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    $\begingroup$ I have added a [sandbox] tag to allow people ignore it more easily (via software support of ignoring tags), and since it seems that we have two sandboxes now, a tag may seem a bit more in place here. $\endgroup$
    – Asaf Karagila Mod
    Jul 18, 2012 at 8:35
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    $\begingroup$ (+1) For thinking outside the (sand)box. $\endgroup$
    – cardinal
    Jul 18, 2012 at 19:40
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    $\begingroup$ At the suggestion of the moderators, I have gone and changed the associated owners of all the answers here to the Community user. This way, the original owners will not receive excess pings for each time another user uses the draft space for their work. Enjoy! $\endgroup$
    – Grace Note StaffMod
    Oct 5, 2012 at 14:45
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    $\begingroup$ To prevent crashes I've found the "Bookmarks to disable/enable MathJax", provided in here, pretty useful. $\endgroup$
    – leo
    Dec 17, 2012 at 18:03
  • 3
    $\begingroup$ PSA: Between the creation of this sandbox (in July 2012) and today (December 2015), technology has advanced. Something like StackEdit (or others, it's simply the only one I know) essentially solves all the limitations of this sandbox. You can have multiple concurrent drafts, you don't have to worry about polluting meta's front page, you can leave your draft untouched for days and expect it to still be there, you don't have to explicitly clear up your draft when you're done... Maybe someday we can get rid of this outdated crutch. $\endgroup$ Dec 2, 2015 at 14:07

17 Answers 17

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If $\mathcal O$ is an order in a set $X$ then for any $\xi$ in $X$ let's we put

$$ (\leftarrow,\xi)_\mathcal O:=\{x\in X:x\prec_\mathcal O\xi\}\quad \text{and} \quad [\leftarrow,\xi)_\mathcal O:=\{x\in X:x\preccurlyeq_\mathcal O\xi\} $$

$$ (\xi,\rightarrow)_\mathcal O:=\{x\in X:\xi\prec_\mathcal O x\} \quad \text{and} \quad [\xi,\rightarrow)_\mathcal O:O=\{x\in X:\xi\preccurlyeq_\mathcal O x\} $$

Moreover, for any $a$ and $b$ in $X$ we can put $$ [a,b]:=\{x\in X:a\preccurlyeq_\mathcal O x\preccurlyeq_\mathcal O b\} \quad\text{and}\quad (a,b):=\{x\in X:a\prec_\mathcal O x\prec_\mathcal O b\} $$

$$ (a,b]:=\{x\in X:a\prec_\mathcal O x\preccurlyeq_\mathcal O b\} \quad\text{and}\quad [a,b):=\{x\in X:a\preccurlyeq_\mathcal O x\prec_\mathcal O b\} $$

Now if $f$ is a function form an order $(X,\mathcal U)$ to an order $(Y,\mathcal V)$ then we say that $f$ directely monotone with respect $\mathcal U$ and $\mathcal V$ whether for any $a$ and $b$ in $X$ the implication $$ (a\preccurlyeq_\mathcal U b)\Longrightarrow\big(f(a)\preccurlyeq_\mathcal V f(b)\big) $$ holds.

So I would like to prove that the following statements are equivalent:

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Reach the solution by applying Discriminant-like method .


Observe that :

$$ \begin{align}P(z):=&\thinspace\thinspace\thinspace z^4+z^3+z^2+2z+3\\ =&\thinspace\thinspace\thinspace \color{#c00}{z^2}(\color{#0a0}{z}^2+\color{#0a0}{z}+1)+2\color{#c00}{z}+3\end{align} $$

Then, you have :

$$ \begin{align}\Delta_{\color{#c00}{z}}&=1-3(\color{#0a0}{z}^2+\color{#0a0}{z}+1)\\ &=-\underbrace {(3\color{#0a0}{z^2}+3\color{#0a0}{z}+2)}_{\Delta_{\color{#0a0}{z}}\thinspace<\thinspace 0}\\ &<0,\thinspace\thinspace\forall z\in\mathbb R\thinspace .\end{align} $$

which completes the answer .


$\rm {Explanation:}$

We rewrote the polynomial "as" a quadratic polynomial respect to $\color{#c00}{z}$ . Then, we analyzed the general discriminant of the quartic with "respect" to $\color{#c00}{z}$ . We have shown that, this discriminant is always negative . Thus, you're done .


$\rm{ Explicit \thinspace\thinspace representation :}$

$$ \begin{align}P(z):=\underbrace{\left(\color{#0a0}{z}^2+\color{#0a0}{z}+1\right)}_{\Delta_{\color{#0a0}{z}\thinspace <\thinspace 0}}\left(\color{#c00}{z}+\frac {1}{\color{#0a0}{z}^2+\color{#0a0}{z}+1}\right)^2+\frac{\overbrace{3\color{#0a0}{z}^2+3\color{#0a0}{z}+2}^{\Delta_{\color{#0a0}{z}\thinspace<\thinspace 0}}}{\underbrace{\color{#0a0}{z}^2+\color{#0a0}{z}+1}_{\Delta_{\color{#0a0}{z}\thinspace<\thinspace 0}}}>0,\thinspace\thinspace \forall z\in\mathbb R\thinspace .\end{align} $$

which is a method of completing the square .

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We can do a little better. Since $f$ is convex and positive, on $[1,3/2]$ it must be below the linear segment connecting $\{(1,1),(3/2,2/3)\}$. This is given by $(5-2x)/3$, so we have $$ -\frac{1}{9}= \int_1^{3/2} \varphi(x) \frac{5-2x}{3}\,dx \le \int_1^{3/2} \varphi(x) f(x)\, dx ;$$ the inequality is reversed because $\varphi$ is negative on this interval. Further, taking the `worst case' scenario of $f(2)=0$ and $f$ piecewise-linear on $[1,2]$ (this violates the monotonicity requirement on $[1,\infty)$, but hey) maximizes the positive...unclear where to go from here.

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Note that the integral converges by Dirichlet's Test. The idea is to use the convexity of $f$ to analyze the integral by looking at piecewise-linear parts.

Since we are guaranteed convergence, we can split the integral into unit-length intervals. For clarity I will write $\infty$ as the upper limit of summation, but really one must take a limit of partial sums, viz $\lim_{M\to\infty}\sum_{k}^M$, as $f(x)=1/x$ shows. Nevertheless, we have: $$I=\int _{1}^{\infty}\varphi(x) f(x)\,dx = \sum_{k=1}^{\infty}\int _{k}^{k+1}\varphi(x) f(x)\,dx$$ Use the periodicity of $\varphi$ and make the substitution $x-(k-1)=t$: $$= \sum_{k=1}^{\infty}\left(\int _{1}^{2}\varphi(t) f(t+k-1)\,dt\right)$$ $$= \sum_{k=0}^{\infty}\left(\int _{1}^{2}\varphi(t) f(t+k)\,dt\right)$$ Put $\varphi(t)=t-3/2$ and split the range of integration. Importantly, $\varphi<0$ on $(1,3/2)$ and $\varphi > 0$ on $(3/2,2)$, while $f$ is always non-negative. $$=\sum_{k=0}^{\infty}\left(\int _1^{3/2}(t-3/2) f(t+k)\,dt+\int _{3/2}^{2}(t-3/2) f(t+k)\,dt\right)$$ Suppose we want to minimize $I$. Then we want to maximize the negative contribution over $(1,3/2)$ and minimize the positive contribution over $(3/2,2)$. Since we know $f$ is convex, on the first part it suffices to assume $f$ is linear. On the second part, an initial lower bound is $f(k+2)$. However, when we move to the next index, this positive contribution will outweigh the next negative contribution. So, in fact we can simplify things by setting $f(x)\equiv 0$ for $x>3/2$. Then our estimate becomes: $$ I >\int _1^{3/2}(t-3/2)((3-2t)f(1)+2(t-1)f(3/2))\,dt $$ This looks a mess, but it's actually easy to integrate: $$ I> -\frac{1}{12}f(1)-\frac{1}{24}f(3/2) =\frac{-1}{9} $$For an upper bound, again we examine $k=0$ and discard further terms of the sum. Here we want to minimize the negative contribution, so on $(1,3/2)$ we will use $f(x)>f(3/2)=2/3$. This will serve as an upper bound on $(3/2,2)$ as well, giving $$ I < \int _1^{3/2}(t-3/2)((3-2t)f(1)+2(t-1)f(3/2))\,dt+\int _{3/2}^2(t-3/2)f(3/2)\,dt $$A quick integration gives $$I < \frac{-1}{36};$$ however, OP claims that $-1/18$ is possible. Perhaps my argument can be refined or perhaps different techniques are needed.

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Proof: a) Suppose $\alpha\in EL_0$. $a_1,...,a_n\in\overline{\mathbb{Q}}$ implies $a_1\alpha,...,a_n\alpha\in EL_0$. Since $f(\alpha,e^{a_1\alpha},...,e^{a_n\alpha})=0$ implies $e^{a_1\alpha},...,e^{a_n\alpha}$ are algebraic over $\mathbb{Q}(\alpha)$, $e^{a_1\alpha},...,e^{a_n\alpha}\in EL_0$. By (2) of Theorem 1, $r_1a_1\alpha+...+r_na_n\alpha=\pi i$ for some non-zero $r_1,...,r_n\in\mathbb{Q}$. This yields $\frac{r_1a_1\alpha+...+r_na_n\alpha}{i}=\pi$ (*). The left-hand side of equation (*) is algebraic, and the right-hand side is transcendent, a contradiction.

b) Suppose $\alpha\in EL_{2N}\setminus EL_{2N-1}$ for some $N\ge 1$. Since $f(\alpha,e^{a_1\alpha},...,e^{a_n\alpha})=0$ implies $e^{a_1\alpha},...,e^{a_n\alpha}$ are algebraic over $\mathbb{Q}(\alpha)$, $e^{a_1\alpha},...,e^{a_n\alpha}\in EL_{2N}$. By (3) of Theorem 1, $r_1a_1\alpha+...+r_na_n\alpha\in \ln EL_{2N-1}$ for some non-zero $r_1,...,r_n\in\mathbb{Q}$ and hence $e^{r_1a_1\alpha+...+r_na_n\alpha}\in EL_{2N-1}$. This means $e^{\alpha(r_1a_1+...+r_na_n)}\in EL_{2N-1}$. Real $\alpha$ implies $\left(e^\alpha\right)^{r_1a_1+...+r_na_n}\in EL_{2N-1}$.

But then $f(\alpha,e^{a_1\alpha},...,e^{a_n\alpha})=0$ implies $\alpha\in EL_{2N-1}$, a contradiction.

c) Suppose $\alpha\in EL_{2N+2}\supset EL_{2N+1}$ and $\alpha\in EL_{2N+1}\setminus EL_{2N}$ for some $N\ge 0$. Since $f(\alpha,e^{a_1\alpha},...,e^{a_n\alpha})=0$ implies $e^{a_1\alpha},...,e^{a_n\alpha}$ are algebraic over $\mathbb{Q}(\alpha)$, $e^{a_1\alpha},...,e^{a_n\alpha}\in EL_{2N+1}$. By (4) of Theorem 1, $\left(e^{a_1\alpha}\right)^{k_1}\cdot ...\cdot\left(e^{a_n\alpha}\right)^{k_n}\in\exp EL_{2N}$ for some non-zero $k_1,...,k_n\in\mathbb{Z}$ and hence $e^{(k_1a_1+...+k_na_n)\alpha}\in\exp EL_{2N}$. This implies $(k_1a_1+...+k_na_n)\alpha\in EL_{2N}$ and $\alpha\in EL_{2N}$, a contradiction.

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Let

$$\small {3x^2+4y^2+4xy-11x-6y=a}$$

Let's rearrange the equation as a quadratic equation with respect to $x$ :

$$\small{3\color{#0a0}{x}^2+\color{#0a0}{x}(4y-11)+(4y^2-6y-a)=0}$$

Then, we determine the discriminant $\Delta_{\color{#0a0}{x}}$ :

$$ \begin{align}\Delta_x&=-32\color{#c00}{y}^2-16\color{#c00}{y}+(121+12a)\\ &≥0\end{align} $$

We recall that, if $\Delta_{\color{#c00}{x}}≥0,\thinspace \forall y\in\mathbb R$, then

$$ \begin{align}&\Delta_{\color{#0a0}{y}}=64+32(121+12 a)≥0\\ \implies &a≥-\frac {41}{4}\thinspace \end{align} $$

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Here's an alternate solution.

Lemma: Let $0\le j,k < n$. Then $$ \sum_{j=0}^{n-1} \cos(2\pi k j/n) = \begin{cases}n,&k=0\\0,&\text{else}\end{cases} $$ This is clear geometrically: the non-unit roots of unity sum to zero, and when $k=0$ there are $n$ terms of $1$.

Take $f(n)$ and multiply through by $2^n$:

$$2^n f(n) = \sum_{j=0}^{2n-1}{\left(2\cos\left( \frac{j \pi}{2n}\right)\right)^n \left( 2\cos \left( \frac{2 j \pi}{n} \right) + 1\right) \cos \left( \frac{j \pi}{2} - \frac{2 j \pi}{n} \right)}$$ $$= \sum_{j=0}^{2n-1}{\left(2\cos\left( \frac{j \pi}{2n}\right)\right)^n \left( 2\cos \left( \frac{4 j \pi}{2n} \right) + 1\right) \cos \left( \frac{nj \pi}{2n} - \frac{4 j \pi}{2n} \right)}$$ By symmetry, it suffices to sum over $0\le j \le n-1$ and then double: $$= 2\sum_{j=0}^{n-1}{\left(2\cos\left( \frac{j \pi}{2n}\right)\right)^n \left( 2\cos \left( \frac{4 j \pi}{2n} \right) + 1\right) \cos \left( \frac{nj \pi-4j\pi}{2n} \right)}$$Denote $X=\frac{j\pi}{2n}$: $$= 2\sum_{j=0}^{n-1}\left(2\cos( X)\right)^n \left( 2\cos (4X) + 1\right) \cos \left( (n-4)X\right)$$Now we are going to use the product to sum rules to turn these products into a sum of frequencies. This is going to be great for two reasons:

  • By the lemma, we need only worry about the constant term, as all the other frequencies will die off.
  • In fact, the constant term is a polynomial in $n$. Once we figure out its degree and sample that many points plus one, we can reverse-engineer it.

Claim: this polynomial is degree five. Then sample six points and interpolate. I need to figure out how to write $(2\cos(X))^n$ as a sum, with $X$ as above, and then multiply through twice. I think I see where the binomial coefficients come from, but not quite...

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$$ k 2^B = rk + 2^B-1 \\ k (2^B-r) = 2^B-1 \\ $$

The first means $n = 2^Ak - 1$. But I don't know how to use the second.

It looks like a Diophantine equation $2ᴺ⁻ᴬn - 3ᴬm = -1$, which could be so

$$(2^N - 3^A)n = 3^A - 2^A \\ (2^N - 3^A)(n+1) = 2^N - 2^A = 2^A ( 2^B -1) \\ \implies n=2^Ak-1 \\ (2^N - 3^A)k = 2^B -1 \\ \implies k \lt 2^B $$ Now let's decode $3^A$ modular in terms of $2^B$: $3^A = m2^B+r$ where $0<r<2^B$ $$ (2^N - m2^B-r)k = 2^B -1 \\ (2^B2^A - m2^B-r)k = 2^B -1 \\ 2^B(2^A - m)k-2^B = rk -1 \\ (2^A - m)k = {rk -1 \over 2^B}+1 = {r \over2^B }k +1 - 1/2^B \\ \text{and in the rhs } \qquad {r \over2^B }k +1 - 1/2^B <k+1 $$ so also the lhs must be smaller than $k+1$ and thus must be $k$ and thus $$ m = 2^A - 1$$ But this means, a perfect power $3^A$ is of the form $(2^A-1)2^B+r$ and in binary representation this means $$ 3^A = \text{'111...1 00...0'}_2 + r $$

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How solve the problem $f(x+2)=f(x)+4x+4$ for any $x$

There are infinitely many real functions which satisfy this recursive property. $\cases{f(2)=0\\f(x+2)=f(x)+4x+4}\tag*{}$

However, all of them have the same value at the point $x=2012$.

Since, you are trying to find the value at an even integer point, it's useful to consider a sequence, $u_n:=f(2n)$ for all $n\in\mathbb N$.

From the recursive property, $$\begin{align}u_n&=f((2n-2)+2)\\ &=f(2n-2)+4(2n-2)+4\\ &= u_{n-1}+8n-4\end{align}$$

Thus, we have: $u_n=\cases{0 & \text{if $n=1$}\\ u_{n-1}+8n-4 &\text{if $n\geq 2$}}\tag*{}$

Is there a nice way to find the $n$-th term of $u_n$?

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$\hbar=m=1,a=\pi,x\mapsto \pi/2-x$

$$-\frac{1}{2}\frac{d^2\psi }{dx^2}+g\,\delta(x)\psi(x) = E\psi(x),\quad x\in (-\pi/2,\pi/2)$$ subject to $\psi(-\pi/2)=\psi(\pi/2)=0$. Note that the delta-potential is an even function of $x$ and therefore $x\to -x$ is a symmetry of the Hamiltonian. Hence its eigenfunctions all have even or odd parity. Moreover, the odd-parity solutions vanish at the origin and so are unperturbed by this potential. Hence for the remainder we focus on the even-parity sector, i.e., we assume $\psi(x)=\psi(-x)$.

The unperturbed eigenfunctions must then satisfy the free Schrödinger equation $$-\frac{1}{2}\frac{d^2\psi }{dx^2} = E\psi(x)$$ for $|x|<\pi/2$ along with $\psi(\pi/2)=0$ and $\psi(x)=\psi(-x)$. This is readily solved, yielding $$\phi_{n}^{(0)}(x)=\sqrt{\frac{2}{\pi}}\cos[nx],\quad E_{n}^{(0)}=\frac12 n^2$$ for odd nonnegative integer $n$. (To simplify notation from here on, all summations from here are only over this domain.) The matrix elements of the perturbation $$V_{mn}=\langle \phi_{m}^{(0)}|\delta (x)|\phi_{n}^{(0)}\rangle=\frac{2}{\pi}.$$ Thus the first few orders of perturbation theory yield

\begin{align} E^{(1)}_1 &= V_{11}=\frac{2}{\pi}\\ E^{(2)}_1 &= \sum_{n\neq1} \frac{V_{1n}^2}{\left(E_1-E_n\right)}\\ &= \frac{8}{\pi^2}\sum_{n\neq 1} \frac{1}{1-n^2} = -\frac{2}{\pi^2} \end{align} \begin{align} E^{(3)}_1 &= \sum_{nm\neq1} \frac{V_{1n}V_{nm}V_{m1}}{\left(E_1-E_n\right)\left(E_1-E_m\right)}-V_{11}\frac{V_{1m}^2}{\left(E_1-E_n\right)^2}\\ E^{(4)}_1 &= \sum_{nmk\neq1} \frac{V_{1n}V_{nm}V_{mk}V_{k1}}{\left(E_1-E_n\right)\left(E_1-E_m\right)\left(E_1-E_k\right)} + V_{11}^2\frac{V_{1n}^2}{\left(E_1-E_n\right)^3} \\&\qquad - V_{11}\frac{V_{1n}V_{nm}V_{m1}}{\left(E_1-E_n\right)^2\left(E_1-E_m\right)}- V_{11}\frac{V_{1n}V_{nm}V_{m1}}{\left(E_1-E_n\right)\left(E_1-E_m\right)^2} \\&- \frac{V_{1n}^2V_{1m}^2}{\left(E_1-E_n\right)\left(2E_1-E_n-E_n\right)\left(E_1-E_m\right)} - \frac{V_{1n}^2V_{1m}^2}{\left(E_1-E_n\right)^2\left(2E_1-E_n-E_n\right)} \end{align}

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$$(1/u)^{u-1/u}=2y$$ $$e^{ln(1/u)(u-1/u)}=2y$$ $u\to e^x$: $$e^{-x(e^x-e^{-x}))}=2y$$

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(Editing just now please do not interfere -------------)

part 4
We'll start with the assumption in (3.4c), ($S = \lceil N \cdot \log_2(3) \rceil $) and for convenience we reformulate this as $$ \alpha \cdot 2^S = \alpha \cdot 3^N + Q(E_{N:S}) \tag {4.1} $$


part 1
First a bit about notation.

The Collatz-transformation is sometimes equivalently expressed in the Syracuse-form : $$ \text{over positive odd integers $a_k$ and }\\ \text{positive integers $A_k$} \tag {Syracuse notation}$$ $$ \mathcal R: \qquad a_{k+1}={ 3a_k+1 \over 2^{A_k} } \tag {1} $$ I use capital letters in the exponents at $2$: $ A_k=\nu_2(3a_k+1)$ and as well $N$ for the number of all odd steps (= length of the orbit) and $S$ for the sum-of-exponents/number-of-divisions-by-$2$.

part 2
An iterated transformation can be memorized by the vector of exponents alone $$E_{N,S}:=[A_k]_{k=1...N} \tag {2.1}$$ and has the well known expression: $$ a_N = a_0 \cdot {3^N \over 2^S} + {Q([A_1,A_2,...,A_N]) \over 2^S} \tag {2.2}$$ where the short-form-notation $Q()$ expanded is the similarly known sum $$ Q([A_1,A_2,...,A_N]) = 3^{N-1} + 3^{N-2}2^{A_1} + 3^{N-3}2^{A_1+A_2} + ... + 3^0 2^{A_1+A_2 + ... + A_{N-1}} \tag {2.3} $$


part 3 Cycles and solutions over rational numbers

From $(2.2)$ we can write the equation for some transformation $a_0 \to a_N$: $$ 2^S a_N = 3^N a_0 + Q(E_{N:S}) \tag {3.1} $$ To form a cycle we demand for this $a_0 = a_N$ . We can derive a determination for $a_0$: $$ a_0 = { Q(E_{N:S}) \over 2^S - 3^N } \tag {3.2} $$ We know only one cycle in the Collatz-problem over positive odd integers ($a_0=1$) and only 3 more in the negative odd integers ($a_0 \in \{-1,-5,-17\}$).
But obviously there are solutions over the rational numbers. To memorize that generalization to rationals we use now the letter $\alpha$ for such a rational or integer solution: $$ \alpha = { Q(E_{N:S}) \over 2^S - 3^N } \tag {3.3} $$ This little equation shows that to have $\alpha$ in the positive numbers we need, that the denominator must be positive, and thus we need $$ 2^S > 3^N \tag {3.4a}$$ and by this, $$S \ge \lceil N \cdot \log_2(3) \rceil \tag {3.4b} $$ or as expressed in the OP $$S \ge \lfloor 1+ N \cdot \log_2(3) \rfloor $$


The cycle- and stoppingtime-discussions usually assume, that in (3.4b) we can always apply the assumption of equality: $$S = \lceil N \cdot \log_2(3) \rceil \tag {3.4c} $$ and the OP puts this assumption in question, targetting the possibility that perhaps $$S = 1+\lceil N \cdot \log_2(3) \rceil \tag {3.4d} $$ to calculate the correct stopping time.

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for real $u,y$:

$$\left(\frac{1}{u}\right)^{u-\frac{1}{u}}=2y$$ $$e^{\ln\left(\frac{1}{u}\right)\left(u-\frac{1}{u}\right)}=2y$$

We can press many expression into a form for applying hyper Lambert W:

$$ue^{f(u)e^u}=e^{g(u)}$$ $$f(u)=\frac{g(u)-\ln(u)}{e^u}$$ $\ $

$$g(u)=\ln\left(\frac{1}{u}\right)\left(u-\frac{1}{u}\right)$$ $$f(u)=\frac{\ln\left(\frac{1}{u}\right)\left(u-\frac{1}{u}\right)-\ln(u)}{e^u}$$ $$ue^\frac{\ln\left(\frac{1}{u}\right)\left(u-\frac{1}{u}\right)-\ln(u)}{e^u}=2y$$ $\ $

$$G\left(\frac{\ln\left(\frac{1}{u}\right)\left(u-\frac{1}{u}\right)-\ln(u)}{e^u};u\right)=2y$$ $$u=HW\left(\frac{\ln\left(\frac{1}{u}\right)\left(u-\frac{1}{u}\right)-\ln(u)}{e^u};2y\right)$$

So we have a closed form for $x$, and the representations of Hyper Lambert W give some hints for calculating $x$.

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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 $$\Delta(N,S) \gt \exp(S \cdot (\ln 2 - 0.1)) \qquad S\gt 27 \tag {\text{EL}}$$

G. Rhin looks at the difference of logarithms $\Lambda(N,S)=| S \ln2 - N \ln 3 |$ and gives $$\Lambda(N,S) \gt {1\over 457 N ^{13.3} }| \tag {\text{RH}}$$ I've looked as well at $\Lambda(N)$ and propose $$\Lambda(N,S) \gt {1\over 10 N \ln N } \tag {\text{HE}}$$ 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.

image1We 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 image2 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. image3The slightly jittering effect in the Ellison-curve is because its function uses the somehow jittering values $S$ instead that of $N$ for its argument.

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$\def\dn{\operatorname{dn}}$

This goal is to understand how to expand inverses of non-elementary functions as a series. For example the Jacobi dn Fourier cosine series from Paramanand’s blogspot: l

$$\dn(u,k)=\frac{a_0}2+\sum_{n=1}^\infty a_n\cos(2nz),a_n=\frac1\pi\int_{-\pi}^\pi\dn(u,k)e^{-2inz}dz,z=\frac{\pi u}{2K(k)}$$

using residue calculus to find $a_n=\frac{2\pi}{K(k)(q^{-n}(k)+q^n(k)}$ with the nome $q(k)$ and complete elliptic integral of the first kind $K(k)$. However, for someone not knowing the residue theorem, it would be hard to derive this result.


For Expressions for the inverse function of $f(x) = \ln(x)e^x$

On some applications of the generalized hyper-Lambert functions mentions a complicated Lagrange inversion series expansion referencing

Saks and Sygmund [8, pp. 201–202])

However their Analytic functions paper does not evaluate the derivatives for the coefficients. The following methods are applicable to solving $za_1^{\cdots^{a_j^z}}=a$.

$$f(z)=z\underbrace{e^{e^{\dots^z}}}_{k “e”\text s}\implies f^{-1}(z)=\sum_{n=1}^\infty\frac{z^ n}{n!}\frac{d^{n-1}}{dw^{n-1}}e^{-n \overbrace{e^{e^{\dots^z}}}^{k-1 “e”\text s}}\bigg|_0$$

Repeated Maclaurin Expansion:

$$\frac{d^{n-1}}{dw^{n-1}}e^{-n \overbrace{e^{e^{\dots^z}}}^{-1 “e”\text s}}\bigg|_0 $$

Stirling Expansion:

$$\frac{d^{n-1}}{dw^{n-1}}e^{-n \overbrace{e^{e^{\dots^z}}}^{-1 “e”\text s}}\bigg|_0 $$

Coefficients for All Branches:

[Evaluate?]

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Free for use. ${}{}{}{}{}{}{}{}$

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    $\begingroup$ I have a question, if two people edit a post at the same time, one of them will undoubtedly lose his work. $\endgroup$
    – user1034536
    Jan 17, 2023 at 4:25
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    $\begingroup$ @user1034536 That is why it is a good idea, to initially edit the posting with one or two lines, saying "This answer box is now in use. Please do not use." Then, you can immediately save this editing. In effect you are placing a temporary do not disturb sign around the answer box. $\endgroup$ Mar 27, 2023 at 14:20
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With a polynomial with real coefficients $p(x)$, what's the largest and smallest imaginary part you can get evaluating $p$ at a quaternion $\vec{q}$ giving $p(\vec{q})$?


What is $\mathbb{Q} \otimes \mathbb{R}$ as Abelian groups?

$q, q'$ are in $\mathbb{Q}$. The rest $a,b,c,d,r,r'$ are in $\mathbb{R}$.

$$ \begin{bmatrix} q & r \end{bmatrix} \begin{bmatrix} a & b \\ c & d \end{bmatrix} \begin{bmatrix} q' \\ r' \end{bmatrix} = \begin{bmatrix} qa+rc & qb+rd \end{bmatrix} \begin{bmatrix} q' \\ r' \end{bmatrix} = qaq' + rcq' + qbr' +rdr' $$

This map $\mathbb{Q} \times \mathbb{R} \to \mathbb{R}$ is multilinear.


Integral domains with exactly two units.

We have $\mathbb{Z}$ and $\mathbb{Z}/3\mathbb{Z}$, but no other finite ones.

Also $\mathbb{Z}[x]$ and $(\mathbb{Z}/3\mathbb{Z})[x]$.


Prove completeness of Routley-Meyer semantics using

equational logic (kinda) with $\Delta$ which sends truth values to $2$, e.g. modus ponens is the following:

$$ \Delta(a) \land \Delta(a \to b) \to \Delta(b) $$

The idea is to make a unifying logic that can express both the inference rules of the relevance logic $B$ and worlds and propositional variables for the semantics. It might not work.


Derivative operator defined in terms of forward difference where you pass in the nominal degree of the function in question.

Also, $\mathbb{Z}/p\mathbb{Z}$ as an almost $\mathbb{Q}$-vector space. Make $1/p$ behave like $0$. Then can we define stuff like $\sin$ using its power series?


Matrix algebras generated by:

$$ \begin{bmatrix} 1 & 1 \\ 1 & 0 \end{bmatrix} \;\; \text{and} \;\; \begin{bmatrix} 1 & 1 & 0 \\ 1 & 0 & 1 \\ 0 & 1 & 0 \end{bmatrix} $$


Four-place distance relation $Sabcd$

with the intuitive meaning $(a, b) \simeq (c, d)$, in prose $a$ is the same distance from $b$ as $c$ is from $d$.

With this you can defined lines and circles ... and triangles and other stuff like that.

How do you axiomatize the Euclidean version?


  • $\varphi : R \to \text{End}(S)$
  • $\psi : S \to \text{End}(T)$

We can make $\alpha : R \times S \to \text{End}(T)$ as follows:

  • $\alpha(r, s, t) = \psi(\varphi(r, s), t)$

Let rings be commutative but not necessarily unital.

Let multiplicatively closed sets do not necessarily contain $1$.

Are there any infinite rings such that all multiplicatively closed sets are finitely generated?

In the infinite ring $\mathbb{Z}$, the ring itself is not finitely generated as a multiplicatively closed set, since there are infinitely many prime numbers by Euclid's theorem.

If we assume temporarily that we have a ring multiplicatively generated by $x$, so the elements are $\{x, x^2, x^3, \cdots\}$ with addition somehow defined, then all of its sets are finitely multiplicatively generated.

Let $F \subset \mathbb{N}$ and consider the set $X$ multiplicatively generated by $\{x^k : k \in F \}$.


Ultranaive compactification.

Add a new point, say that its only open neighborhood is the entire space.

Now, any open cover of the space MUST include $\varnothing^c$.

Is this a valid compactification?


A logic where $\varphi \to \psi$ maps to $\bigwedge_{v \in \text{Vars}(v)} \lozenge(v) \land \square(\varphi \to \psi) $ in the classical modal logic S5.

The idea is that this logic is potentially badly behaved and interacts poorly with substitution.


We work in NFU with choice, infinity, and pairs.

$$ [\forall xyzw]((x, y) = (z, w) \leftrightarrow (x=z \land z=w)) $$

Let $P(z)$ hold of $z$ if and only if it is a pair:

$$ P(z) \;\; \text{is defined as} \;\; [\exists r][\exists w](z = (r, w)) $$

Let $R(x)$ hold of $x$ if it is a relation:

$$ R(x) \;\;\text{is defined as}\;\; [\forall z \in x](P(x)) $$

Let $B(x)$ hold of $x$ if and only if it is a bijection of the whole space:

$$ B(x) \;\;\text{is defined as}\;\; R(x) \land [\forall z](\cdots) $$


What are the maximal ideals of this ring?

Let $X$ be an infinite set with a distinguished point $x_0$.

Let $F$ be a family of sets $\{ a : a \in 2^X \;\text{and $a$ is finite} \}$ .

Let $a + b$ be $\{ w : w \in a \not\leftrightarrow w \in b \}$.

Let $a\cdot b$ be $(a \cap b) \setminus \{x_0\}$.

$0$ is the annihilator for $\cdot$.

Additionally, $(a+b)\cdot(c + d) = ac + bc + ad + cd$ and the other distributive properties hold.


Best effort substructure

One somewhat obvious fact is that given an arbitrary subset of the domain of a structure, one cannot always take a substructure with that subset as its domain. Relation symbols do not present an obstacle here, but function symbols do.

(Also, what to do about partial congruences?) Partial congruences are interesting, they also give us partial ideals.

Let $L$ be a language and let $A$ be an $L$-structure. Let $B$ be a subset of $A$.


green's relations on semigroups.


Fake Burnside ring.

This is based on intentionally misunderstanding how the product (defined on $G$-sets) in the Burnside ring is defined.

Suppose we have $\boxed{1}$ and $\boxed{3}$ and

$\boxed{1}\boxed{1}$ is $\boxed{1}$

$\boxed{1}\boxed{3}$ is $\boxed{3}$

$\boxed{3}\boxed{1}$ is $\boxed{3}$

$\boxed{3}\boxed{3}$ is $9\,\boxed{1}$

multiplication is associative.

Equivalent to ring $\mathbb{Z}[b] / (b^2 - 9)$.

It gets weirder if we get rid of $\boxed{1}$.

Which rules correspond to actual groups?

The split complexes have a fake Burnside representation if we allow $\boxed{-1}$.

The complex numbers however do not have a fake Burnside representation.


https://hsm.stackexchange.com/questions/6256/did-gauss-know-jacobis-four-squares-theorem/12593

Where does this function $P(x, y) = \sum_{k} x^{(k^2)}(y^k + y^{-k})$ come from?


Annihilator as a maximal ideal

Let all rings be commutative and unital.

Lemma 51: Let $G$ be a finite group. If $I$ is an ideal of $R$, then $I[G]$ is an ideal of $R[G]$.

Proof. Suppose $I$ is an ideal of $R$.

Consider an element of $I[G]$. It always has the form $\sum_i a_ig_i$ where $a_i$ is an element of $I$ and $g_i$ is an element of $G$. $I[G]$ is obviously closed under subtraction.

Consider an element of $R[G]$ it halways has the form $\sum_j r_jg_j$ where $r_j$ is an element of $R$.

Their product is $\sum_{ij}r_ja_ig_ig_j$.

I note that $r_ja_i$ is in $I$.

Therefore $I[G]$ is closed under multiplication by $R[G]$, as desired.


Lemma 61: Let $G$ be a finite group. If $I$ and $J$ are ideals of $R$ and $I \subsetneq J$, then $I[G] \subsetneq J[G]$ and furthermore $J \setminus I[G]$ is nonempty.

$G$ contains an identity element $e$. Let $j$ be an element of $J$ that isn't in $I$. $je$ is in $J[G]$ but not $I[G]$, as desired.

$je = j$, additionally, is in $J$.


Note that $S$ is an $R$-module in addition to being an $R[G]$-module.

$I$ is the annihilator in $R$ of $S$, i.e.

$$ I = \{ r : rS = 0 \} $$

Suppose that $I$ is not a maximal ideal of $R$, i.e. there exists a $J$ such that $I \subsetneq J \subsetneq R$.

$(I[G])S$ is $0$.

$(R[G])S$ is the whole module $S$ because $1$ is in there.

Now let's look at $JS$.

$JS$ cannot be $0$, because $J$ is not a subset of the annihilator of $S$.


ways of distinguishing standard topology and half-open topology

  1. half-open is a disjoint union of open sets.

Ideals of cylindric algebras.

If you tweak the signature of a cylindric algebra you can get the following:

  1. The constants $0$ and $1$.
  2. xor $+$ and and $*$
  3. The universal quantifiers $u_k$ corresponding to $[\forall x_k](\cdots)$ in the traditional syntax of first-order logic.
  4. The constants $d_{kl}$ corresponding to $x_k = x_l$ in the traditional syntax.

Neglecting 4, this makes a cylindric algebra into a boolean ring.

Let $I$ be an ideal of a cylindric algebra.

Let an ideal $I$ be universally-closed if and only if $u_k(I) \subset I$ for all $k$.


Constructing a point set for a topology given by a lattice.

Let $\Lambda$ be a lattice with arbitrary meets and finite joins. Is there a way to convert it to a topology whose lattice of closed sets is isomorphic to it?


multi-valued algebra.

Let $f, g, h, \cdots$ be multi-valued functions ($n+1$-place relations).

Let $f(g(x)) \approx h(x)$ insist that $h(x)$ and $g(x)$ are equal at all values and that, for all values of $x$ $h(x)$ takes on zero or one values and likewise for $f(g(x))$.

This is one possible idea for a multi-valued alegbra.


this is wrong. we want the rank of $\kappa$

How do you axiomatize ZFC + Worldly Cardinals?

How do I tell when I've defined ZFC + Worldly Cardinals instead of something weaker like ZFC + Con(ZFC)?


Given the definition of a worldly cardinal, I can see how to extend ZFC with enough axioms to make a constant symbol $\kappa$ a worldly cardinal.

Let $M$ be a $(\kappa, \in)$-structure that we're considering as a possible model of ZFCW.

Let $\kappa$ be a constant symbol, intended to name our worldly cardinal.

Here's an axiomatization of ZFCW.

Let $P$ be a list of parameters.

Let $\varphi(P)$ be a sentence with parameters that's an axiom of ZFC (possibly instantiated from an axiom schema).

Let $\varphi^*(P)$ be the following $P \in \kappa \to \varphi'(P)$ where $\varphi'$ is $\varphi$ but with $\forall x (\cdots)$ replaced with $\forall x (x \in \kappa \to \cdots)$ and likewise for $\exists$. The notation $P \in k$ abbreviates $(P_1 \in \kappa \land P_2 \in \kappa \land \cdots \land P_n \in \kappa)$.

At this point, we only know that $\kappa$ is a standard model of ZFC.

If I keep going and insist that $\kappa$ is transitive and totally ordered by the elementhood relation, then I think I've only insisted that $\kappa$ is a standard and transitive model.

At this point, I'm out of ideas for what additional axioms to toss in to make $\kappa$ worldly enough.


I changed this question completely since I first wrote it. This answer completely solved my original question of how parameters work in the axiomatization of ZFC. Basically, when assessing the truth of $M \models \text{ZFC}$ for some $(\in)$-structure $M$, the parameters used in expansions of axiom schemas of $\text{ZFC}$ are taken directly from the carrier of $M$.


transfinite induction:

Let $A$ be a set of ordinals.

$$ f(0) = \cdots \\ f(\alpha^+) = \cdots \\ f(\lim A) = \cdots $$

Definition of $+$. Let $\kappa + B$ abbreviate $\{x : x \in B \land x+\kappa \}$

$$ 0+\kappa = \kappa \\ \kappa+0 = \kappa \\ \kappa + \beta^+ = (\kappa+\beta)^+ \\ \kappa + \lim(B) = \lim(B + \kappa)$$


Is the "ordinalification" of ZFC equiconsistent with ZFC?

Let's define $T$, a two-sorted theory ($E, F$) with the following relations:

  • $ \in : E \times E \to 2 $
  • $ \in : E \times F \to 2 $

and the following function:

  • $ (,) : E \times E \to E $

Let $\Delta$ consist of all well-formed formulas $\varphi(\vec{x})$ with parameters in the language of ZFC such that:

  1. Every parameter in $\vec{x}$ is an ordinal.
  2. All bound variables introduced by quantifiers are restricted to be ordinals, i.e. $\forall x ( \cdots )$ does not occur outside of $\forall x (\mathrm{ordinal}(x) \to \cdots)$ and likewise for $\exists$.
  3. $\varphi(\vec{x})$ is a theorem of ZFC.

Let $T$ consist of $\Delta$ together with:

  1. Comprehension: For all well-formed formulas with parameters $\varphi(x, \cdots)$, it holds that $\exists f \mathop ( x \in f \leftrightarrow \varphi(x, \cdots))$.
  2. Extensionality for the sort $F$.
  3. The defining property of ordered pairs (and no other properties about ordered pairs). $\forall x \forall y \forall z \forall w ((x, y) = (z, w) \leftrightarrow (x = z) \land (y = w))$

These axioms are inspired by the first-order theory called second-order arithmetic.

Consistency of ZFC implies consistency of $T$.

Assume ZFC is consistent.

ZFC has a model $A$.

Let's construct a new model of $T$, $B$.

Take $A$ and attach an arbitrarily chosen bijection between $\text{Ord} \times \text{Ord}$ and $\text{Ord}$ as the interpretation of $(,)$.

Take the powerset of $\text{dom}(A)$ as the interpretation of the sort $F$.

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