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Answer 1

With $n+k= {\lceil n \log_23\rceil}<n \log_23+1$ we can use the Rhin bound (page 160: $|µ_1\log 2+µ_2\log 3|\geq H^{-13.3}$ with $H=max(|µ_1|,|µ_2|)$)

$$|(n+k) \log2 - n \log3|>\frac 1{(n+k)^{13.3}}>\frac1{(n \log_23+1)^{13.3}}>\frac 1{n^a}$$ If we choose $a=15$ the above is true for $n>41$ (note that $(\log_23)^{13.3}\sim457$)

Now (growth of $x^a$ vs $a^x$) and with $\mu = e^{\log(2)-.1}=1.80967483607...$,

$$\frac 1{n^a}>\frac{\mu}{(\frac{3}{\mu^{\log_2(3)}})^n}$$ is true for $n>\frac{W_{-1}\Big(\frac{-\log\frac{3}{\mu^{\log_2(3)}}}{a\cdot \mu^\frac{1}{a}}\Big)}{\Big(\frac{-log\frac{3}{\mu^{\log_2(3)}}}{a}\Big)}$ or $n>610$ with chosen $a=15$ ($W$ is the productlog), and we also have (using $x>log(1+x)$) $$\frac{\mu}{(\frac{3}{\mu^{\log_2(3)}})^n}=\frac{\mu^{n\log_2(3)+1}}{3^n}>\frac{\mu^{\lceil n\log_2(3)\rceil}}{3^n}>\log(1+\frac{\mu^{\lceil n\log_2(3)\rceil}}{3^n})$$ So for $n>610$ we have $$|(n+k)\log2 - n \log3|=\log(\frac{2^{n+k}}{3^n})>\log(1+\frac{\mu^{\lceil n\log_2(3)\rceil}}{3^n})$$ or $$\frac{2^{n+k}}{3^n}>1+\frac{\mu^{\lceil n\log_2(3)\rceil}}{3^n}$$ $$2^{\lceil n \log_23\rceil}-3^n>\mu^{\lceil n\log_2(3)\rceil}$$

This leads to ($n>610$):

$$\begin{array}{|c|}\hline 2^{\lceil n \log_23\rceil}-3^n>\mu^{\lceil n\log_2(3)\rceil}\\\hline\end{array}$$ and with manual checking, inequality holds for ${\lceil n \log_23\rceil}>27$

Similarly, ....

Answer 2

The lower bounds of distances of perfect powers $\Delta=\mid a^A-b^B \mid$ is one of those attractive, simple-to-state, but difficult-to-solve problem. For me it occured in the question of cycles in the Collatz-problem, where we ask for the lower bound of distances of perfect powers of $2$ and $3$: $$\Delta(N,S)=|2^S - 3^N|$$ or $$\Lambda(N,S)=|S \cdot \ln2-N \cdot \ln 3| \qquad .$$
What I've found so far is

• the estimate by J.W.Ellison formulated in the 1970'ies (adressing $\Delta(N,S)$), and
• the estimate by G.Rhin formulated in 1987 (adressing $\Lambda(N,S)$), as reported in J.Simons & deWeger[2007] on the existence of "m-cycles" in the Collatz-problem.

The lower bounds from those estimates are -in the view of an amateur- astonishingly far away from the actual lowest distances -as checked for $N$ of about $1$ million digits. (Of course, the only relevant $N$ are selected according to the convergents of the continued fraction of $\log_2(3)$.)

Often discussions of this topic state the general form of estimation-formulae, but don't give explicite values for the constants involved - for instance in the well-visited blog of T. Tao on the "separation of powers of 2 and 3". Naive functional estimates for such lower bounds based on heuristics would suggest small values for such constants, for instance while G.Rhin gives something like $1/457/N^{13.3}$ a naiive fitting suggests something like $.../N^1$ or $.../N^{1.01}$ instead, valid for $N$ in my mentioned range of search.
A recent guess, using $.../N/\ln(N)$ instead, leaving aside any $1+\varepsilon$-exponent at all seems to be a really good estimate for the lower bounds. However, the famous estimate for the number of primes below some number $n$ as value of the integrallogarithm $Li()$ as proposed by the young K.F.Gauss has later been proven to be not really a lower bound, but where the first counterexample could only be upper-limited by something like $10^{10^{10^{34}}}$(WP:Skewes-Number) (modern computation reduced that upper bound to $1.39 \cdot 10^{316}$, see there).

Actual computations with large exponents diverge extremely far from that given estimates so I once tried to find an own algebraical pathway to such an lower-bound estimate. My current attempt provides a lower bound which is much nearer to the empirical values than the two other estimates - unfortunately this is not based on an analytical argument but only on heuristic finding.
Here I intend to create a "biglist" of immediately usable such lower bounds and invite other readers to add information and / or alternatives to the two established estimates and to try to improve my proposed estimate with analytical arguments.

Preliminaries:
While Ellison immediately looks at $\Delta(N,S) = |2^S - 3^N|$ and gives $$\Delta(N,S) \gt \exp(S \cdot (\ln 2 - 0.1)) \qquad S\gt 27 \tag {EL}$$

G. Rhin looks at the difference of logarithms $\Lambda(N,S)=| S \ln2 - N \ln 3 |$ and gives $$\Lambda(N,S) \gt {1\over 457 N ^{13.3} }| \tag{RH}$$ I've looked as well at $\Lambda(N)$ and propose $$\Lambda(N,S) \gt {1\over 10 N \ln N } \tag{Lb}$$ To compare that bounds, we must norm for the data of $\Delta()$ or for that of $\Lambda()$. The best adaption is surely dependent on the question that one has - either using a formula involving the exponential expression or one involving the logarithmic expression. So the discussion of this should be done in the single answers of the big-list where we reference some question in MSE.
To trigger the interest, here some pictures, how the three estimates behave in contrast to the empirical values. Here I adapted the Ellison-estimate to a $\Lambda()$-version.

We see, that the empirical $\Lambda()$ jitter between $0$ and $1/2 \cdot \ln 2$ (blue dots), the values at $N=\{2,5,12,53,...\}$ show very small distances, and even decreasing towards zero. The idea is to have a continuous function $f(N)$ which can supply a lower bound, such that we can say $\Lambda(N,S) \gt f(N)$ .
The red curve for the Ellison-estimate is below of the empirical values only for $N \gt 17$ while the green curve for the Rhin-estimate is so near the zero-line, that we barely see it. My estimate shown by the brown curve is always below the $\Lambda()$ .
To see a bit more detail, and to get aware of the basic characteristics of the estimates, a picture in log()-scalings is more appropriate. Picture 2: x,y logarithmic scaling Here we see the characteristics better. All three methods of estimates work for $N$ towards $=200$. Ellison took a much different shape compared with Rhin's, and while Rhin's lower bound is unfortunate small over the whole range, the Ellison's is initally too large, then better than Rhin's but from about $N \approx 700$ it decreases much faster than Rhin's. So, for the larger $N$, Rhin's lower bound should be preferred, but for the moderate values in $N$ Ellison's estimate gives a simpler formulation and a better lower bound.
But the "LBLam"-function has the best characteristic; it is nearly parallel to a lower hullcurve, and also is valid from the smallest $N=2$ on. That this property holds forever, that means the formula is analytically useful for all $N$, has not been proved, but empirical heuristic show that it holds for all $N$ up to $1$ million decimal digits and is always tight to the most extreme small values of $\Lambda()$...
A last picture, for the aesthetical impression is the following. I rescaled the values of $\Lambda(N,S)$ to the open interval $-1[...]1$ by $w(N,S)=4 \cdot \Lambda(N,S) / \ln 2 -1$ and then show the values $Y(N,S)=\tanh^{-1}(w(N,S))$ and the accordingly rescaled values for the three lower-bound-estimates. The slightly jittering effect in the Ellison-curve is because its function uses the somehow jittering values $S$ instead that of $N$ for its argument.

Answer 3

Consider the function $f(x,t) = xe^{\lambda t}$. Let $Y_t = f(X_t, t)$; using Itô's lemma, we get: $$ \begin{align*} dY_t &= f_t(X_t, t) dt + \frac{1}{2}\underbrace{f_{xx}(X_t, t)}_{=0} (dX_t)^2 + f_x(X_t, t) dX_t \\ &= \lambda X_t e^{\lambda t} dt + e^{\lambda t} dX_t \\ &= \sqrt{2c\lambda}e^{\lambda t} dB_t \end{align*}$$

Thus, assuming $X_0 = 0$ as an initial condition, we have $$Y_t = X_t e^{\lambda t} = \sqrt{2c\lambda } \int_0^t e^{\lambda s}dB_s$$ Since an Itô integral of a deterministic function is a Gaussian process, we have that $Y_t$ is a Gaussian process with mean zero and variance $$Var(Y_t) = 2c\lambda \int_0^t e^{2\lambda s}ds = 2c\lambda \frac{e^{2\lambda t } - 1}{2\lambda} = c(e^{2\lambda t } - 1)$$ After dividing by $e^{\lambda t}$, we find that $X_t$ is a Gaussian process with mean zero and variance $c(1 - e^{-2\lambda t})$

Answer 4

I'm trying to get a better understanding of ultraproducts and their typical uses.

To that end, I'm wondering what happens if we fix an ultrafilter $U \in 2^{2^{\mathbb{N}}}$ and look at $\prod_{i \in \mathbb{N}}$ where $S_i$ is the symmetric group associated with $i \in \mathbb{N}$.

The result is a group, and by the intution behind Łoś's theorem (proofwiki), it should satisfy the properties that "most" (with heavy scare quotes) symmetric groups satisfy.

However, being a symmetric group in the first place is not a first-order property, so I'm curious what group you actually end up with (and whether it's sensitive to the choice of $U$).

---

Suppose $I$ is a poset.

$F \in 2^{2^I}$ is a filter if and only if

• $F$ is nonempty.
• It holds that $\forall x \in F \mathop. \forall y \in F \mathop. \exists z \in F \mathop. z \le x \land z \le y$.
• It holds that $\forall x \in F \mathop. \forall y \in I \mathop. x \le y \to y \in F$.

$F$ is a proper filter if and only if $F \neq 2^I$.

$U$ is an ultrafilter if and only if

• $U$ is a filter on $I$.
• $U$ is not equal to $2^I$.
• Every proper filter $F$ extending $U$ is equal to $U$.

An ultrafilter is principal if and only if it is the upward closure of a single element of $I$.

Let $(I, \le)$ be $(\mathbb{N}, \le^\mathbb{N})$.

Let $U$ be a fixed non-principal ultrafilter on $I$.

Let $S_k$ be the $n$th symmetric group as usual.

I'm curious about $\prod_{i \in I} S_k / U$ (which I'll be writing as $\prod^U_{i \in I} S_k$ for convenience.

If we restrict our attention to algebraic theories, then we can define the truth of a sentence in $\prod^U_{i \in I} S_k$ as follows.

$\prod^U_{i \in I} S_k \models t_1 = t_2$ if and only if if the set of indices for which the $i$th component of $t_1$ and $t_2$ are equal is in $U$. By the nonemptiness of $U$ and the upward-closedness property of $U$, it holds that $I \in U$.

The interpretation of $f(\vec{a})$ is the equivalence class associated with $\langle f(a_0), f(a_1), \cdots \rangle$ where $a_i$ is the $i$th component of the canonical representative of the equivalence class $\vec{a}$.

I'm pretty sure $\prod^U_{i \in I} S_k$ and that this follows from the fact that $\prod_{i \in I} S_k $ is a group and modding out by the equivalence relation associated with $U$ is a group homomorphism.

$\prod_{i \in I} S_k $ is a Cartesian product of models of an algebraic theory consisting entirely of $\forall$-sentences, so it is again a model of that theory (the theory of groups, not the theory of symmetric groups).

I'm using the non-principalness of $U$ to insist that there isn't a proper subset of indices that completely determines the behavior of $\prod^U_{i \in I} S_k $. Although if I'm being honest, I'm mostly insisting on this because I've heard "non-principal ultrafilter" before.

As an object, $\prod^U_{i \in I} S_k$ is a bit odd.

I'm pretty sure it's infinite, but all of its factors are finite, so we can't prove its infiniteness by constructing an infinite sequence of examples that differ at every index.

Let $G$ be a finite symmetric group, let $g_k$ be $(k \mathop{\text{mod}} |G|)$-th element of the group, where the elements are assigned an index in some arbitrary fashion.

Let $b_i$ be equal to $(S_0)_i, (S_1)_i, (S_2)_i, \cdots$.

I'm pretty sure that $b_0, b_1, b_2, \cdots$ contains infinitely many distinct elements of $\prod^U_{i \in I} S_k$.

I think you can show this by collecting the index sets between identical elements and saying that they intersect to the empty set when there are infinitely many of them ... and therefore that $U = 2^I$ ... but I'm really not sure.

Answer 5

Let $g$ be a continuous function on $\mathbb R$ with the following property:

• $0\le g\le 1$,
• there are positive sequences $(a_n),(x_n),(y_n),(b_n)$ all converging to zero and $b_{n-1}<a_n<x_n<y_n<b_n<a_{n+1}$ so that $g =1$ in $[x_n, y_n]$, $g=0$ outside $[a_n, b_n]$ and $$\sum b_n-a_n <+\infty.$$

Define $f(x) =\int_0^x g(s)ds$. Then $$\frac{f(y_n)-f(x_n)}{y_n-x_n}=1$$ But $f'(0)=0$: to see this, note that for any $\epsilon >0$

Answer 6

Still editing as of 20th of Dec, 2021

To Lee David Chung Lin: If you are wondering why this lost a vote, I used an upvote to mark it as the one I normally edit, and I'm simply using another post to store a draft now. Happy holidays :)

The divergence of the integral as stated in the question is due to the presence of an atom (discrete point mass), as pointed out by BGM's comment. There are "proper" ways to deal with the point mass, but this answer will present the common technique that allows for an elementary solution.

Most of the analysis will be in the first section, and most of the calculation is left to the second. Two appendices come after the code block (numerical verification) to provide technical details.

Strategy and Analysis

The analysis presented here applies to general $n$ and not just the requested $n = 6$.

Denote $W$ as the largest of the set of $n-1$ iid in contrast to $Z$ being the max of $n$. Here's let's say $W$ excludes $X_1$. $$W \equiv \max\{X_2,\,X_3,\,\ldots,\, X_6\} $$ Clearly $W$ is independent to $X_1,\,$ and its marginal density $f_W$ has the same form as $f_Z$ except with $n-1$ in place of $n$. The distribution of $Z$ as the max of $n$ iid exponential is well-known, as are the moments. Same for $W$.

Knowing that $X_1$ is interchangeable with any of the $n-1$ other $X_i$, we have the "$\frac1n$ and$\frac{n-1}n$" split that allows $W$ to come in. \begin{align*} &\phantom{{}={}} E\bigl[ ZX_1] \\ &= E\bigl[ZX_1~\big|\, Z = X_1 \bigr] \Pr\{Z = X_1\} + E\bigl[ZX_1~\big|\, Z > X_1\bigr] \Pr\{Z > X_1\} \\ &= E\left[Z^2~\middle|\, Z = X_1\right] \frac1n + E\bigl[ZX_1~\big|\, Z > X_1 \bigr] \frac{n-1}n \tag{1.a} \label{Eq_conditional_split} \\ &= E\left[Z^2 \right] \frac1n + E\bigl[WX_1~\big|\, W > X_1 \bigr] \frac{n-1}n \tag{1.b} \label{Eq_conditional_split_transfer} \end{align*} With the moments of $Z$ and $W$ well-known, at this point the analysis is already done, and one may wish to skip directly to the next section Calculation and Results. Here allow me to elaborate the intracacies below.

Going from the second line \ref{Eq_conditional_split} to the third line \ref{Eq_conditional_split_transfer} is trivial yet at the same time profound.

The first term $E\left[Z^2~\middle|\, Z = X_1 \right] = E\left[Z^2 \right]$ is effectively unconditional, because $Z = X_i$ for exactly one of the $i = 1 \sim n$ almost surely (ties has measure zero) while all $X_i$ play the same role. To be specific, due to $E\left[Z^2~\middle|\, Z = X_i \right] = E\left[Z^2 \,\middle|\, Z = X_j \right]$ for all $i \neq j$, we have $$\begin{align} E\left[Z^2\right] &= \sum_{i=1}^n \frac1n E\left[Z^2\,\middle|\, Z = X_i \right] \\ &= n \cdot \frac1n E\left[Z^2~\middle|\, Z = X_1 \right] = E\left[Z^2~\middle|\, Z = X_1 \right]\end{align}$$ From the same line of reasoning, in fact we have identical distribution $\left(Z^2~\middle|\, Z = X_1 \right) \overset{d}{=} Z^2$. Note that this is conditioning on $Z = X_1$ and NOT conditioning on a given value of $Z = X_1 = x_1$.

As for the second term with $X_1 < Z$ going from Eq\eqref{Eq_conditional_split} to Eq\eqref{Eq_conditional_split_transfer}, by definition $Z = \max\{W, X_1\}$ so upon conditioning on $Z > X_1$ we have $W = Z$ as an identity between random variables (and not just identical in distribution).

---

Summary of the Strategy: diagonal split

The space can be split into disjoint cases as done in Eq\eqref{Eq_conditional_split}. For region $Z = X_1$ we bypass the joint density $f_{X_1Z}$ that is formidable by utilizing the marginal density $f_Z$ that is easy. For the other region, $Z > X_1$, we bypass the joint $f_{X_1Z}$ by recognizing its identical replacement $f_{X_1W}$, which is a joint density easily handled due to $W \perp X_1$.

---

This "diagonal split" applies to expectation of any well-behaving function $g(X_1, Z)$ in general and not just the product $g(X_1, Z)=X_1 Z$. Namely, we can compute $E[g(X_1, Z)]$ while not having to deal with $f_{X_1Z}$ which is cumbersome.

Nonetheless, $\, f_{X_1Z}$ is rather interesting and worth examining. Formally, we have the Heaviside step function (one form of indicator function), its derivative in the form of $\delta(\cdot)$ the Dirac delta function, and the first derivative of the Dirac delta function $\delta'(\cdot)$. Please consult wiki or Eq.(10) of the Wolfram page to see how $\delta'(\cdot)$ works. See Appendix.2 for an outline of the derivation of the full form of $f_{X_1Z}$ as in Eq\eqref{Eq_joint_density_full_form}.

Calculation and Results

For the contribution from the "boundary" $Z = X_1$, one can obtain the moments with minor effort (see Appendix.1 for details if needed) \begin{align*} \text{At}~n = 6\,,& & E[Z] &= \frac{49}{20} \frac1{\lambda} & E\left[Z^2 \right] &= \frac{13489}{1800}\frac1{\lambda^2 } &&\qquad \tag{2} \label{Eq_Z_2nd_moment_n=6} \end{align*} For what it's worth, $13489 = 7\cdot 41 \cdot 47$ and $13489/1800 \approx 7.49388888\ldots$

In the $Z > X_1$ region, since $W \perp X_1$ the calculation of $E\bigl[WX_1~\big|\, W > X_1 \bigr]$ is easy to setup. The joint density is just $f_{X_1W}(x, w) = f_X(x)\, f_W(w)$ where the marginal density for $X_1$ is $f_X$ that is common to all $X_i$.

$$ E\bigl[WX_1~\big|\, W > X_1 \bigr] = \frac{ E\bigl[W X_1\cdot \mathcal{I}_{W>X_1} \bigr] }{ \Pr\bigl\{ W > X_1 \bigr \} } = \frac{ \int\limits_{x=0}^{\infty} \int\limits_{w=x}^{\infty} \displaystyle ~x \, w\, f_{X_1W}(x, w) \,\mathrm{d}w \,\mathrm{d}x }{ (n-1)/n }$$ where the $\mathcal{I}_{\text{statement}}$ is the indicator function, as in $\mathcal{I}_{\text{statement}} = 1$ wherever "statement" is true and $\mathcal{I}_{\text{statement}} = 0$ otherwise.

Again, the denominator $\Pr\{ W > X_1 \}$ equals $(n-1)/n$ due to symmetry (note that ties have zero probability marginally). To be pedantic: $$\Pr\{ W > X_1 \} = 1 - \Pr\{ W \leq X_1 \} = 1 - \Pr\{ X_1~\text{is max} \} = 1 - \frac1n $$ For the record, below is the integral that one needs to evaluate. Note that the leading factor $n/(n-1)$ will cancel with that in Eq\eqref{Eq_conditional_split_transfer}, and this is not a coincidence. $$\begin{align} &E\bigl[WX_1~\big|\, W > X_1 \bigr] \\ & \qquad {}= \frac{n}{n-1} \int\limits_{x=0}^{\infty} \int\limits_{w=x}^{\infty} \lambda^2 (n-1) x\,w\,e^{-\lambda(w+x)} \left(1 - e^{-\lambda w} \right)^{n-2} \,\mathrm{d}w \,\mathrm{d}x \tag{3.a} \label{Eq_E[XZ_on_X-smaller]_general_n} \end{align}$$

The smaller $n$ cases are borderline manageable by hand, while in general one should not attempt this without CAS (computer algebra system) see the later half of Appendix.1. At $n = 6$, this evaluates to

$$E\bigl[WX_1~\big|\, W > X_1 \bigr] = \frac65 \frac{17381}{10800} \frac1{\lambda^2} \tag{3.b} \label{Eq_E[XZ_on_X-smaller]_n=6} $$ For what it's worth, $17381 = 7 \cdot 13 \cdot 191$ and $17381/10800 \approx 1.609351851851851\ldots~$, while $6/5$ times that is $\approx 1.93122222\ldots$

Take Eq\eqref{Eq_E[XZ_on_X-smaller]_n=6} and Eq\eqref{Eq_Z_2nd_moment_n=6} into Eq\eqref{Eq_conditional_split}, we have for $n = 6$:

\begin{align*} E[Z X_1] &= \frac16 E\left[ Z^2 ~\middle|\, X_1\right] + \frac56 E\bigl[ Z X_1~\big|\, Z > X_1 \bigr] \\ &= \frac16 E\left[ Z^2 \right] + \frac56 E\bigl[WX_1~\big|\, W > X_1 \bigr] \\ &= \frac{13489}{10800}\frac1{\lambda^2 } + \frac{17381}{10800} \frac1{\lambda^2}\\ &\hspace{-7pt} \bbox[4pt,border:2pt solid #55BB11]{ \begin{aligned}[t] &=\frac{343}{120} \frac1{\lambda^2} \\ &\approx \frac{ 2.8583333 }{\lambda^2} \end{aligned} } \end{align*} From here, one readily obtain the covariance using $E[Z]$ above and $E[X_1] = 1 / \lambda$. The covariance for $n = 6$ is $49/(120 \lambda^2)$. Using $E\left[ Z^2 \right]$ to get $V[Z]$ and do similar with $E\left[ X_1^2 \right] = 1/\lambda^2$, one obtains the correlation coefficient as $\frac72\sqrt{7\,/\,767} \approx 0.3343639$, moderately positive as expected.

Here's a quick and dirty Mathematica code verifying the results numerically (Monte Carlo).

ClearAll@SimulateEZX;
SimulateEZX[itrIn_: 17, InN_: 10^5, nIn_: 6, prcIn_: 20] := Module[{n (* # of iid *), itr, vN (* vector size *), itbyN, GTotal (* grand total*), data, Z, X, sumZX, prec, XisZ, sumBound, LenBound, bdN(* boundary case count *), bdTotal}, Timing[{itr, vN, n, prec} = Ceiling@{Norm@itrIn, Norm@InN, Norm@nIn, Norm@prcIn};  data = ConstantArray[0, {vN, n}]; Z = ConstantArray[0, vN];  X = Z; sumZX = ConstantArray[0, itr]; sumBound = sumZX;  LenBound = sumZX; XisZ = sumZX;
Do[ data = RandomVariate[ExponentialDistribution@1, {vN, n}, WorkingPrecision -> prec];
 Z = Max /@ data; X = data[[;; , 1]];     sumZX[[k]] = Total[X Z];
 XisZ[[k]] = Pick[X, MapThread[SameQ, {X, Z}]]; (* boundary case when X_1 \[Equal] Z *)
 LenBound[[k]] = Length@XisZ[[k]];    sumBound[[k]] = Total[ XisZ[[k]]^2 ]; 
 Print[{Row@{k, "th loop: "}, Row@{"sample_E[\!\(\*SubscriptBox[\(ZX\), \(1\)]\)] = ", sumZX[[k]]/vN}, Row@{"  sample_E[\!\(\*SubscriptBox[\(ZX\), \(1\)]\) | \ Z=\!\(\*SubscriptBox[\(X\), \(1\)]\)] = ", sumBound[[k]]/LenBound[[k]]}}]
 , {k, itr}];
{itbyN, GTotal, bdN, bdTotal} = {itr  vN, Total@sumZX, Total@LenBound, Total[sumBound]};
{ Row@{"Sample size = ", itbyN},  Row@{"  sample_E[\!\(\*SubscriptBox[\(ZX\), \(1\)]\)] = ",      GTotal/itbyN} , Row@{"boundary case: count = ", bdN, " ,freq = ", N[bdN/itbyN, prec]}, Row@{"  sample_E[\!\(\*SubscriptBox[\(ZX\), \(1\)]\) | \ Z=\!\(\*SubscriptBox[\(X\), \(1\)]\)] = ", bdTotal/bdN},  Row@{"  sample_E[\!\(\*SubscriptBox[\(ZX\), \(1\)]\) | \ \!\(\*SubscriptBox[\(X\), \(1\)]\)<Z] = ", (GTotal - bdTotal)/(itbyN - bdN)}
} ]  ];
SimulateEZX[] (* default quick run  *)
SimulateEZX[20, 4.5 10^5, 6, 15] (* run with your own parameters *)

$$\color{white}.$$

Appendix.1 : Distribution and Moments of Max of iid Exponential

Denote $Z$ as the maximum of $n$ random variables with iid exponential distribution with rate parameter $\lambda$. \begin{align*} F_Z(t) &= \left( 1 - e^{-\lambda t} \right)^n &&& f_Z(t) &= \lambda n e^{-\lambda t} \left( 1 - e^{-\lambda t} \right)^{n-1} \tag{A.1.1} \label{Eq_CDF_and_density_of_max} \end{align*} The moments can be obtained via routine (not necessarily easy) integration to be \begin{align*} E[Z] &= \frac1{\lambda} \mathcal{H}(n) & E\left[Z^2 \right] &= \frac1{ \lambda^2} \left( \frac{ \pi^2 }6 + \mathcal{H}(n)^2 -\psi'(n+1) \right) \tag{A.1.2} \label{Eq_Z_moments_general_n} \end{align*} where $\mathcal{H}(n)\equiv \sum_k^n \frac1k$ is the harmonic number and $\psi'(\cdot)$ is the first derivative of the digamma function (a.k.a polygamma with order 1).

A related calculation is the indispensable Eq\eqref{Eq_E[XZ_on_X-smaller]_general_n}. Ignoring the leading $n/(n-1)$ factor, this integral evaluates to a sequence $S_n$ that satisfies the following 2nd order difference equation

$$S_{n+1} = \frac{ n+ 1}{ (n-1) (n+2)^3 } \biggl[ n-4 + (n+2) \Bigl( 2 S_{n+1} \left(n^2+n-1 \right) - S_n + n^2 \Bigr) \biggr]$$

for $n \geq 2$, with the initial two terms $S_2 = 1/2$ (obvious) and $S_3 = 47/54$ (not so obvious). The first few terms are $\{\frac12, \frac{ 47 }{ 54 }, \frac{ 335 }{ 288 }, \frac{ 12641 }{ 9000 }, \frac{ 17381 }{ 10800 }, \frac{ 1103219 }{ 617400 }, \ldots\}$. Here for $n = 6$ what goes into Eq\eqref{Eq_E[XZ_on_X-smaller]_n=6} is $S_6 = 17381/10800$.

$$\color{white}.$$

Appendix.2 : Derivation of Unconditional Joint Density $f_{ZX_1}$

I shall NOT include a sketch of the $X_1$-$W$ plane, with $X_1$ as the horizontal axis. Please make the sketch yourself as it will be helpful. The rectangular region that $F_{X_1Z}(x,z)$ covers is always tall-and-thin (height is as least as large as the width) but never short-and-fat.

Keep in mind that $Z = \max\{W, X_1\}$. From the sketch it should be easy to see that \begin{align*} F_{X_1Z}( x \, ,\, z ) &= \Pr\bigl\{ W \leq z ~~\& ~~X_1 \leq x \bigr\} &&\\ &= \Pr\bigl\{ W \leq z \bigr\} \cdot \Pr\bigl\{X_1 \leq x \bigr\} \qquad \because W \perp X_1 \\ &= F_W(z) \cdot F_X(x) \tag{A.2.1} \label{Eq_joint_CDF_elementary} \end{align*} where again $F_W$ is as seen in Eq\eqref{Eq_CDF_and_density_of_max} with $W$ instead of $Z$ and $n-1$ place of $n$.

Of course, $F_{X_1Z}$ should be different from $F_{X_1W}$, and one needs more explicit expressions.

$$ F_{X_1Z}( x,\, z ) = F_W(z) \cdot F_X(x) \cdot \mathcal{I}_{z \geq x} \tag{A.2.2} \label{Eq_joint_CDF_indicator}$$

Even in its full indicator-form (with indicators regarding the range of $X$ and $W$), $F_{X_1W} = F_W(z) \cdot F_X(x)$ doesn't have this particular indicator $\mathcal{I}_{z \geq x}$.

Moving on. The indicator is a legitimate function, and the derivative of Eq\eqref{Eq_joint_CDF_indicator} follows the product rule as usual. \begin{align*} f_{X_1Z}( x \, ,\, z ) &= \frac{ \partial^2 F_W(z) F_X(x)}{ \partial x \partial z } \mathcal{I}_{z \geq x} + F_W(z) F_X(x)\frac{ \partial^2 \mathcal{I}_{z \geq x}}{ \partial x \partial z } \\ &\hspace{36pt} + \frac{ \partial F_W(z) F_X(x) }{ \partial z } \frac{ \partial \mathcal{I}_{z \geq x} }{ \partial x } + \frac{ \partial F_W(z) F_X(x) }{ \partial x } \frac{ \partial \mathcal{I}_{z \geq x} }{ \partial z } \end{align*} Apply the relations between the $\{ \mathcal{I}_{\cdot},\delta(\cdot),\delta'(\cdot) \}$ to eventually arrive at \begin{align*} f_{X_1Z}( x \, ,\, z ) &= \color{magenta}{ f_W(z)\, f_X(x)\, \mathcal{I}_{z \geq x} } + \delta(z-x) \biggl( \frac{ \partial }{ \partial z } \Bigl( F_W(z) F_X(x) \mathcal{I}_{z \geq x} \Bigr) \\ & \hspace{102pt} + F_W(z) f_X(x) - f_W(z) F_X(x) \biggr) \tag{A.2.3} \label{Eq_joint_density_full_form} \end{align*} The CDFs $F_W, F_X$ and their densities $f_W$, $f_X$ are established. One can pair up the densities in Eq\eqref{Eq_joint_density_full_form} to make it into a nice "matching" form if so desired.

The leading term (highlighted in magenta) corresponds to the region $Z > X_1$ that is identical to $f_{X_1W}$ conditionally.

Answer 7

This answer is free for anyone to use.

Answer 8

Question: Berry curvature and a discontinuous wavefunction

Body: Apparently, the wavefunction of quantum mechanics can have discontinuities at a point, because when conjugate squared, these contribute nothing to the overall probability.

However, the phase of a wavefunction has been found to matter in [recent years] in the form of Berry curvature. Insert briefest description of Berry curvature here.

What would happen if the wavefunction were discontinuous at a point, and we wanted to calculate the Berry curvature in the vicinity? Would the discontinuioty still contribute nothing to the physics, and only be a mathematical curiosity, or would it have an observable effect?

Tags: berry-pancharatnam-phase wavefunction quantum-mechanics

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Suppose I have a solid (as opposed to hollow) steel pole, which is initially straight, but is then bent into a torus. There is now an inner and outer radius, indicating that two line of atoms previously of the same length are now different. I am referring to the lines describing inner and outer radii.

What happens to the atoms intersecting these lines as the pole is continuously deformed into the new toroidal shape? Initially there would be about the same number of atoms intersected by each line when the steel was in a pole shape, but once deformed the numbers would disagree. Where do these atoms go when the pole is deformed?

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My initial guess would be that some of them go into the bulk of the pole (from the inner radius), and others move along the surface away from the inner radius (as opposed to just getting closer together). The atoms on the outer radius would seperate from another, while some atoms from the bulk, and others from the nearby surfaces would come in to fill the gaps.

I am assuming the pole does not tear in such a deformation.

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Suppose I have a triangular lattice, currently containing two atoms near absolute zero which are very close to one another.

There are two spin states available (up and down). In the two atom system, the configurations with the equally lowest energy are those with atom up and the other one down. If a third atom is added to the triangular lattice, there are more configurations of the system of the system with equally lowest energy than there were previously, and the average energy per atom has increased.

A phase transition is classically defined as when the free energy of the system suddenly changes. It can also be defined as when spontaneous symmetry breaking of an order parameter occurs, or in more recent years when a change in the systems topology occurs.

I want to know if adding the third atoms constitutes a phase transition of sorts, and if so whether it is a topological one?

Answer 9

This answer is free for anyone to use.

Answer 10

Note from Tyma Gaidash:

Hello, you can use this space.

Here is another post about uncommon standard special functions in the form of Elliptic Integrals which will be special cases of the Lauricella D function. Most elliptic integrals can be put in terms of a double hypergeometric functions like the Jacobi functions and others like the Arithmetic Geometric Mean, Elliptic Logarithm, and Legendre Ellipctic functions can all be put into closed form with a single or double hypergeometric series, but the Carlson Elliptic functions and Bulirsch’s Elliptic Integrals. Let’s start with some definitions which will allow integral and sum representations for the more complex elliptic integrals of a triple hypergeometric function with the Pochhammer Symbol. The Lauricella D function is also in Wolfram functions

$$\text F_\text D^{(n)}(a,b_1,…,b_n,c;x_1,…,x_n)=\sum_{m_1\ge0}\cdots\sum_{m_n\ge0}\frac{(a)_{\sum_{j=1}^\infty m_j}}{(c)_{\sum_{j=1}^\infty m_j}}\prod_{j=1}^\infty \frac{(b_j)_{m_j}x_j^{m_j}}{m_j!}=\frac{\Gamma(c)}{\Gamma(a)\Gamma(c-a)}\int_0^1t^{a-1}(1-t)^{c-a-1}\prod_{j=1}^n (1-x_j t)^{-b_j}dt$$

When the Hyperelliptic hypergeometrc function from DLMF can give any elliptic integral as a special case:

$$\text R_a(b_1,..,b_n;z_1,…,z_n)=\frac 1{\text B(-a,(-a)’)}\int_0^\infty t^{(-a)’-1}\prod_{j=1}^n(t+z_j)^{-b_j}dt,(-a)’=a+\sum_{j=1}^n b_j$$

The problem here is that the integral bounds for the Lauricella function, so some substitutions help fix the bounds to put the Hyperelliptic function into Lauricella form: $$\frac{\Gamma(a)\Gamma(c-a)}{\Gamma(c)}\text F^{(n)}_\text D(a,b_1,…,b_n,c;x_1,…,x_n)=\int_0^1t^{a-1}(1-t)^{c-a-1}\prod_{j=1}^n (1-x_j t)^{-b_j}dt \mathop=^{t\to\frac1t}-\int_\infty ^1t^{-a+1}(t-1)^{c-a-1}t^{-(c-a-1)}\prod_{j=1}^n (t-x_j )^{-b_j}t^{b_j}t^{-2}dt =\int_1^\infty t^{-c} (t-1)^{c-a+1}t^{b_1}\cdots t^{b_n}\prod_{j=1}^n (t-x_j)^{-b_j} dt =\int_1^\infty t^{-c+\sum\limits_{j=1}^n b_j}(t-1)^{c-a-1}\prod_{j=1}^n(t-x_j)^{-b_j}\mathop=^{t\to t+1}_{t-1\to t} \int_0^\infty (t+1)^{-c+\sum\limits_{j=1}^n b_j}t^{c-a-1}\prod_{j=1}^n(t+1-x_j)^{-b_j} $$

Now assume the two hypergemoetric functions are equal:

$$\frac{\Gamma(a)\Gamma(c-a)}{\Gamma(c)}\text F^{(n)}_\text D(a,b_1,…,b_n,c;x_1,…,x_n) = \int_0^\infty (t+1)^{-c+\sum\limits_{j=1}^n b_j}t^{c-a-1}\prod_{j=1}^n(t+1-x_j)^{-b_j}= \text B(-a,(-a)’) \text R_a(b_1,..,b_n;z_1,…,z_n)=\int_0^\infty t^{(-a)’-1}\prod_{j=1}^n(t+z_j)^{-b_j}dt $$

Therefore the following is implied. Note that we had the same $a$ variable in both hypergeometric functions, so let’s represent one variable for the hyperelliptic function as $a=\alpha$:

$$t^{\alpha+\sum\limits_{j=1}^n b_j-1}\prod_{j=1}^n(t+z_j)^{-b_j} = (t+1)^{-c+\sum\limits_{j=1}^n b_j}t^{c-a-1}\prod_{j=1}^n(t+1-x_j)^{-b_j} \implies z_j=1-x_j,c=\sum_{j=1}^n b_j=(-a)’-a,c-a-1= \alpha+\sum\limits_{j=1}^n b_j-1 =\alpha +c-1\implies -a\to\alpha,a=-\alpha$$

and:

$$\implies \frac{\Gamma(-\alpha)\Gamma\left(\sum\limits_{j=1}^n b_j+\alpha\right)}{\Gamma\left(\sum\limits_{j=1}^n b_j\right)}\text F^{(n)}_\text D\left(-\alpha,b_1,…,b_n, \sum\limits_{j=1}^n b_j;1-x_1,…,1-x_n \right)= \text B\left(-\alpha,\alpha+\sum_{j=1}^n b_j\right) \text R_\alpha(b_1,..,b_n;z_1,…,z_n)\implies \text R_\alpha(b_1,..,b_n;z_1,…,z_n) = \frac{\Gamma(-\alpha)\Gamma\left(\sum\limits_{j=1}^n b_j+\alpha\right)}{\Gamma\left(\sum\limits_{j=1}^n b_j\right) \text B\left(-\alpha,\alpha+\sum\limits_{j=1}^n b_j\right)} \text F^{(n)}_\text D\left(-\alpha,b_1,…,b_n, \sum\limits_{j=1}^n b_j;1-x_1,…,1-x_n \right) $$

The coefficients cancel. I have derived this simple looking formula before, and got the same result, but am not sure if it is correct: $$\boxed{\text F^{(n)}_\text D\left(-a,b_1,…,b_n, \sum\limits_{j=1}^n b_j;1-x_1,…,1-x_n \right)=\text R_a(b_1,…,b_n;x_1,…,x_n) =\frac{\int_0^\infty t^{a-1}\prod\limits_{j=1}^n(1+tx_j)^{-b_j} dt}{\text B\left(-a,a+\sum\limits_{j=1}^nb_j\right)}= \sum_{m_1\ge0}\cdots\sum_{m_n\ge0}\frac{(-a)_{\sum_{j=1}^\infty m_j}}{\left(\sum\limits_{j=1}^nb_j\right)_{\sum_{j=1}^\infty m_j}}\prod_{j=1}^\infty \frac{(b_j)_{m_j}(1-x_j)^{m_j}}{m_j!} }$$

This is the basic expression of the conversion formula, but is it correct? We can now put any elliptic integral into a Lauricella hypergeometric form and have many new decomposition formula for the Lauricalla D function which is a standard function.

Answer 11

The objective is to find $f(x)$ such that $$f(h(x)) = j(f(x)) \tag 1$$ where $h(x)$ and $j(x)$ are known functions. In order to get rid of the $h(x)$ inside, you want to somehow transform this into the equation $$g(t+1) = j(g(t)) \tag 2$$ This reveals a system of equations: $$\begin{align}f(h(x)) &= g(t+1)\tag 3\\ f(x) &= g(t) \tag 4\end{align}$$

$x$ is now a function of $t$ that we must find. In order to find $x(t)$, you can plug in $t+1$ to $(4)$ to get $$f(x(t+1)) = g(t+1)$$ which must be equal to $f(h(x(t))$. That is, you have that $$f(x(t+1)) = f(h(x(t))$$

Assuming $f$ is one-to-one, you can apply the inverse to get $$x(t+1) = h(x(t)) \tag 5$$

Note how $(2)$ and $(5)$ are pretty much the same. So the same techniques for solving $(5)$ can be applied to $(2)$. The question is, how is $(5)$ solved?

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Starting from a fixed point, $x(t_0) = c_0$, you can find that $x(t_0+1) = h(c_0)$, $x(t_0+2) = h(h(c_0)), \cdots,$ and $$x(t_0+m) = \underbrace{h(h(...h}_{h \text{ applied } m \text{ times}}(c_0)...)) \tag 6$$

With $g(t_1) = c_1$

$$g(t_1+n) = \underbrace{j(j(...j}_{j \text{ applied } n \text{ times}}(c_1)...)) \tag 7$$

If you assume everything's sufficiently nice, you can say $t = t_0+m$ to find an equation for $x$ in terms of $t, t_0, c_0$ and $t = t_1+n$ to find an equation for $g$ in terms of $t, t_1, c_1$.

Then from $(4)$, you got $$f(x(t)) = g(t) \to f(x(x^{-1}(u))) = f(u) = g(x^{-1}(u)) \tag 8$$

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With $h(x) = 10x$ and $j(x) = x+1$, $$x(t_0+m) = 10^m c_0 \to x(t) = 10^{t-t_0}c_0 = 10^{t-C_0}$$ and $$g(t_1+n) = n+c_1 \to g(t) = t-t_1+c_1 = t-C_1$$

(we're gonna ignore that this doesn't cover negative $x$)

Using $(8)$, $$f(10^{t-C_0}) = t-C_1 \to f(10^{\log_{10}(u)+C_0-C_0}) = \log_{10}(u)+C_0-C_1$$ Which yields $$f(u) = \log_{10}(u) + C$$

Answer 12

(Deutsch: MathJax: LaTeX Basic Tutorial und Referenz)

1. To see how any formula was written in any question or answer, including this one, right-click on the expression and choose "Show Math As > TeX Commands". (When you do this, the '$' will not display. Make sure you add these. See the next point.

Or click the Edit link at the bottom o a post to view the source code.

2. For inline formulas, enclose the formula in $...$. For displayed formulas, use $$...$$.
   These render differently. For example, type
   $\sum_{i=0}^n i^2 $
   to show $\sum_{i=0}^n i^2$ (which is inline mode) or type
   $$\sum_{i=0}^n i^2$$
   to show $$\sum_{i=0}^n i^2 $$ (which is display mode).

3. There are lower case Greek letters, $\alpha, \beta, \ldots, \omega$ and uppercase, $\Gamma, \Delta, \ldots, \Omega$. Some Greek letters have variant forms: $\epsilon$, $\varepsilon$, $\phi$, $\varphi$ and others.

4. For superscripts and subscripts, use ^ and _. For example $x_i^2$, $x^2_i$, $x_{i^2}$, ${x_i}^2$, $\log_2 x$, $x_{i,j}$.

5. Groups. Superscripts, subscripts, and other operations apply only to the next “group”. A “group” is either a single symbol, or any formula surrounded by curly braces {…}. If you do 10^10, you will get a surprise: $10^10$. But 10^{10} gives what you probably wanted: $10^{10}$. Use curly braces to delimit a formula to which a superscript or subscript applies: x^5^6 is an error; {x^y}^z is ${x^y}^z$, and x^{y^z} is $x^{y^z}$. Observe the difference between x_i^2 $x_i^2$ and x_{i^2} $x_{i^2}$.

6. Parentheses $(2+3)[4+4]$,$\{a,b,c\}$.

   These do not scale with the formula in between, so if you write (\frac{\sqrt x}{y^3}) the parentheses will be too small: $(\frac{\sqrt x}{y^3})$. Using \left(…\right) will make the sizes adjust automatically to the formula they enclose: \left(\frac{\sqrt x}{y^3}\right) is $\left(\frac{\sqrt x}{y^3}\right)$.

   \left and\right apply to all the following sorts of parentheses: $(x)$, $[x]$, $\{ x \}$, $|x|$, $\Vert x \Vert$, $\langle x \rangle$, $\lceil x \rceil$, and $\lfloor x \rfloor$.

7. Sums and integrals the subscript is the lower limit and the superscript is the upper limit, $\sum_1^n$. Don't forget {…} if the limits are more than a single symbol. For example, \sum_{i=0}^\infty i^2 is $\sum_{i=0}^\infty i^2$. Similarly, \prod $\prod$, \int $\int$, \bigcup $\bigcup$, \bigcap $\bigcap$, \iint $\iint$, \iiint $\iiint$, \idotsint $\idotsint$.

8. Fractions and binomials $\frac 17 23 \frac{17}{23}$,$\frac ab$,$\frac{a+1}{b+1}$$\binom{n+1}{2k}$

9. Different Fonts

• $\mathbb{C}$, $\mathbb{R}$, $\mathbb{Q}$, $\mathbb{Z}$,
• $\mathcal{CHNQRZ}$
• $\mathscr{CHNQRZ}$
• $\mathfrak{CHNQRZ}$.

10. Radical signs / roots Use sqrt, which adjusts to the size of its argument: \sqrt{x^3} $\sqrt{x^3}$; \sqrt[3]{\frac xy} $\sqrt[3]{\frac xy}$. For complicated expressions, consider using {...}^{1/2} instead.

11. Some special functions such as $$\sin,\cos, \tan, \cot, \arcsin,\arccos, \arctan, \arccot, \arcctg,\sec, \csc$$ $$\sinh,\cosh, \tanh, \coth$$ more complicate $ \DeclareMathOperator\arctanh{arctanh} \DeclareMathOperator\arcsinh{arcsinh} \DeclareMathOperator\arccosh{arccosh} \DeclareMathOperator\arccoth{arccoth} $

$$\arctanh, \arcsinh,\arccosh, \arccoth$$

$$\ln, \log,\lg,\log_2,\log_{16}$$ Use subscripts to attach a notation to \lim: \lim_{x\to 0} $$\lim_{x\to 0}$$ Nonstandard function names can be set with \operatorname{foo}(x) $\operatorname{foo}(x)$.

12. There are a very large number of special symbols and notations, too many to list here; see this shorter listing, or this exhaustive listing. Some of the most common include:

• $\lt$, $\gt$, $\le$, $\leq$, $\leqq$, $\leqslant$, $\ge$, $\geq$, $\geqq$, $\geqslant$, $\neq$. You can use \not to put a slash through almost anything: \not\lt $\not\lt$ but it often looks bad.
• $\times$, $\div$, $\pm$, $\mp$. \cdot is a centered dot: $x\cdot y$
• $\cup$, $\cap$, $\setminus$, $\subset$, $\subseteq$, $\subsetneq$, $\supset$, $\in$, $\notin$, $\emptyset$, $\varnothing$
• {n+1 \choose 2k} or \binom{n+1}{2k} ${n+1 \choose 2k}$
• $\to$, $\rightarrow$, $\leftarrow$, $\Rightarrow$, $\Leftarrow$, $\mapsto$
• $\land$, $\lor$, $\lnot$, $\forall$, $\exists$, $\top$, $\bot$, $\vdash$, $\vDash$
• $\star$, $\ast$, $\oplus$, $\circ$, $\bullet$
• $\approx$, $\sim $, $\simeq$, $\cong$, $\equiv$, $\prec$, $\lhd$, $\therefore$
• $\nabla$, $\partial$ \Im \Re $\Im$, $\Re$
• For modular equivalence, use \pmod like this: a\equiv b\pmod n $a\equiv b\pmod n$.
• For the binary mod operator, use \bmod like this: a\bmod 17 $a\bmod 17$.
• Avoid using \mod, as it produces extra space: compare the above with a\mod 17 $a\mod 17$.
• \ldots is the dots in $a_1, a_2, \ldots ,a_n$ \cdots is the dots in $a_1+a_2+\cdots+a_n$

Detexify lets you draw a symbol on a web page and then lists the $\TeX$ symbols that seem to resemble it. These are not guaranteed to work in MathJax but are a good place to start. To check that a command is supported, note that MathJax.org maintains a list of currently supported $\LaTeX$ commands, and one can also check Dr. Carol JVF Burns's page of $\TeX$ Commands Available in MathJax.

13. Spaces MathJax usually decides for itself how to space formulas, using a complex set of rules. Putting extra literal spaces into formulas will not change the amount of space MathJax puts in: a␣b and a␣␣␣␣b are both $a b$. To add more space, use \, for a thin space $a\,b$; \; for a wider space $a\;b$. \quad and \qquad are large spaces: $a\quad b$, $a\qquad b$.

To set plain text, use \text{…}: $\{x\in s\mid x\text{ is extra large}\}$. You can nest $…$ inside of \text{…}, for example to access spaces.

14. Accents and diacritical marks Use \hat for a single symbol $\hat x$, \widehat for a larger formula $\widehat{xy}$. If you make it too wide, it will look silly. Similarly, there are \bar $\bar x$ and \overline $\overline{xyz}$, and \vec $\vec x$ and \overrightarrow $\overrightarrow{xy}$ and \overleftrightarrow $\overleftrightarrow{xy}$. For dots, as in $\frac d{dx}x\dot x = \dot x^2 + x\ddot x$, use \dot and \ddot.

15. Special characters used for MathJax interpreting can be escaped using the \ character: \\\$ $\$$, \{ $\{$, \_ $\_$, etc. If you want \ itself, you should use \backslash (symbol) or \setminus (binary operation) for $\backslash$, because \\ is for a new line.

ta.stackexchange.com/a/29979/676335

Answer 13

$\nearrow \quad \rightarrow$ applies to odd numbers;

$\searrow_\alpha \quad \rightarrow$ applies to numbers of the form $2^mn$, with $m> \alpha$ and $n$ odd;

$\searrow_{\widehat{\alpha}} \quad \rightarrow$ applies to numbers of the form $2^{\alpha}n$, with $n$ odd;

$3 \rightarrow \quad \nearrow \searrow_{\widehat{1}}\nearrow \searrow_{\widehat{4}}$

$2^{D(n) + k + 1} \cdot a + n \rightarrow n$

$D(n)$ is the number of times you need to divide $n$ to reach 1;

$k \in \mathbb{N} + \{0\}$

$a \in \mathbb{N}$

Answer 14

There is a very simple formula to get an upper bound for steps $N$ compared with the largest $a$, such that all values below $a$ are proven to fall down to $1$.
Consider a $1$-step transformation: $b = {3a+1\over2^A}$ (where $A=v_2(3a+1)$). If this is a cycle, then $b=a$ and we have by reorganizing $2^A=(3+1/a)$. If we have a 2-step transformation, $b = {3a+1\over2^A}, c={3b+1\over2^b}$ and if this is a cycle we have $c=a$. Then we can do the product $b\cdot a = {3a+1\over2^A}{3b+1\over2^S}$ Rearranged this gives $2^{A+B}=(3+1/a)(3+1/b)$ Usually I write for the sum-of-exponents at $2$ the letter $S$ and the number of $3x+1$-steps with the letter $N$, so we can write $2^S=(3+1/a)(3+1/b)$ . Now $a$ and $b$ differ and we can assume a value $a_m$ between $a$ and $b$ such that as well we have $2^S=(3+1/a_m)^N$ from where we also have $$ 2^{S/N} = 3 + 1/a_m $$

Answer 15

This answer is free for anyone to use.

Answer 16

The post was closed as a duplicate, but it is not. In the duplicate target, it is assumed that $M$ is a metric space with a metric topology. In this post I am asking for a general topological space. The only answer in the duplicate used the metric in an essentially way, that I don't think that can generalized here.

I was trying to solve the following problem.

If every bijection $f:M\to M$ is a homeomorphism, then every subset of $M$ is clopen.

I tried to prove the equivalent statement,

if every bijection is homeomorphism, then every singleton is open (as if every singleton is open, then every subset is union of open sets and therefore open, hence all subsets are clopen, and converse holds trivially).

I tried to argue by contradiction. Assume that every bijection is homeomorphism and there exists some singleton that is not open, say $\{a\}$. Then, we can construct a bijection that maps every element to itself, but maps this singleton to some singleton which is open, say {b}. Then, this function is not a homeomorphism and so it's not true that every bijection is homeomorphism, contradiction. However, this argument assumes that at least some singleton is open, and I cannot see what goes wrong if none of the singletons were open. Did I even go in the right direction, and if so what could go wrong if none of the singletons were open?

Answer 17

this box is free for everyone to use