Sandbox for drafts of long, complex posts

Question

This sandbox is intended for saving drafts of long, complex posts, especially posts whose composition takes a long time. It serves to localize to one thread the front-page "bumps" caused by edits to drafts of such posts, so that they may be easily ignored. Also, it helps to guard against losing longly-composed posts due to system crashes.

When you are happy with your draft here, you may simply copy the code and paste it to the desired location.

Proper Use of the Sandbox

1. Do not post a new answer! We wish all the answers on this page to be owned by the Community user (so that only a non-sentient bot is informed of edits to these answers). Posting a new answer will make you the owner, meaning that you will be notified whenever another user makes an edit to that answer.

   The sandbox has been "wiki locked" to prevent the creation of new answers. There are more than enough existing answers for users to edit over, and this will greatly reduce the frequency at which we request that the answers be disassociated from specific users.

2. Do not delete answers! Deleting seems like a reasonable option, but there are no "hard deletions" on Stack Exchange, and users with sufficient privileges will still see your supposedly deleted postings. Deleted answers will be undeleted and cleared for the use of others.

3. Do look for an answer which indicates that it is free and then edit it to your heart's content. If none appears available, take over the one that has been left unchanged the longest (which will appear at the bottom of the page if you order answers by "activity").

4. Do not expect your draft to remain untouched for days. There are no guarantees that your draft will be the latest revision if you return days later. While users will try not to step over others' toes, it may happen that an unfinished draft is edited out. Your draft will, however, still exist as a revision of the answer it was made in. If your drafting is expected to take place over a longer period of time, either

   • take note of the URL of the answer provided by clicking the share button, or
   • save a copy of your draft locally (or even "in the cloud").

5. Do clear your draft when you are finished. This includes removing all $\LaTeX$ from your answers. Replacing all code with a simple statement like

   This answer is free for anyone to use

   is sufficient. Periodically users may go through and free up answer slots that have not been edited in, say, over one month. But you can aid in the smooth running of this sandbox by clearing away your drafts when you are finished with them.

6. Do not "claim" multiple answers concurrently. Since this post is closed, the answers are a limited resource. If you really must compose several long, complex posts at the same time, you can still use a single answer, separating the different drafts using Markup: horizontal rules (---) and/or headings (# Header 1 #) are natural choices.

7. Do not create new such sandboxes. The point of having a unique such sandbox is that it minimizes the noise on the front page when the sandbox is edited. If there were multiple sandboxes they will frequently occupy numerous front page slots, pushing other topics off the front page, and increasing noise.

Answer 1

This answer is free for anyone to use.

Answer 2

Calculation for transformationrule $T_7:=a_{k+1}=\{7 \cdot a_k+1 \}_{2,3,5}$
The exponents at $2,3,5$ had been summed in $50 \times 3$ boxes for documenting the sum of their occurences in each of the (at most) $ 50$ iterations. We have less than 50 iterations when the transformed number $a_k$ becomes lower than 100. The $a_0$ are taken from numbers larger than 10^10 and only of the residueclasses $(1,7,11,13,17,19,23,29) \pmod {30}$

$ \small \begin{array} {r|rrr|rc} it\# & \hat S_2& \hat S_3 & \hat S_5 & n_{it} \\ \hline 1 & 2.0000040 & 0.75000400 & 0.31250000 & 1000000 \\ 2 & 1.9999940 & 0.72083200 & 0.32364560 & 1000000 \\ 3 & 2.0000402 & 0.70273872 & 0.33434304 & 1000000 \\ 4 & 2.0001205 & 0.69197638 & 0.34290131 & 1000000 \\ 5 & 2.0002148 & 0.68582518 & 0.34908932 & 1000000 \\ 6 & 2.0002937 & 0.68244162 & 0.35323792 & 999984 \\ 7 & 2.0003345 & 0.68066224 & 0.35585501 & 999963 \\ 8 & 2.0003499 & 0.67979198 & 0.35743384 & 999320 \\ 9 & 2.0003712 & 0.67942422 & 0.35836046 & 994438 \\ 10 & 2.0004165 & 0.67931762 & 0.35889442 & 993295 \\ 11 & 2.0004784 & 0.67932262 & 0.35919258 & 984903 \\ 12 & 2.0005343 & 0.67934158 & 0.35934571 & 980992 \\ 13 & 2.0005689 & 0.67931087 & 0.35940904 & 963011 & :\\ 14 & 2.0005905 & 0.67919550 & 0.35942030 & 946996 &: \\ 15 & 2.0006349 & 0.67898869 & 0.35940702 & 941772 & \text{less than 10% missing}\\ \hline 16 & 2.0007545 & 0.67871026 & 0.35938796 & 889158 & \text{more than 10% missing} \\ 17 & 2.0010039 & 0.67840071 & 0.35937325 & 869017 & : \\ 18 & 2.0014249 & 0.67811162 & 0.35936579 & 849907 & : \\ 19 & 2.0020399 & 0.67789472 & 0.35936445 & 836866 \\ 20 & 2.0028501 & 0.67779270 & 0.35936792 & 791487 \\ 21 & 2.0038388 & 0.67783381 & 0.35937770 & 771699 \\ 22 & 2.0049763 & 0.67803069 & 0.35939946 & 738338 \\ 23 & 2.0062224 & 0.67838250 & 0.35944229 & 706518 \\ 24 & 2.0075287 & 0.67887889 & 0.35951678 & 687055 \\ 25 & 2.0088383 & 0.67950394 & 0.35963252 & 645856 \\ 26 & 2.0100864 & 0.68023926 & 0.35979616 & 623550 \\ 27 & 2.0112020 & 0.68106548 & 0.36001049 & 599977 \\ 28 & 2.0121113 & 0.68196257 & 0.36027448 & 578882 \\ 29 & 2.0127427 & 0.68290927 & 0.36058429 & 500821 \\ 30 & 2.0130325 & 0.68388236 & 0.36093452 & 466451 \\ 31 & 2.0129306 & 0.68485623 & 0.36131946 & 444705 \\ 32 & 2.0124052 & 0.68580304 & 0.36173380 & 433647 \\ 33 & 2.0114447 & 0.68669346 & 0.36217292 & 391434 \\ 34 & 2.0100588 & 0.68749796 & 0.36263259 & 373936 \\ 35 & 2.0082765 & 0.68818831 & 0.36310835 & 336386 \\ 36 & 2.0061431 & 0.68873919 & 0.36359492 & 290319 \\ 37 & 2.0037159 & 0.68912961 & 0.36408560 & 256745 \\ 38 & 2.0010601 & 0.68934398 & 0.36457211 & 232435 \\ 39 & 1.9982443 & 0.68937292 & 0.36504478 & 207889 \\ 40 & 1.9953373 & 0.68921357 & 0.36549317 & 193511 \\ 41 & 1.9924057 & 0.68886956 & 0.36590696 & 161354 \\ 42 & 1.9895117 & 0.68835066 & 0.36627706 & 143606 \\ 43 & 1.9867130 & 0.68767208 & 0.36659681 & 123702 \\ 44 & 1.9840623 & 0.68685348 & 0.36686295 & 118139 \\ 45 & 1.9816077 & 0.68591789 & 0.36707639 & 111627 \\ 46 & 1.9793931 & 0.68489031 & 0.36724254 & 100572 \\ 47 & 1.9774584 & 0.68379639 & 0.36737122 & 95169 \\ 48 & 1.9758400 & 0.68266104 & 0.36747609 & 86931 \\ 49 & 1.9745702 & 0.68150721 & 0.36757363 & 78950 \\ 50 & 1.9736773 & 0.68035486 & 0.36768178 & 75124 \\ --- & ------- & ------- & ------ & ---- \\ est & 2.0..... & 0.68054... & 0.35964... & \infty \end{array} $

Answer 3

This post if free to use for any one.

Answer 4

$$ T = \small \begin{bmatrix} 1/2 & 2/3 \\ 1/2 & 1/3 \\ \end{bmatrix} $$ From the idea that the probabilities $P$ equal before AND after transformation we formulate with P as columnvector: $$ T \cdot P = P $$ $\qquad \qquad$ (for easier notation we don't note the condition $[1,1] \cdot P = [1]$ here)
and procede by reformulating $$ \begin{array} {ll} T \cdot P = P \to T \cdot P = I \cdot P \to (T-I) \cdot P = [0,0]^t \end{array}$$

$$ T - I = \small \begin{bmatrix} -1/2 & 2/3 \\ 1/2 & -2/3 \\ \end{bmatrix}$$ $$ \small \begin{bmatrix} 2.7251141 & -3.6334855 \\ -2.7251141 & 3.6334855 \\ \end{bmatrix} $$

Answer 5

This answer is free for anyone to use.

Answer 6

This post if free to use for any one.

Answer 7

This post if free to use for any one.

Answer 8

This post if free to use for any one.

Answer 9

Assume the (Abel-) equation $$ F(\exp(x))=F(x)+1 $$ Using matrix-system with Carlemanmatrices $$ V(x) \cdot E \cdot F = V(x) \cdot F \cdot P $$ We assume that this equation is possible and there is a continuous range in $x$ giving consistent results, then the coefficients on the lhs and rhs must be equal and it is possible to note $$ E \cdot F = F \cdot P $$ $$ g \cdot \left[\small \begin{array} {r} 1&1&1&1 & \cdots \\ 0&1&2&3 & \cdots \\ 0&1&2^2&3^2 & \cdots \\ 0&1&2^3&3^3 & \cdots \\ \vdots &\vdots &\vdots &\vdots & \ddots \end{array} \right] \times \small \left[\begin{array}{} 1 &a_{0,1}&a_{0,2} & \cdots\\ 0 &a_{1,1}&a_{1,2} & \cdots\\ 0 &a_{2,1}&a_{2,2} & \cdots\\ 0 &a_{3,1}&a_{3,2} & \cdots\\ \vdots &\vdots &\vdots &\ddots \end{array} \right] = \left[ \small \begin{array}{} 1 &a_{0,1}&a_{0,2} & \cdots\\ 0 &a_{1,1}&a_{1,2} & \cdots\\ 0 &a_{2,1}&a_{2,2} & \cdots\\ 0 &a_{3,1}&a_{3,2} & \cdots\\ \vdots &\vdots &\vdots &\ddots \end{array} \right] \times \left[\small \begin{array} {r} 1&1&1&1 & \cdots \\ \cdot&1&2&3 & \cdots \\ \cdot&\cdot&1&3 & \cdots \\ \cdot&\cdot&\cdot&1 & \cdots \\ \vdots &\vdots &\vdots &\vdots & \ddots \end{array} \right] $$ Because these all are Carleman-matrices, the first column is known by definition ($=V(0)$) and we need only the second column for the coefficients $a_{r,1}$ in the powerseries: $$ E \cdot F_{,1} = F_{,0..1} \cdot P_{0..1,1}= F_{,0..1} \cdot [1,1]^\tau = 1 \cdot F_{,0}+1 \cdot F_{,1} = V(0)+I \times F_{,1} $$ $$ g \cdot \left[\small \begin{array} {r} 1&1&1&1 & \cdots \\ 0&1&2&3 & \cdots \\ 0&1&2^2&3^2 & \cdots \\ 0&1&2^3&3^3 & \cdots \\ \vdots &\vdots &\vdots &\vdots & \ddots \end{array} \right] \times \left[ \small \begin{array}{} a_{0,1} \\ a_{1,1} \\ a_{2,1} \\ a_{3,1} \\ \vdots \end{array} \right] = \left[ \small \begin{array}{} 1&a_{0,1} \\ 0& a_{1,1} \\ 0&a_{2,1} \\ 0 & a_{3,1} \\ \vdots \end{array} \right] \times \left[\small \begin{array} {r} 1 \\ 1 \end{array} \right] $$

$$ E \cdot F_{,1} = F_{,0}+I \cdot F_{,1} \\ (E - I) \cdot F_{,1} = V(0)$$

$$ g \cdot \left[\small \begin{array} {r} 1-1&1&1&1 & \cdots \\ 0&1-1&2&3 & \cdots \\ 0&1&2^2-2&3^2 & \cdots \\ 0&1&2^3&3^3-6 & \cdots \\ \vdots &\vdots &\vdots &\vdots & \ddots \end{array} \right] \times \left[ \small \begin{array}{} a_0 \\ a_1 \\ a_2 \\ a_3 \\ \vdots \end{array} \right] = \left[ \small \begin{array}{} 1 \\ 0 \\ 0 \\ 0 \\ \vdots \end{array} \right] $$

Next step we want to solve by matrix-inversion, but the system we've arrived at doesn't allow to invert the lhs-matrix. This is what we would like to have the system solvable: $$ F_{,1} = (E-I)^{-1} \cdot V(0) = (E-I)^{-1}_{,0} $$ $$ fS2F \cdot Q \cdot F = F \cdot Q $$

Answer 10

@mjd is drafting a question here. If you read it, please pretend that you haven't.

---

MJD: Before posting this question look at this and see if it helps:

How do you extend the topological semantics for intuitionistic propositional logic (IPC) to a semantics for intuitionistic first order logic (IFOL)?

Previously the draft was at: https://math.stackexchange.com/questions/4988253/how-do-we-construct-counterexample-models-for-statements-of-intuitionistic-fol

---

One way to prove that a formula of logic is intuitionistically invalid is by using the topological semantics. We assign values to propositional variables that are subsets of $\Bbb R^n$. We interpret conjunction and disjunction as $\cap$ and $\cup$, and negation of $F$ as $\operatorname{Int}(\Bbb R^n - \operatorname{value}(F))$. Then a formula is valid if and only if its interpretation is exactly $\Bbb R^n$ for all possible valuations of its variables.

Suppose I have a formula which I suspect is not intuitionistically valid. I would like to construct open subsets of $\Bbb R^n$, one for each variable, that witness the invalidity. For example, if the formula was $P\lor \lnot P$ I might select $\operatorname{value}(P) = (\infty, 0)$. Then the value of $P\lor\lnot P$ is $\Bbb R-\{0\}$ which proves that the formula is invalid.

In the past I have constructed these witnesses by trial and error and ad-hoc reasoning. This recent post by Math SE user Zhen Lin suggests that it can be done more systematically. Something is happening here that I can't quite see.

1. Can someone elaborate on Zhen Lin's argument, not just on the steps themselves, but on how they were found? For example, Zhen Lin says:

   Instead, consider the topological interpretation: imagine that we have some open subset ϕ⊆Γ×X and an open subset ψ⊆Γ…

   Why $\Gamma \times X$? What's the intuition that leads us to consider a product space? Is Zhen's choice of the letter $X$ intended to suggest a connection with the quantified variable $x$ in the formula? What exactly is meant by “$\forall_X(\phi)$”?

2. Are there general, mechanical methods for computing these counterexamples? Perhaps a tableau method or something analogous?

Answer 11

This post if free to use for any one.

Answer 12

This post if free to use for any one.

Answer 13

This post is free to use for anyone.

Answer 14

This answer is free for anyone to use.

Answer 15

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

Answer 16

$\def \F{\text F}$ To derive a power series for the inverse of $f(z)=\,_2\F_1(a,b,c,z)$, start with the hypergeometric differential equation, apply the inverse substitution $y\to x,y’\to\frac1{y’},y’’\to-\frac{y’’}{y’^3}$:

$$z(1-z)y’’+(c-(a+b+1) z)y’-a b y=0,y(0)=1,y’(0)=\frac{ab}c$$$$\implies y(y-1)y’’+(c-(a+b+1)y)y’^2-a b zy'^3=0,y(1)=0,y’(1)=\frac c{ab}\tag1$$

for $y=f^{-1}(z)$

since $y(0)=-\infty$, we expand around $z=1$ by substituting $\frac c{ab}(y-1)\to y$. We then have a series in $v=\frac c{ab}(z-1)$:

$$(abv+c)y’^3+((a+b+1)y-c)y’^2+yy’’-y^2y’’=0,y(0)=0,y’(0)=1$$

Substituting $y=\sum\limits_{n=1}^\infty a_nv^n$, gathering powers of $v$, and rearranging gives the recurrence relation for $a_n$, so:

$$\,_2{\F_1}^{-1}(a,b,c,z)=\sum_{n=1}^\infty a_n\left(\frac c{ab}(z-1)\right)^n\\\\n(c+n-1)a_{n+1}=ca_1^2na_n-\sum_{k=2}^{n-1}k(k-1)a_ka_{n-k+1}-\sum_{k=1}^{n-1}\sum_{m=1}^{n-k}a_ma_{n-k-m+1}((n-k-m+1)(abkma_k+(a+b+1)ka_k+cm (k+1)a_{k+1})-m(m-1)a_k),a_1=1$$

---

To derive a Puisex like series for $y=f^{-1}(z)$, start with its differential equation $(1)$, notice $y\sim \frac cb\left(1-z^{-\frac1a}\right),|z|\to0$, and let $z=(1-\frac bcv)^{-a}[]$:

$$y(y-1)(c-bv)y’’-y((c-bv)y’((a+b+1)y-c)+(a+1)by(y-1)+(c-bv)^2y’^2)=0$$

to get a series $y=\sum_{n=1}^\infty a_nv^n$ in $v=\frac cb \left(1-z^{-\frac1a}\right)$. Substituting it and finding the recurrence relation via the same process gives:

$$$$

---

---

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

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:

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

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.

Answer 17

This post is free for anyone to use.