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Answer 1

This answer is free for anyone to use.

Answer 2

How can you solve for alternate trigonometric functions in general? Besides the regular circle functions.

I was inspired to return to an old problem I came up with after seeing

this question.

This was the problem of finding the analogues of other trigonometric functions which would parametrize a certain graph f(x,y)=1 or y=f(x) taking the polar curve angle $\theta=\theta_0$ and finding the length of the curve for one intersection of f(x,y)+0:

This just boils down to finding trigonometric functions which parametrize the curve f(x,y). This is easy, but the goal is have the distance from the intersection of the angle with the curve to the x,y-axis be the these functions $(x(\theta), y(\theta))$. Here is how to find such functions. For convention, let $\mathrm{x(\theta)=cos_{f(x,y)=0}(\theta)=cos_f(x),y(\theta)=sin_{f(x,y)=0}(\theta)=sin_f(x)}$. Sorry for the notation, but it is intuitive. Here is how to find such functions easily:

$$\mathrm{f(x,y)=0,y=x\,tan(\theta)\implies f(x,x\,tan(\theta))=0\implies x=cos_{f(x,y)=0}(\theta),y=sin_{f(x,y)=0}(\theta)}$$

There is not an easy inverse function for both arguments. However, when y is given explicitly you can solve for the alternate trigonometric functions which sometimes works through recursion.

$$\mathrm{\frac{f(x)}{x}=tan(\theta)\implies x=f^{-1}(x\,tan(\theta))= f^{-1}(f^{-1}(f^{-1}(…x…\,tan(\theta))\,tan(\theta))\,tan(\theta))=cos_f(\theta)\implies y=sin_f(\theta)=tan(\theta) f^{-1}(f^{-1}(f^{-1}(…x…\,tan(\theta))\,tan(\theta))\,tan(\theta))}$$

Unless you use an inversion theorem, then you cannot always solve for the inverse of y=f(x), unless if there is a closed form inverse. For example if you modeled the trigonometric equations with y=x tan(θ) and f(x)=cos(x), you would find that $\mathrm{x(\theta)=\mathrm{cos_{cos(x)}(θ)}=cos_f(x), y(\theta)=x(\theta)tan(\theta)}$ would be the parametrization for cos(x) for $(x(\theta),y(\theta))$. This uses the cardinal sine/ sinc function. Note $x_0$ here is just a parameter as a consequence of the nesting:

$$\mathrm{\frac{sin(x)}{x}=sinc(x)=tan(\theta)\implies cos_f(x)=sin^{-1}(sin^{-1}(…x_0tan(\theta))tan(\theta)…),\frac{cos(x)}x=tan(\theta)\implies cos_f(x)=cos^{-1}(cos^{-1}(…x_0tan(\theta))tan(\theta)…)}$$

Here are the results in this graph. Notice that lengths of the legs of the right triangle, like in the diagram, will give the derived modified cosine and sine functions. These new trigonometric functions are now based on sine and cosine graph instead of the circle.

Of course we can just use an inversion theorem, in the link above, to find the alternate trigonometric functions. Here is a formula that will not need to be used very much as we can usually find these alternate trigonometric functions. This is the general result for a function with explicit y=f(x) with the Lagrange Inversion Theorem. This uses the Bell polynomials and the rising factorial

$$\mathrm{\frac {f(x)}x=tan(\theta)\implies x(\theta)=cos_f(x)\left(Inverse\ of\ \frac yx\right)[tan(\theta)]=a+\sum_{n=1}^\infty\lim_{x\to a}\frac{d^{n-1}}{dx^{n-1}}\left[\left(\frac{x-a}{\frac{f(x)}{x}-\frac{f(a)}{a}}\right)^n\right]\frac{\left(tan(\theta)-\frac{f(a)}{a}\right)^n}{n!}=\sum_{k=0}^\infty c^{-n}_1\sum_{K=1}^{k-1}(-1)^K k^{(K)}B_{k-1,K}\left(\frac{c_2}{2c_1}, \frac{c_3}{3c_1},…, \frac{c_{n-k+1}}{(n-k+1)c_1}\right),\frac{f(x)}x=\sum_{n=0}^\infty c_n\frac{x^n}{n!},n\ge 2}$$

All I did to solve for x was use the linked Lagrange Inversion theorem formulas.

As a side note, I will leave you with the alternate trigonometric functions for $a|x|^m+b|x|^n=1$. Here is an interactive graph The analogue for these trigonometric functions is just $$\mathrm{x(\theta)=\pm\frac{1}{\sqrt[n]{a+b|tan(\theta)|^m}},y(\theta)= \pm\frac{tan(\theta)}{\sqrt[n]{a+b|tan(\theta)|^m}}}$$.

The secant and cosecant analogues are just $\frac{1}{x(\theta)}$ and $\frac{1}{y(\theta)}$. The tangent and cotangent analogues are almost always tan(θ) and cot(θ) because $\mathrm{y(\theta)=x(\theta)tan(\theta)}$. Please tell me if any part of this is confusing.

I tried the best I could to propose new notations for the trigonometric analogues. See the d-analogue and the q-analogue as ways to back up my $\cos_f(\theta)$ and $\sin_f(\theta)$ notations where f is the function we would like the alternate trigonometric functions based on as seen in the graphic. I also could have easily made a typo.

The question now is how to solve $$\mathrm{y=x\,tan(\theta),f(x,y)=0\implies f(x,x\,tan(\theta))=0}$$ for x in exact form, not necessarily closed form, using any method except approximation. Please correct me and give me feedback!

Answer 3

Wikipedia on Collatz Conjecture in Binary

         111
        1111
       10110
      10111
     100010
    100011
    110100
   11011
  101000
 1011
10000

After 3x+1... For certain x:

With all "1" digits (or if it ends in just 3 "1"s), result in ending in 10.

With ending in "011", it ends in 00010.

With ending in "01", it ends in 00.

With ending "101", it ends in 1000.

At most, 3x+1 can increase an odd number number by ~3x, and must result in a division by 2... So can increase at most ~3/2...

I guess something like this, but more focused...

Something all "1"s leads to something where there must be a 0 in it, which leads to something with more 0s, on and on, until it hits 1?

A.k.a if 3x +1 for 2^n - 2^y 2^z -1, for all n, y, and z, except z can be 0, result in something of the same form except -2^a, or more terms for any "a".

A counter-argument would be something all 1s except for one "0", when 3x + 1 = all ones... Sorta. Then you have to prove that that strings of 1s doesn't turn into more zeros...

... A.k.a, is there any number ((2^n - 1) - 1) / 3

A.k.a any odd number (2^n - 2) / 3... No... Even / Odd = Even. There must be a 2 in the divisors...

So, there are no strings of "1", for the Collatz, that preserve 1111...

Answer 4

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 5

Is there an example Hamiltonian that is uncomputable?

In a paper from 2015 Toby S. Cubitt et al showed that the problem of determining the existence of a band gap in the excitation spectrum of a quantum many-body system, was undecidable. This result applied to atoms interacting via nearest neighbour interactions in a 2D lattice, and it was followed up by another publication which applied to 1D lattices as well.

These results were extended by Johannes Bausch et al in a 2021 paper, which demonstrated that there exist phase diagrams of many-body quantum systems which were also uncomputable.

Insofar, none of the papers mentioned provide a constructive proof of these results. I was wondering if an example Hamiltonian of a quantum many-body system had been identified yet, for which the existence of a spectral band gap was undecidable? My motivation for asking is that such a Hamiltonian could be constructed in experiment, and then a measurement could be made of the systems spectrum that which provably couldn't be solved by theory.

Tags: mp.mathematical-physics lo.logic

Answer 6

This answer is free for anyone to use

Answer 7

This answer is free for anyone to use.

Answer 8

This answer is free for anyone to use.

Answer 9

$$ \lambda^{\circ 5}(\;^44) = \lambda \lambda 4 + \lambda\left(1+{\lambda \lambda 4 + \lambda\left( 1+ { \lambda\lambda4 + \lambda \left( 1+{ \lambda\lambda4 \over \;^24 \lambda4} \right) \over \;^14 \lambda4 }\right) \over \lambda 4} \right)\\ $$

$$ \lambda^{\circ 4}(\;^44) = \lambda 4 + \lambda \lambda 4 + \lambda \left( 1+ { \lambda\lambda4 + \lambda \left( 1+{ \lambda\lambda4 \over \;^24 \lambda4} \right) \over \;^14 \lambda4 } \right) \\ = \lambda 4\left(1+{\lambda \lambda 4 + \lambda\left( 1+ { \lambda\lambda4 + \lambda \left( 1+{ \lambda\lambda4 \over \;^24 \lambda4} \right) \over \;^14 \lambda4 }\right) \over \lambda 4} \right)\\ $$

For the following, let's denote for the logarithm $\lambda (x) := \log(x)$, and $ \lambda^{\circ h}(x):=\log^{\circ h}(x)$. Moreover, we allow to omit the parentheses for a simple argument like $\lambda m := \log(m)$, $\lambda\lambda m := \log(\log(m))$ and so on.

Then for $m=3$ we have $$ \lambda^{\circ 2}(\;^33) = \;^13 \lambda3 \left( 1+ { \lambda\lambda3 \over \;^13 \lambda3 } \right) = 3\lambda3+ \lambda\lambda 3 $$

For $m=4$ we have $$ \lambda^{\circ 3}(\;^44) = \;^14 \lambda4 \left( 1+ { \lambda\lambda4 + \lambda \left( 1+{ \lambda\lambda4 \over \;^24 \lambda4} \right) \over \;^14 \lambda4 } \right) \\ = 4 \lambda4 + \lambda \lambda 4+ \lambda\left(1+ {\lambda\lambda4\over \;^24 \lambda 4}\right) $$

For $m=5$ we have $$ \lambda^{\circ 4}(\;^55) = \;^15 \lambda5 \left( 1+ { \lambda\lambda5 + \lambda \left( 1+{ \lambda\lambda5 + \color{red}{ \lambda\left( 1+ {\lambda\lambda5\over \;^35 \lambda5} \right)} \over \;^25 \lambda5} \right) \over \;^15 \lambda5 } \right) \\ = 5 \lambda5 + \lambda \lambda 5+ \lambda\left(1+ {\lambda\lambda5 + \varepsilon \over \;^25 \lambda 5}\right) \qquad \varepsilon \lt 10^{-2000}$$ Here, the $\color{red} {\text{red}}$ part might be set to zero - although Pari/GP can evaluate it to the small number $1.5472633343566660957 \cdot 10^{-2185}$.

It is obvious, how this recursive expression extends for the case $m=6$ and $m \gt 6$.

So in general we could -with very good approximation to thousands of digits- denote the general case for $m \ge 4$: $$ \lambda^{\circ m-1}(\;^mm) = \;^1m \lambda m \left( 1+ { \lambda\lambda m + \lambda \left( 1+{ \lambda\lambda m \color{red}{+ \varepsilon} \over \;^2 m \lambda m} \right) \over \;^1m \lambda m } \right) \qquad \phantom {aaaaaaaaaaaaaaaaaaaaaaa}\\ = m \lambda m + \lambda \lambda m+ \lambda\left(1+ {\lambda\lambda m + \varepsilon \over \;^2m \lambda m}\right) \qquad m \ge 4,\varepsilon \lt 10^{-2000} $$

Answer 10

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?

---

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 $$f(x(t)) = g(t) \to f(x(x^{-1}(u))) = f(u) = g(x^{-1}(u)) \tag 8$$

---

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 11

(Sum of the form $r+r^2+r^4+\dots+r^{2^k} = \sum_{i=1}^k r^{2^i}$)

Although this question has already an answer even with a bounty assigned, it might be interesting to see another ansatz, which puts it in relation to a wider field.

This is the area of iteration of functions, series-of-iterated-functions, and the decomposition of the problem into the terms of the Schroeder-mechanism for functional iteration.

The terms of the series can be understood as iterated squares of the first element. The ansatz for a Schroeder-mechanism implies to find a function which decodes one number of the form $r^{2^k}$ into its "index" $2^k$, let's call it $g()$ then to multiply by $2$, and then apply the inverse function, say $f()$ to arrive $r^{2^{k+1}}$. (...)

Answer 12

New form of the question

---

I'm still fiddling with the question of existence/non-existence of 2-step-cycles in the generalized Collatz-problem with $m x+1$.

For a $2$-step cycle in such a problem, with positive odd $a,b$ we can write $ a \to b \to a \to \cdots$ with the transformations $$ b = { m \cdot a +1\over 2^A } \qquad , \qquad a = { m \cdot b +1\over 2^B } \tag 1$$ We are only interested in cases where $m \gt 1$ positive odd integer _{(using $m=3$ we have the Collatz-problem)} and $ (a,b) $ positive odd integers with $ab \gt 1$. Moreover, $A,B$ are written for the 2-adic valuations of $(m \cdot a +1), (m \cdot b +1)$ respectively.

---

Few solutions in small numbers are known, for instance $$(S,m,a,b) \in \{ (5,5,1,3),(15,181,27,611),(15,181,35,99)\}$$ but it seems to be a hard problem. $ \qquad \\ \qquad \\ $

• To get it accessible at all I looked at it with the restriction that $a=1$. This can then be reformulated such that -hinted by a helpful comment in MO - I could relate it to an article, which allowed a derivation of the solution for that simplified version. By this $(S,m,a,b)=(5,5,1,3)$ is the only possible solution with $a=1$ and $b \gt 1$.
• The version with $a \gt 1$ and indeterminate is still out of reach for me. But just recently I noticed, that a reformulation gives a heuristic which has exponential growth on the lhs of a certain equation and a smaller growth on the rhs - which when correctly quantified would give a (likely relative small!) upper limit for the involved variables $(S,m,a,b)$. ^{(I've added pictures at the end of the question, which show the numerical heuristic/ evidence for the different growing rates upper limit.)}

The questions are now:

• can the property of the difference of growthrates be well established?
• can an argument in the style of Steiner/Simons/deWeger in their disproof of the "1-cycle" in the Collatzproblem ($m=3$) be found that can be applied to my equation?

---

Remaining part of the old version ...

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Although it is easy to find few solutions in small numbers, it seems to be very hard to finda exponential diophantine problem hard to solve for the generality. I've attempted this problem with a simplified question: fixing $a=1$ and leaving only $b$ indeterminate. Using a more general expression for the inherent diophantine problem I got in [MathOverflow][2] a link to some article, from where as a consequence the proof for the nonexistence of such 2-step-cycles with $m \gt 5$ comes out *($m=5$ has the 2-step-cycle $(a,b)=(1,3)$ as is well known)* .

I've now taken another attempt to the problem with $(a,b)$ indeterminate, and get now a tendency for unsolvability for increasing $m$ which looks heuristically very strong, namely exponential growth for a required coefficient against linear growth of the sum $S=A+B$ in the exponents in (1).

Question:

The question is then: is such an exponential/linear relation known by some results of diophantine approximation? (description see below)

^{Note: I cannot apply the Rhin-bound for the disproof of the 1-cycle in the Collatz-problem as done by R. Steiner and J. Simons, because this Rhin-bound depends on the varying $N$ for the length of the supposed 1-cycle in $3x+1$, while I have here the fixed $N=2$ with the $2$-step-cycle in $mx+1$.}

Derivation:

Combining the two equations in (1) by inserting and expansion give the determinations: $$ a = { m + 2^A \over 2^S - m^2} \qquad , \qquad b = { m + 2^B \over 2^S - m^2} \tag 2 $$ Here I use $S=A+B$ and $m = \lfloor 2^{S/2} \rfloor$ . Conditions:

• If $a=b$ we have also $A=B$ and thus $S=2A=2B$ is even. It is easy to show, that then $a=b=1$ (not done here), so
• we deal here with $a \lt b$ and $A \lt B$ and $S \gt 1$ being odd.
• From $(a,b)\gt 1$ it follows that $m = \lfloor 2^{S/2} \rfloor $ .
• Requiring this and that $m$ is odd we get a subset of $S$ which is only to be considered:
  $(S,m) \in \{ (1,1), (5,5), (7,11), (11,45), (15,181), (27,11585), (33,92681), (35,185363),... \} \tag 3 $

While my problem so far has been, that I could never make the numerators in $(2)$ diophantinely significant in any way - prominently that numerators muust be evenly be divisibly by the denominator $D = 2^S - m^2$ , the difference $b-a$ gives an interesting expression: $$ \begin{align} b - a &= { 2^B - 2^A \over 2^S - m^2} \\ &= 2^A{ 2^C -1 \over D} \qquad \text{where } C=S-2A \lt S\end{align} \tag 4$$

Here $C= \lambda_2(D) \cdot k$ must be a multiple of the cyclic order of $D$ to base $2$. For instance if $D=7$ then $C=3k$ because $ \lambda_2(7)=3$.

Heuristic:

Heuristically, the growth of $C$ over $S$ is exponential over linear which produces a contradiction to the condition $C=S-2A \lt S$ and the last possible case is $(S,m)=(15,181)$ which indeed has (even) two $2$-step-cycles.

See the two following pictures:
Overview:

and the detail for small $S$:

The exponential growth of $\lambda_2(D)$ tells me, that possibly there is some established knowledge about such a relation, or a known upper bound in such a diophantine relation. I don't really think that this can be derived here in an answer, so I think the only chance is that there is an established result from some related known problem.

A longer description is in a draft at my homepage

Answer 13

(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 14

Classically, the fractional derivative of order $\alpha$ is given by: $$ D^{\alpha}f(x) = \frac{1}{\Gamma(1-\alpha)}\frac{d}{dx}\int_0^x \frac{f(t)}{(x-t)^{\alpha}}\,dt$$

As $\alpha\to 1^-$, this reduces to $D^1 f(x)= \frac{d}{dx} f(x)$ and for $\alpha=0$ this becomes (for continuous $f$, anyway) $D^0 f(x) = f(x)$. Further, note that $D^1 e^x = e^x$, as can be seen directly: $$ \lim_{\alpha \to \ 1^-} \frac{1}{\Gamma(1-\alpha)}\frac{d}{dx}\int_0^x \frac{e^t}{(x-t)^{\alpha}}\,dt = e^x \lim_{\alpha \to \ 1^-} \frac{\Gamma(1-\alpha)-\Gamma(1-\alpha,x)}{\Gamma(1-\alpha)} = e^x(1-0)=e^x $$ In other words, any continuous function is an eigenfunction of $D^0$ for $\alpha=0$ and $e^x$ is an eigenfunction of $D^1$.

Answer 15

$b=3a+1$ \ on odd a then the Syracuse-equivalent transformation is $C(b)= \{b\}_2 \cdot 3 +1 $ , then $C°^2(b)= \{\{b\}_2 \cdot 3 +1\}_2\cdot 3+1 $ , and so on.

Then given that $b$ is $4$-rooted, all $ 4^kb$ are $4$-rooted. Next, using $b_k = 4^kb$ then $ (b_k-1)$
$$ \begin{bmatrix}1&5\end{bmatrix} \cdot \begin{bmatrix} 1&1 \\0 &4 \end{bmatrix}$$ $$ \begin{bmatrix}1&5\end{bmatrix} \cdot \begin{bmatrix} 1&-1/3 \\0 &2/3 \end{bmatrix}$$

This answer is free for anyone to use.

Answer 16

@Zach Siegel

Thanks for your anwer. With all due respect, I do not think that is appropriate.

At first, please allow me to use $X^-$ to replace $X^+$ because we are discussing $Var(X|X<a)$. In fact, $X^-$ in the answer is a function of $X$, not the variable after truncation we are concerned. Note that $E(X|X<a)$ and $Var(X|X<a)$ depend on the truncated pdf function $f(x|X<a)=\frac{f(x)I(x<a)}{\int_{-\infty}^a \ \ f(x)dx}$. However, $E(X^+)=\int xI(x<a)f(x)dx$ and $Var(X^+)=\int (xI(x<a)-E(X^+))^2f(x)dx$ still rely on the distribution of $X$.

Therefore, $Var(X^-)$ is not equivalent to $Var(X|X<a)$, and the given inequality may need more conditions.

Answer 17

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?