilon)\mu(K_0)$. Cover each element of $K$ with a centered open ball contained in $U$. Apply the Besicovitch Covering Theorem to this cover. A disjoint subfamily with maximal $\mu$-measure will have measure at least $5^{-d}\mu(U_0)$. Since $\mu(K)<\infty$, there exists a finite disjoint subfamily - call it $C_0$ - with $\mu(C_0)\geq 6^{-d}\mu(U_0)$.
Given $(K_i,U_i,C_i)_{i=0}^n$ we continue the iteration by setting $K_{n+1}=K_n\setminus C_n$ and applying outer regularity to $K_{n+1}$ to find an open set $U_{n+1}$ containing $K_{n+1}$ with $\mu(U_{n+1}) \leq (1+\epsilon)\mu(K_{n+1})$. In fact, since $K_{n+1}\subseteq U_n$ we can further choose $U_{n+1}$ to be a subset of $U_n$ (by intersecting with $U_n$ if necessary). Cover $K_{n+1}$ with open balls in $U_{n+1}$ and apply Besicovitch to obtain a disjoint finite subcover $C_{n+1}$ with $\mu(C_{n+1})\geq 6^{-d}\mu(U_{n+1})$. These inequalities imply that
$$
\mu(K_{n+1})\leq \mu(U_n)-\mu(C_n)\leq (1-6^{-d})\mu(U_n)\leq (1-7^{-d})\mu(K_n),
$$
provided that $(1+\epsilon)(1-6^{-d})\leq (1-7^{-d})$ - which we can ensure by taking $\epsilon$ sufficiently small. Thus, we find that $\mu(K_n)$ decreases exponentially in $n$, and in fact
$$
\sum_{n=0}^{\infty}\mu(K_n)\leq 7^d\mu(K).
$$
Now it is time to bound the overlap between the $C_i$. Observe that
$$
C_n\cap (C_0\cup\ldots\cup C_{n-1})\subseteq U_n\setminus K_n,
$$
from which it follows by induction that
$$
\mu(C_0)+\cdots+\mu(C_n)\leq\mu(C_0\cup \cdots\cup C_n)+\mu(U_1\setminus K_1)+\cdots+\mu(U_n\setminus K_n),
$$
and we can bound the right side by $\mu(U_0)+\epsilon\sum_n \mu(K_n)$ (using the monotonicity $U_i\subseteq U_0$ for all $i$) to obtain an upper bound of
$$
\mu(C_0)+\cdots+\mu(C_n)\leq (1+\epsilon+\epsilon 7^d)\mu(K).\qquad (\star)
$$
We stop the iteration when $n$ becomes large enough so that $(1-7^{-d})^n\leq \epsilon$, in which case $\mu(K_n)\leq \epsilon\mu(K)$ and $\mu(U_n)\leq \epsilon(1+\epsilon)\mu(K)$. Rather than taking $C_{n+1}$ to be a disjoint subfamily as before, we allow overlaps in this final step. First, apply Besicovitch to the balls in $U_{n+1}$ covering $K_{n+1}$ to obtain $m\leq 5^d$ subfamilies whose union still covers $K_{n+1}$. Then apply compactness of $K_{n+1}$ to extract a finite subcover $C_{n+1}'$. Observe that even though this finite subcover is not disjoint, thanks to Besicovitch we still have a weak bound of
$$
\sum_{B\in C_{n+1}'}\mu(B)\leq 5^d\mu(U_{n+1}).\qquad (\star\star)
$$
Finally, we form the cover $C_0\cup\cdots\cup C_n\cup C_{n+1}'$ of $K$, which we now reindex in terms of the underlying open balls as $B_1\cup\cdots\cup B_k$. Combining $(\star)$ and $(\star\star)$ then yields
$$
\sum_{i=1}^k\mu(B_i)\leq (1+\epsilon+\epsilon 7^d+5^d\epsilon(1+\epsilon))\mu(K).
$$
Taking $\epsilon$ to $0$ yields the inequality
$$
\mu(K)\geq \inf\sum_{i=1}^k \mu(B_i),
$$
completing the proof of the proposition. $\square$


text:
\begin{definition}[Definition:Game/Rules]
A [[Definition:Game|game]] is specified completely by its '''rules'''.
They prescribe for each [[Definition:Player|player]] a [[Definition:Model of Rational Choice|model of rational choice]]:
{{:Definition:Model of Rational Choice}}
\end{definition}

config: wiki


text:
:: Consequences of Regularity Axiom
::  by Andrzej Trybulec

environ

 vocabularies TARSKI, XBOOLE_0;
 notations TARSKI, XBOOLE_0, ENUMSET1;
 constructors TARSKI, XBOOLE_0, ENUMSET1;
 registrations XBOOLE_0;
 requirements BOOLE;
 theorems TARSKI, XBOOLE_0, XBOOLE_1, ENUMSET1;
 schemes XFAMILY;

begin

 reserve x,y,X1,X2,X3,X4,X5,X6,Y,Y1,Y2,Y3,Y4,Y5,Z,Z1,Z2,Z3,Z4,Z5 for set;
 reserve X for non empty set;

:: To jest po prostu "poprawne" sformulowanie aksjomatu regularnosci
:: W TARSKI 'misses' jest rozekspandowane, zeby uniknac definiowania 'misses'

theorem Th1:
  ex Y st Y in X & Y misses X
proof
  consider x being object such that
A1:  x in X by XBOOLE_0:def 1;
  consider Y such that
A2: Y in X and
A3: not ex x be object st x in X & x in Y by A1,TARSKI:3;
  take Y;
  not ex x being object st x in X & x in Y by A3;
  hence thesis by A2,XBOOLE_0:3;
end;

theorem
  ex Y st Y in X & for Y1 st Y1 in Y holds Y1 misses X
proof
  defpred P[set] means $1 meets X;
  consider Z such that
A1: for Y holds Y in Z iff Y in union X & P[Y] from XFAMILY:sch 1;
  consider Y such that
A2: Y in X \/ Z and
A3: Y misses X \/ Z by Th1;
  assume
A4: not thesis;
  now
    assume
A5: Y in X;
    then consider Y1 such that
A6: Y1 in Y and
A7: not Y1 misses X by A4;
    Y1 in union X by A5,A6,TARSKI:def 4;
    then Y1 in Z by A1,A7;
    then Y1 in X \/ Z by XBOOLE_0:def 3;
    hence contradiction by A3,A6,XBOOLE_0:3;
  end;
  then Y in Z by A2,XBOOLE_0:def 3;
  then Y meets X by A1;
  hence contradiction by A3,XBOOLE_1:70;
end;

theorem
  ex Y st Y in X & for Y1,Y2 st Y1 in Y2 & Y2 in Y holds Y1 misses X
proof
  defpred P[set] means ex Y1 st Y1 in $1 & Y1 meets X;
  consider Z1 such that
A1: for Y holds Y in Z1 iff Y in union X & P[Y] from XFAMILY:sch 1;
  defpred Q[set] means $1 meets X;
  consider Z2 such that
A2: for Y holds Y in Z2 iff Y in union union X & Q[Y] from XFAMILY:sch 1;
  consider Y such that
A3: Y in X \/ Z1 \/ Z2 and
A4: Y misses X \/ Z1 \/ Z2 by Th1;
A5: now
    assume
A6: Y in Z1;
    then consider Y1 such that
A7: Y1 in Y and
A8: Y1 meets X by A1;
    Y in union X by A1,A6;
    then Y1 in union union X by A7,TARSKI:def 4;
    then Y1 in Z2 by A2,A8;
    then Y1 in X \/ Z1 \/ Z2 by XBOOLE_0:def 3;
    hence contradiction by A4,A7,XBOOLE_0:3;
  end;
  assume
A9: not thesis;
A10: now
    assume
A11: Y in X;
    then consider Y1,Y2 such that
A12: Y1 in Y2 and
A13: Y2 in Y and
A14: not Y1 misses X by A9;
    Y2 in union X by A11,A13,TARSKI:def 4;
    then Y2 in Z1 by A1,A12,A14;
    then Y2 in X \/ Z1 by XBOOLE_0:def 3;
    then Y2 in X \/ Z1 \/ Z2 by XBOOLE_0:def 3;
    hence contradiction by A4,A13,XBOOLE_0:3;
  end;
  Y in X \/ (Z1 \/ Z2) by A3,XBOOLE_1:4;
  then Y in Z1 \/ Z2 by A10,XBOOLE_0:def 3;
  then Y in Z2 by A5,XBOOLE_0:def 3;
  then Y meets X by A2;
  then Y meets X \/ Z1 by XBOOLE_1:70;
  hence contradiction by A4,XBOOLE_1:70;
end;

theorem
  ex Y st Y in X &
   for Y1,Y2,Y3 st Y1 in Y2 & Y2 in Y3 & Y3 in Y holds Y1 misses X
proof
  defpred P[set] means ex Y1,Y2 st Y1 in Y2 & Y2 in $1 & Y1 meets X;
  consider Z1 such that
A1: for Y holds Y in Z1 iff Y in union X & P[Y] from XFAMILY:sch 1;
  defpred R[set] means $1 meets X;
  defpred Q[set] means ex Y1 st Y1 in $1 & Y1 meets X;
  consider Z2 such that
A2: for Y holds Y in Z2 iff Y in union union X & Q[Y] from XFAMILY:sch
  1;
  consider Z3 such that
A3: for Y holds Y in Z3 iff Y in union union union X & R[Y]
    from XFAMILY:sch 1;
  consider Y such that
A4: Y in X \/ Z1 \/ Z2 \/ Z3 and
A5: Y misses X \/ Z1 \/ Z2 \/ Z3 by Th1;
A6: now
    assume
A7: Y in Z2;
    then consider Y1 such that
A8: Y1 in Y and
A9: Y1 meets X by A2;
    Y in union union X by A2,A7;
    then Y1 in union union union X by A8,TARSKI:def 4;
    then Y1 in Z3 by A3,A9;
    then Y1 in X \/ Z1 \/ Z2 \/ Z3 by XBOOLE_0:def 3;
    hence contradiction by A5,A8,XBOOLE_0:3;
  end;
A10: now
    assume
A11: Y in Z1;
    then consider Y1,Y2 such that
A12: Y1 in Y2 and
A13: Y2 in Y and
A14: Y1 meets X by A1;
    Y in union X by A1,A11;
    then Y2 in union union X by A13,TARSKI:def 4;
    then Y2 in Z2 by A2,A12,A14;
    then Y2 in X \/ Z1 \/ Z2 by XBOOLE_0:def 3;
    then Y meets X \/ Z1 \/ Z2 by A13,XBOOLE_0:3;
    hence contradiction by A5,XBOOLE_1:70;
  end;
  set V = X \/ Z1 \/ Z2 \/ Z3;
A15: V = X \/ (Z1 \/ Z2) \/ Z3 by XBOOLE_1:4
    .= X \/ (Z1 \/ Z2 \/ Z3) by XBOOLE_1:4;
  assume
A16: not thesis;
  now
    assume
A17: Y in X;
    then consider Y1,Y2,Y3 such that
A18: Y1 in Y2 & Y2 in Y3 and
A19: Y3 in Y and
A20: not Y1 misses X by A16;
    Y3 in union X by A17,A19,TARSKI:def 4;
    then Y3 in Z1 by A1,A18,A20;
    then Y3 in X \/ Z1 by XBOOLE_0:def 3;
    then Y3 in X \/ Z1 \/ Z2 by XBOOLE_0:def 3;
    then Y3 in X \/ Z1 \/ Z2 \/ Z3 by XBOOLE_0:def 3;
    hence contradiction by A5,A19,XBOOLE_0:3;
  end;
  then Y in Z1 \/ Z2 \/ Z3 by A15,A4,XBOOLE_0:def 3;
  then Y in Z1 \/ (Z2 \/ Z3) by XBOOLE_1:4;
  then Y in Z2 \/ Z3 by A10,XBOOLE_0:def 3;
  then Y in Z3 by A6,XBOOLE_0:def 3;
  then Y meets X by A3;
  hence contradiction by A15,A5,XBOOLE_1:70;
end;

theorem
  ex Y st Y in X &
   for Y1,Y2,Y3,Y4 st Y1 in Y2 & Y2 in  Y3 & Y3 in Y4 & Y4 in Y
    holds Y1 misses X
proof
  defpred P[set] means ex Y1,Y2,Y3 st Y1 in Y2 & Y2 in Y3 & Y3 in $1 & Y1
  meets X;
  consider Z1 such that
A1: for Y holds Y in Z1 iff Y in union X & P[Y] from XFAMILY:sch 1;
  defpred S[set] means $1 meets X;
  defpred R[set] means ex Y1 st Y1 in $1 & Y1 meets X;
  defpred Q[set] means ex Y1,Y2 st Y1 in Y2 & Y2 in $1 & Y1 meets X;
  consider Z2 such that
A2: for Y holds Y in Z2 iff Y in union union X & Q[Y] from XFAMILY:sch
  1;
  consider Z4 such that
A3: for Y holds Y in Z4 iff Y in union union union union X & S[Y] from
  XFAMILY:sch 1;
  consider Z3 such that
A4: for Y holds Y in Z3 iff Y in union union union X & R[Y] from
  XFAMILY:sch 1;
  consider Y such that
A5: Y in X \/ Z1 \/ Z2 \/ Z3 \/ Z4 and
A6: Y misses X \/ Z1 \/ Z2 \/ Z3 \/ Z4 by Th1;
A7: now
    assume
A8: Y in Z3;
    then consider Y1 such that
A9: Y1 in Y and
A10: Y1 meets X by A4;
    Y in union union union X by A4,A8;
    then Y1 in union union union union X by A9,TARSKI:def 4;
    then Y1 in Z4 by A3,A10;
    then Y1 in X \/ Z1 \/ Z2 \/ Z3 \/ Z4 by XBOOLE_0:def 3;
    hence contradiction by A6,A9,XBOOLE_0:3;
  end;
A11: now
    assume
A12: Y in Z1;
    then consider Y1,Y2,Y3 such that
A13: Y1 in Y2 & Y2 in Y3 and
A14: Y3 in Y and
A15: Y1 meets X by A1;
    Y in union X by A1,A12;
    then Y3 in union union X by A14,TARSKI:def 4;
    then Y3 in Z2 by A2,A13,A15;
    then Y3 in X \/ Z1 \/ Z2 by XBOOLE_0:def 3;
    then Y meets X \/ Z1 \/ Z2 by A14,XBOOLE_0:3;
    then Y meets X \/ Z1 \/ Z2 \/ Z3 by XBOOLE_1:70;
    hence contradiction by A6,XBOOLE_1:70;
  end;
A16: X \/ Z1 \/ Z2 \/ Z3 \/ Z4 = X \/ (Z1 \/ Z2) \/ Z3 \/ Z4 by XBOOLE_1:4
    .= X \/ (Z1 \/ Z2 \/ Z3) \/ Z4 by XBOOLE_1:4
    .= X \/ (Z1 \/ Z2 \/ Z3 \/ Z4) by XBOOLE_1:4;
A17: now
    assume
A18: Y in Z2;
    then consider Y1,Y2 such that
A19: Y1 in Y2 and
A20: Y2 in Y and
A21: Y1 meets X by A2;
    Y in union union X by A2,A18;
    then Y2 in union union union X by A20,TARSKI:def 4;
    then Y2 in Z3 by A4,A19,A21;
    then Y2 in X \/ Z1 \/ Z2 \/ Z3 by XBOOLE_0:def 3;
    then Y meets X \/ Z1 \/ Z2 \/ Z3 by A20,XBOOLE_0:3;
    hence contradiction by A6,XBOOLE_1:70;
  end;
  assume
A22: not thesis;
  now
    assume
A23: Y in X;
    then consider Y1,Y2,Y3,Y4 such that
A24: Y1 in Y2 & Y2 in Y3 & Y3 in Y4 and
A25: Y4 in Y and
A26: not Y1 misses X by A22;
    Y4 in union X by A23,A25,TARSKI:def 4;
    then Y4 in Z1 by A1,A24,A26;
    then Y4 in X \/ Z1 by XBOOLE_0:def 3;
    then Y4 in X \/ Z1 \/ Z2 by XBOOLE_0:def 3;
    then Y4 in X \/ Z1 \/ Z2 \/ Z3 by XBOOLE_0:def 3;
    then Y4 in X \/ Z1 \/ Z2 \/ Z3 \/ Z4 by XBOOLE_0:def 3;
    hence contradiction by A6,A25,XBOOLE_0:3;
  end;
  then Y in Z1 \/ Z2 \/ Z3 \/ Z4 by A16,A5,XBOOLE_0:def 3;
  then Y in Z1 \/ (Z2 \/ Z3) \/ Z4 by XBOOLE_1:4;
  then Y in Z1 \/ (Z2 \/ Z3 \/ Z4) by XBOOLE_1:4;
  then Y in Z2 \/ Z3 \/ Z4 by A11,XBOOLE_0:def 3;
  then Y in Z2 \/ (Z3 \/ Z4) by XBOOLE_1:4;
  then Y in Z3 \/ Z4 by A17,XBOOLE_0:def 3;
  then Y in Z4 by A7,XBOOLE_0:def 3;
  then Y meets X by A3;
  hence contradiction by A16,A6,XBOOLE_1:70;
end;

theorem
  ex Y st Y in X &
   for Y1,Y2,Y3,Y4,Y5 st Y1 in Y2 &  Y2 in Y3 & Y3 in Y4 & Y4 in Y5 & Y5 in Y
    holds Y1 misses X
proof
  defpred P[set] means ex Y1,Y2,Y3,Y4 st Y1 in Y2 & Y2 in Y3 & Y3 in Y4 & Y4
  in $1 & Y1 meets X;
  consider Z1 such that
A1: for Y holds Y in Z1 iff Y in union X & P[Y] from XFAMILY:sch 1;
  defpred T[set] means $1 meets X;
  defpred S[set] means ex Y1 st Y1 in $1 & Y1 meets X;
  defpred R[set] means ex Y1,Y2 st Y1 in Y2 & Y2 in $1 & Y1 meets X;
  defpred Q[set] means ex Y1,Y2,Y3 st Y1 in Y2 & Y2 in Y3 & Y3 in $1 & Y1
  meets X;
  consider Z2 such that
A2: for Y holds Y in Z2 iff Y in union union X & Q[Y] from XFAMILY:sch
  1;
  consider Z5 such that
A3: for Y holds Y in Z5 iff Y in union union union union union X & T[Y]
  from XFAMILY:sch 1;
  consider Z3 such that
A4: for Y holds Y in Z3 iff Y in union union union X & R[Y] from
  XFAMILY:sch 1;
  consider Z4 such that
A5: for Y holds Y in Z4 iff Y in union union union union X & S[Y] from
  XFAMILY:sch 1;
  set V = X \/ Z1 \/ Z2 \/ Z3 \/ Z4 \/ Z5;
  consider Y such that
A6: Y in V and
A7: Y misses V by Th1;
A8: V = X \/ (Z1 \/ Z2) \/ Z3 \/ Z4 \/ Z5 by XBOOLE_1:4
    .= X \/ (Z1 \/ Z2 \/ Z3) \/ Z4 \/ Z5 by XBOOLE_1:4
    .= X \/ (Z1 \/ Z2 \/ Z3 \/ Z4) \/ Z5 by XBOOLE_1:4
    .= X \/ (Z1 \/ Z2 \/ Z3 \/ Z4 \/ Z5) by XBOOLE_1:4;
A9: now
    assume
A10: Y in Z1;
    then consider Y1,Y2,Y3,Y4 such that
A11: Y1 in Y2 & Y2 in Y3 & Y3 in Y4 and
A12: Y4 in Y and
A13: Y1 meets X by A1;
    Y in union X by A1,A10;
    then Y4 in union union X by A12,TARSKI:def 4;
    then Y4 in Z2 by A2,A11,A13;
    then Y4 in X \/ Z1 \/ Z2 by XBOOLE_0:def 3;
    then Y meets X \/ Z1 \/ Z2 by A12,XBOOLE_0:3;
    then Y meets X \/ Z1 \/ Z2 \/ Z3 by XBOOLE_1:70;
    then Y meets X \/ Z1 \/ Z2 \/ Z3 \/ Z4 by XBOOLE_1:70;
    hence contradiction by A7,XBOOLE_1:70;
  end;
A14: now
    assume
A15: Y in Z2;
    then consider Y1,Y2,Y3 such that
A16: Y1 in Y2 & Y2 in Y3 and
A17: Y3 in Y and
A18: Y1 meets X by A2;
    Y in union union X by A2,A15;
    then Y3 in union union union X by A17,TARSKI:def 4;
    then Y3 in Z3 by A4,A16,A18;
    then Y3 in X \/ Z1 \/ Z2 \/ Z3 by XBOOLE_0:def 3;
    then Y3 in X \/ Z1 \/ Z2 \/ Z3 \/ Z4 by XBOOLE_0:def 3;
    then Y3 in V by XBOOLE_0:def 3;
    hence contradiction by A7,A17,XBOOLE_0:3;
  end;
A19: now
    assume
A20: Y in Z3;
    then consider Y1,Y2 such that
A21: Y1 in Y2 and
A22: Y2 in Y and
A23: Y1 meets X by A4;
    Y in union union union X by A4,A20;
    then Y2 in union union union union X by A22,TARSKI:def 4;
    then Y2 in Z4 by A5,A21,A23;
    then Y2 in X \/ Z1 \/ Z2 \/ Z3 \/ Z4 by XBOOLE_0:def 3;
    then Y2 in V by XBOOLE_0:def 3;
    hence contradiction by A7,A22,XBOOLE_0:3;
  end;
A24: now
    assume
A25: Y in Z4;
    then consider Y1 such that
A26: Y1 in Y and
A27: Y1 meets X by A5;
    Y in union union union union X by A5,A25;
    then Y1 in union union union union union X by A26,TARSKI:def 4;
    then Y1 in Z5 by A3,A27;
    then Y1 in V by XBOOLE_0:def 3;
    hence contradiction by A7,A26,XBOOLE_0:3;
  end;
  assume
A28: not thesis;
  now
    assume
A29: Y in X;
    then consider Y1,Y2,Y3,Y4,Y5 such that
A30: Y1 in Y2 & Y2 in Y3 & Y3 in Y4 & Y4 in Y5 and
A31: Y5 in Y and
A32: not Y1 misses X by A28;
    Y5 in union X by A29,A31,TARSKI:def 4;
    then Y5 in Z1 by A1,A30,A32;
    then Y5 in X \/ Z1 by XBOOLE_0:def 3;
    then Y5 in X \/ Z1 \/ Z2 by XBOOLE_0:def 3;
    then Y5 in X \/ Z1 \/ Z2 \/ Z3 by XBOOLE_0:def 3;
    then Y meets X \/ Z1 \/ Z2 \/ Z3 by A31,XBOOLE_0:3;
    then Y meets X \/ Z1 \/ Z2 \/ Z3 \/ Z4 by XBOOLE_1:70;
    hence contradiction by A7,XBOOLE_1:70;
  end;
  then Y in Z1 \/ Z2 \/ Z3 \/ Z4 \/ Z5 by A8,A6,XBOOLE_0:def 3;
  then Y in Z1 \/ (Z2 \/ Z3) \/ Z4 \/ Z5 by XBOOLE_1:4;
  then Y in Z1 \/ (Z2 \/ Z3 \/ Z4) \/ Z5 by XBOOLE_1:4;
  then Y in Z1 \/ (Z2 \/ Z3 \/ Z4 \/ Z5) by XBOOLE_1:4;
  then Y in Z2 \/ Z3 \/ Z4 \/ Z5 by A9,XBOOLE_0:def 3;
  then Y in Z2 \/ (Z3 \/ Z4) \/ Z5 by XBOOLE_1:4;
  then Y in Z2 \/ (Z3 \/ Z4 \/ Z5) by XBOOLE_1:4;
  then Y in Z3 \/ Z4 \/ Z5 by A14,XBOOLE_0:def 3;
  then Y in Z3 \/ (Z4 \/ Z5) by XBOOLE_1:4;
  then Y in Z4 \/ Z5 by A19,XBOOLE_0:def 3;
  then Y in Z5 by A24,XBOOLE_0:def 3;
  then Y meets X by A3;
  hence contradiction by A8,A7,XBOOLE_1:70;
end;

:: Opuszczone dwa przypadki, ktore odpowiadaja irrefleksywnosci
:: i asymetri

theorem
  not ( X1 in X2 & X2 in X3 & X3 in X1)
proof
A1: X2 in { X1,X2,X3 } & X3 in { X1,X2,X3 } by ENUMSET1:def 1;
A2: X1 in { X1,X2,X3 } by ENUMSET1:def 1;
  then consider T being set such that
A3: T in { X1,X2,X3 } and
A4: { X1,X2,X3 } misses T by Th1;
  T = X1 or T = X2 or T = X3 by A3,ENUMSET1:def 1;
  hence thesis by A2,A1,A4,XBOOLE_0:3;
end;

theorem
  not ( X1 in X2 & X2 in X3 & X3 in X4 & X4 in X1)
proof
A1: X2 in { X1,X2,X3,X4 } & X3 in { X1,X2,X3,X4 } by ENUMSET1:def 2;
A2: X4 in { X1,X2,X3,X4 } by ENUMSET1:def 2;
A3: X1 in { X1,X2,X3,X4 } by ENUMSET1:def 2;
  then consider T being set such that
A4: T in { X1,X2,X3,X4 } and
A5: { X1,X2,X3,X4 } misses T by Th1;
  T = X1 or T = X2 or T = X3 or T = X4 by A4,ENUMSET1:def 2;
  hence thesis by A3,A1,A2,A5,XBOOLE_0:3;
end;

theorem
  not ( X1 in X2 & X2 in X3 & X3 in X4 & X4 in X5 & X5 in X1)
proof
  assume that
A1: X1 in X2 and
A2: X2 in X3 and
A3: X3 in X4 and
A4: X4 in X5 and
A5: X5 in X1;
  set Z = { X1,X2,X3,X4,X5 };
A6: for Y st Y in Z holds Z meets Y
  proof
    let Y such that
A7: Y in Z;
    now
      per cases by A7,ENUMSET1:def 3;
      suppose
A8:     Y = X1;
        take y = X5;
        thus y in Z & y in Y by A5,A8,ENUMSET1:def 3;
      end;
      suppose
A9:     Y = X2;
        take y = X1;
        thus y in Z & y in Y by A1,A9,ENUMSET1:def 3;
      end;
      suppose
A10:    Y = X3;
        take y = X2;
        thus y in Z & y in Y by A2,A10,ENUMSET1:def 3;
      end;
      suppose
A11:    Y = X4;
        take y = X3;
        thus y in Z & y in Y by A3,A11,ENUMSET1:def 3;
      end;
      suppose
A12:    Y = X5;
        take y = X4;
        thus y in Z & y in Y by A4,A12,ENUMSET1:def 3;
      end;
    end;
    hence thesis by XBOOLE_0:3;
  end;
  X1 in { X1,X2,X3,X4,X5 } by ENUMSET1:def 3;
  hence contradiction by A6,Th1;
end;

theorem
  not ( X1 in X2 & X2 in X3 & X3 in X4 & X4 in X5 & X5 in X6 & X6 in X1)
proof
  assume that
A1: X1 in X2 and
A2: X2 in X3 and
A3: X3 in X4 and
A4: X4 in X5 and
A5: X5 in X6 and
A6: X6 in X1;
  set Z = { X1,X2,X3,X4,X5,X6 };
A7: for Y st Y in Z holds Z meets Y
  proof
    let Y such that
A8: Y in Z;
    now
      per cases by A8,ENUMSET1:def 4;
      suppose
A9:     Y = X1;
        take y = X6;
        thus y in Z & y in Y by A6,A9,ENUMSET1:def 4;
      end;
      suppose
A10:    Y = X2;
        take y = X1;
        thus y in Z & y in Y by A1,A10,ENUMSET1:def 4;
      end;
      suppose
A11:    Y = X3;
        take y = X2;
        thus y in Z & y in Y by A2,A11,ENUMSET1:def 4;
      end;
      suppose
A12:    Y = X4;
        take y = X3;
        thus y in Z & y in Y by A3,A12,ENUMSET1:def 4;
      end;
      suppose
A13:    Y = X5;
        take y = X4;
        thus y in Z & y in Y by A4,A13,ENUMSET1:def 4;
      end;
      suppose
A14:    Y = X6;
        take y = X5;
        thus y in Z & y in Y by A5,A14,ENUMSET1:def 4;
      end;
    end;
    hence thesis by XBOOLE_0:3;
  end;
  X1 in { X1,X2,X3,X4,X5,X6 } by ENUMSET1:def 4;
  hence contradiction by A7,Th1;
end;

subset_name: curated


TITLE: 2 How do I solve this equation involving a logarithm?
QUESTION [2 upvotes]: 2 How do I solve this equation involving a logarithm? 3

REPLY [1 votes]: $$\log_2\left(\frac{x}{2}\right) = \log_3\left(\frac{2+x}{3}\right)$$
$$\log_2(x)-\log_2(2) = \log_3(2+x)-\log_3(3)$$
$$\log_2(x)-1 = \log_3(2+x)-1$$
$$\log_2(x) = \log_3(2+x)$$
$$\log_3(2+x)=\frac{\log_{10}(2+x)}{\log_{10}(3) }$$
$$\log_2(x) = \frac{\log_{10}(x)}{\log_{10}(2)}$$
$$\therefore \frac{\log_{10}(2+x)}{\log_{10}(3)} = \frac{\log_{10}(x)}{\log_{10}(2)}$$
From now on, we'll write $\log_{10}(a)$ as $\log(a)$.
$$\log(2+x)\log(2) = \log(3)\log(x)$$
$$\log(2+x) = \frac{\log(3)}{\log(2)}\log(x)$$
$$ \frac{\log(3)}{\log(2)} \approx 1.585\ldots$$
$$\log(2+x) = 1.585\log(x)$$
$$\log(2+x) = \log(x^{1.585})$$
I don't have the time to finish off the answer right now, I'll do it later. Hopefully it should be obvious, just subtract and multiply out.