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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 Jul 18 '12 at 8:35
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    $\begingroup$ (+1) For thinking outside the (sand)box. $\endgroup$ – cardinal Jul 18 '12 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 Oct 5 '12 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 '12 at 18:03
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    $\begingroup$ 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 Maybe someday we can get rid of this outdated crutch. $\endgroup$ – Najib Idrissi Dec 2 '15 at 14:07

17 Answers 17

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First of all to follows we remember some elementary definitions and results about manifolds.

Definition

A function $f$ defined in a subset $S$ of $\Bbb R^k$ is said of class $C^r$ if it can be extended to a function $\phi$ (said $C^r$-extension) that is of class $C^r$ in a open neighborhood of $S$.

Lemma

If $f$ is a function defined in a subset $S$ of $\Bbb R^n$ such that for any $x\in S$ there exist a function $f_x$ defined in a neighborhood of $x$ that is of class $C^r$ and compatible with $f$ on $U_x\cap S$ then $f$ is of class $C^r$.

Lemma

If $U$ is an open set of $H^n_k:=\Bbb R^{n-k}\times[0,+\infty)^k$ for any $k\le n$ then the derivatives of two different extensions $\phi$ and $\varphi$ of a $C^r$-function $f$ agree in $U$.

Definition

A $k$-manifold with boundary/corners in $\Bbb R^n$ of calss $C^r$ is a subspace $M$ of $\Bbb R^n$ whose points have a neighborhood $V$ in $M$ that is the immage of a homeomorphism $\phi$ of calss $C^r$ defined an open set $U$ of $\Bbb R^k$ or of $H^k_1/H^k_m$ and whose derivative has rank $k$.

So now let be $M$ a $(n-1)$-manifold with boundary of class $C^r$ in $\Bbb R^n$ and thus let be $\gamma$ a injective curve defined in the unit interval $I:=[0,1]$ such that $\gamma(0)=0$ and such that the unit tangent vector not lies -for each $t\in I$- to the tangent space at any point of $M$. So we call cylindroid $C$ of trajectory $\gamma$ and of section $M$ the set $$ C:=\bigcup_{t\in I}\big(M+\gamma(t)\big) $$ that is obtained moving along $\gamma$ the points of $M$. Click here to see an example. So if $\xi\in C$ then there exist a coordinate patch $\alpha:U\rightarrow V$ and $t\in I$ such that $$ \xi=\alpha(x)+\gamma(t) $$ for any $x\in U$ and thus let be $\phi$ the function from $U\times I$ to $\Bbb R^n$ defined through the equation $$ \phi(x,t):=\alpha(x)+\gamma(t) $$ for any $(x,t)\in U\times I$ and then we prove that any restriction of this function is a coordinate patch about $\xi$. So we observe that the immage of $\phi$ is open in $C$: indeed the immage $V$ of $\alpha$ is an open set of $M$ so that there exist an open set $W$ of $\Bbb R^n$ whose intersection with $M$ is $V$ and so remembering that the translation is a homeomorphism we observe that $$ \phi[U\times I]=\bigcup_{t\in I}\big(V+\gamma(t)\big)=\bigcup_{t\in I}\Big((W\cap M)+\gamma(t)\Big)=\bigcup_{t\in I}\Big(\big(W+\gamma(t)\big)\cap\big(M+\gamma(t)\big)\Big)=\\ \Biggl(\bigcup_{t\in I}\big(W+\gamma(t)\big)\Biggl)\cap\Biggl(\bigcup_{t\in I}\big(M\cap\gamma(t)\big)\Biggl)=\Biggl(\bigcup_{t\in I}\big(W+\gamma(t)\big)\Biggl)\cap C $$ and thus we conclude that the immage of $\phi$ is open in $C$ because $W+\gamma(t)$ is open for each $t\in I$. Now if $\overset{°}M$ and $\partial M$ are the sets of the interior and boundary points of $M$ then we observe that $C$ is union of the sets $$ \bigcup_{t\in\text{int}\, I}\big(\overset{°}M+\gamma(t)\big)\,\,\,\text{and}\,\,\,\bigcup_{t\in\text{bd}\, I}\big(\overset{°}M+\gamma(t)\big)\,\,\,\text{and}\,\,\,\bigcup_{t\in I}\big(\partial M+\gamma(t)\big) $$ so that we analyse separately the case where $\xi$ is an element of the first set and the case where $\xi$ is an element of the second or either of the third: in particular this means to analyse separately the case where the set $ U $ is open in $ \Bbb R^{n-1} $ and $ t $ is an element of $ \text{int} \, I $ and the case where this is not and so just this is what we will do to follows -clearly this can be done independentely from the definition of the three mentioned sets nevertheless we decided to define them because we thought it makes more clear the following argumentetions, that's all. In particular in the example posted above these sets are respectively the invisible part, the red part, the green and balck parts. So first of all we observe that if $\xi$ is an element of the first set then $\alpha$ is a coordinate patch of $M$ defined in an open set $U$ of $\Bbb R^{n-1}$ and so the map $\phi$ defined above is a diffeomorphism in a neighborhood at any point of $U\times\text{int}\,I$ where is contained $\phi^{-1}(\xi)$, because if the unit tangent vector of $\gamma$ not lies to the tangent space at any point of $M$ then the derivative of $\phi$ is an isomorphism and so the statement follows directely from the inverse function theorem. So we conclude that the set $$ \bigcup_{t\,\in\,\text{int}\,I}\big(\overset{°}M+\gamma(t)\big) $$ is a $n$-manifold without boundary. In particular in this way we proved that the invisible part of the linked example is a manifold without boundary. Now if $\xi$ is a not an element of the first set then the previous argumentations hold only with some efforts that we show to follow. So the functions $\alpha$ and $\gamma$ can be extended to two $C^r$-functions $\beta$ and $\psi$ defined in a open neighborhood of $U$ and $I$ respectively so that the function $\phi$ can be extended to a function $\varphi:=\beta+\psi$ defined in a open neighborhood of $U\times I$ and in particular at any point $(x,t)$ of $U\times I$ this function has not singular derivative so that by the inverse function theorem there exist a (rectangular) open neighborhood $W_x\times W_t$ where $\varphi$ is a diffeomorphism. Now the set $\varphi[W_x\times W_t]$ is open in $\Bbb R^n$ and it is not disjoint from $C$ -indeed $W_x\times W_t$ is not disjoint from $U\times I$ and $\varphi$ is compatible with $\phi$ in $U\times I$ and the immage of $\phi$ is contained in $C$- so that by the continuity of $\phi$ the set $$ \phi^{-1}[\varphi[W_x\times W_t]] $$ is (not empty and) open in $U\times I$ and contains $W_x\times W_t\cap U\times I$ where $\phi$ is a diffeomorphism. So if we prove that the immage of $W_x\times W_t\cap U\times I$ through $\phi$ is open in $C$ then we conclude that the restriction of $\phi$ to this set is a coordinate patch defined in a open set of $\Bbb R^{n-1}\times[0,+\infty)$ or in a open set of $\Bbb R^{n-2}\times[0,+\infty)^2$ if $U$ is open in $\Bbb R^{n-1}$ and $t$ is not an element of $\text{int}\, I$ or if $U$ is not open in $\Bbb R^{n-1}$ and $t$ is an element of $\text{int}\, I$ or either if $U$ is not open in $\Bbb R^{n-1}$ and $t$ is not an element of $\text{int}\, I$ respectively. So how do this? Could someone help me, please?

To follows some observations that I tried to use: naturally you are not constrained to read it if you do not desidre.

OBSERVATION

If the function $\phi$ was injective in $U\times I$ the statement follows immediately because in particular the function $\varphi$ would be injective in $W_x\times W_t\cup U\times I$ so that $$ \phi[W_x\times W_t\cap U\times I]=\varphi[W_x\times W_t\cap U\times I]=\varphi[W_x\times W_t]\cap\varphi[U\times I]=\\ \varphi[W_x\times W_t]\cap\phi[U\times I] $$ having remembered that $\varphi$ and $\phi$ are compatible in $U\times I$. So in particular we observe that the injectivity of $\phi$ follows immediately if the set $M+\gamma(t_0)$ and $M+\gamma(t_1)$ was disjoint for any $t_0,t_1\in I$ such that $t_0\neq t_1$ and in particular I tried to prove this using the injectivity of $\gamma$ that above I did not use. Moreover since $\alpha$ is a homeomorphism and since $W_x$ is open in $\Bbb R^{n-1}$ then $\alpha[W_x\cap U]$ is open in $M$ and so exist a open set $W$ of $\Bbb R^n$ whose intersection with $M$ is $\alpha[W_x\cap U]$ and thus implementing the argumentations used above it follows that $$ \phi[W_x\times W_t\cap U\times I]=...=\Biggl[\bigcup_{t\in W_t\cap I}\big(W+\gamma(t)\big)\Biggl]\cap\Biggl[\bigcup_{t\in W_t\cap I}\big(M+\gamma(t)\big)\Biggl] $$ so that if the set $\Biggl[\bigcup_{t\in W_t\cap I}\big(M+\gamma(t)\big)\Biggl]$ and $\Biggl[\bigcup_{t\in I\setminus W_t}\big(M+\gamma(t)\big)\Biggl]$ was disjoint then $\phi[W_x\times W_t\cap U\times I]$ was open in $C$ and this surely happens if the set $M+\gamma(t_0)$ and $M+\gamma(t_1)$ was disjoint for any $t_0,t_1\in I$ such that $t_0\neq t_1$.

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Comment My initial thoughts go in the following direction, but by second read I see that this concerns likely only a subset of impossible solutions. But perhaps the ansatz itself helps as initial idea...


Error detected - needs update

(taken from my two earlier comments)
By subtracting $8$ on each side one gets $7⋅(7^X−1)=3⋅(3^Y−1)+5⋅(5^Z−1)$ and $(X,Y,Z)=(x−1,y−1,z−1)$. Then divide all by $7⋅3⋅5$.

For instance we get on the lhs $(7^X−1)/15$ and on the rhs equivalent terms.

This gives restrictions for the modular classes for the exponents, and possibly one can proceed with further modular considerations (for instance on primefactorizations in lhs and rhs).

For instance the lhs has then always $2^5⋅5$ as factor, the terms in the rhs always $2^4$ and $2^3$ as factor. Since the constant occurence of $2^5$ as minimum means a strong selection for the $5^Z−1$-term: only $Z=24j$ is possible and by this the selected have all the primefactors $3⋅13⋅31⋅31$. This can similarly be checked for the other terms and often it is possible to derive from this contradictions.


By the corresponding primefactors of 2 and 13 in the lhs and rhs I find that in the lhs we can only have valid solutions in steps of 12 in the exponent. The primefactor decomposition of the first two exponents give

(7^12k-1)/15:
 X,    value,                primefactors
--------------------------------------------------------------
[12, 922752480,            "2^5.3.13 .5.19.43.181 "]
[24, 12772082092037760960, "2^6.3.13 .5.19.43.181  xxx.73.193.409.1201"]

This means, with all following admissible exponents we shall have the set of primefactors of the first row.

Similarly we look at the $5^z$ term, in already required $24$ steps:

(5^24j-1)/21:
 Z    value                primefactors
--------------------------------------------------------------
[24, 2838316417875744,     "2^5.3.13   .31.313.601.390001"]
[48, 16917684184764290..., "2^6.3.13   .31.313.601.390001.17.11489.152587500001"]

Again, in all following admissible exponents we shall have the set of primefactors in the first row. But interesting: $7^{12k}-1 $ as well as $5^{24j}-1$ have the primefactor $3$ - but of course this cannot happen in the third term:

(3^24i-1)/35:
 Y    value                primefactors
--------------------------------------------------------------
[24, 8069415328,           "2^5  .13      .41.73.6481"]
[48, 227904123076...     , "2^6  .13      .41.73.6481   .17.97.193.577.769"]

So, if I've not made a silly mistake, this is a proof of nonexistence of solutions aside $(X,Y,Z)=(0,0,0)$ resp $(x,y,z)=(1,1,1)$. This answer space is free to use :)

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  • $\begingroup$ @MakotoKato Could you please delete the content of your answer except the remark "... is free to use" - so that other users can use this space without worry? $\endgroup$ – IV_ Jul 13 at 20:00
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$$\frac 1x + \frac 1y + \frac 1z = \frac 56$$

$$\frac 1x + \frac 1y = \frac 56 - \frac 1z$$

$$\frac{x+y}{xy} = \frac {5z-6}{6z}$$

$$$\frac{x+y}{xy} ​= \frac {5z-6}{6z}$6

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  • $\begingroup$ You should remove "This answer is free for anyone to use" at the beginning. I almost overwrote this. $\endgroup$ – robjohn Jun 6 at 22:08
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I have a simple inversion routine for lower triangular matrices (assumed to be invertible, no errorchecks) in Pari/GP

I get, dependent on counting, for multiplications either 1/6*x^3 + 1/2*x^2 + 1/3*x (where x means dimension $n$) or 1/6*x^3 - 1/3*x ; see below.

 \\ invert a lower triangular matrix; no errorchecks! 
{triinv(m)=local(tmp,rs=rows(m),cs=cols(m),su_m,su_d);
  tmp=matrix(rs,cs);
  su_m=su_d=0; \\ inserted for documenting the multiplications & divisions
  for(c=1,cs,
       tmp[c,c]=1/m[c,c]; su_d++;
       for(r=c+1,rs,
               tmp[r,c]=-sum(k=c,r-1,m[r,k]*tmp[k,c])
                   / m[r,r];
               su_m+= (r-c);
               su_d++;
           );
      );
  \\ return(m);  \\ commented out, we need the counters of mult. & divs!
  return([cs,su_m,su_d]);}

Document:

\\ document the multiplications, divisions in a list for m=1..20   
for(m=1,20,print( triinv(PPow(1,m))));   \\ PPow gives a Pascalmatrix of size m x m

Protocol:

\\ protocol
\\ dim mu di  "mu"=multiplications "di"=divisions
\\ [1  0  1]
\\ [2  1  3]
\\ [3  4  6]
\\ [4 10 10]
\\ [5 20 15]
\\ [6 35 21]

Find formula:

\\ for multiplications we get the sequence [0,1,4,10,20,...]
\\ for divisions [1,3,6,10,...]
polinterpolate([1,4,10,20]) \\ check formula for multiplications
 
\\ %7 = 1/6*x^3 + 1/2*x^2 + 1/3*x   \\ result by Pari/GP

polinterpolate([0,1,4,10,20])

\\ %8 = 1/6*x^3 - 1/6*x      \\ result by Pari/GP

This space is now free for anyone to use

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  • $\begingroup$ @AbhigyanSaha Could you please delete the content of your answer except the remark "This space is now free for anyone to use" - so that other users can use this space without worry? $\endgroup$ – IV_ Jul 13 at 20:00
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image

For the general case a multistep-transformation, be it a cycle or not, of $N$ steps and the sum-of-exponents $S=A_1+A_2+...+A_N$ where $a_{k+1}={ 3a_k+1\over 2^{A_k} }$ and $A_k=\nu_2(3a_k+1)$ might be written as $$ a_{N+1}=T(a_1;[A_1,A_2,...,A_N]) $$ or - a so to say "canonical form" -: $$ a_{N+1}={3^N \over 2^S} a_1 + T(0;[A_1,A_2,...,A_N]) $$ The latter form I've once named "canonical form" because here $T()$ is free from any variable $a_1$ and the constant $0$ is inserted for all cases, and the $T()$-notation hides your explicite $\sum()$-notation of powers of products of $3$ and of $2$.
For example we can take the example $(N,S)=(3,5)$ and $E_N:=[A_1,A_2,A_3]=[1,2,2]$ then a first solution is via $$T(0;E_N)= {9+3\cdot 2 + 2^3\over 2^5}={23\over 32} $$ $$ a_4 = { 27\over 32} a_1 + {23\over 32} \\ \implies \\ a_1= 43 + k\cdot 2 \cdot 2^S \\ a_4= 37 + k\cdot 2 \cdot 3^N $$ Note, that the $+k \cdot 2 \cdot 2^S$-term (and following the $ \cdot 3^N$ term) occur by the modularity condition such that the rhs must be integral.


That notation shows modularity to the modulus $2 \cdot 2^S$: a second solution $b_1 \to b_4$ to such a transformation is in the modulus to $2 \cdot 2^S$; such that, for instance, $$ b_1 = a_1 + 2 \cdot 2^S$$ and then $$ b_{N+1}={3^N \over 2^S} b_1 + T(0;E_N) $$ $$ b_{N+1}={3^N \over 2^S} (a_1+2 \cdot 2^S) + T(0;E_N) $$ Then follows $$ b_{N+1}=a_{N+1}+ {3^N \over 2^S} (2 \cdot 2^S) \\ b_{N+1}=a_{N+1}+ 2 \cdot 3^N $$ So $$b_1 - a_1 = 2 \cdot 2^S \\ b_{N+1} - a_{N+1} = 2 \cdot 3^N $$ Another view, which is immediately derivable, is that $$ 2^S \cdot a_{N+1} - 3^N a_1 = 2^S \cdot T(0;E_N) \\ 2^S \cdot b_{N+1} - 3^N b_1 = 2^S \cdot T(0;E_N)$$


Of course we can as well subtract $2 \cdot 2^S$ from $a_1$ to get $c_1=a_1-2 \cdot 2^S$ and then $$ c_{N+1}=a_{N+1}- {3^N \over 2^S} (2 \cdot 2^S) = a_{N+1}- 2 \cdot 3^N $$ such that

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Question title: Can first order logic be extended to include infinite conjunctions?

Can first order logic (FOL) be extended in some way where infinite conjunctions are permissible? Specifically, can it be extended and still refute statements in a finite number of steps?

I would like this question to be answered using Tarskian semantics, where names refer to objects external to the logic.



Self answer:

Suppose that an infinite conjunction is legal syntax in a first order theory. In first order logic, we are free to negate any formula we can reason about, so a theory which can reason about infinite disjunctions will necessarily reason about their negation.

\begin{equation}\tag{1} \{P_a,P_b,\dots\} \end{equation}

\begin{equation}\tag{2} \lnot(P_a\lor P_b\lor\dots) \end{equation}

Consider a first order theory which contains the infinite set of formulae $(1)$ and the negation of an infinite disjunction $(2)$. Any interpretation which satisfies $(2)$ will necessarily make the set of formulae $(1)$ unsatisfiable. However, every finite subset of formulae will be satisfiable, which by compactness means that the whole thing is.

This is a contradiction, so it must not be the case that an infinite conjunction can be made legal syntax in a first order theory.



Question title: What did Hilary Putnam mean by this following quote of his?

In Putnam's paper "The logic of quantum mechanics", he states:

There is nothing really answering to the Copenhagen idea that two kinds of description (classical and quantum mechanical) must always be used for the description of physical reality (one kind for the ways to be used for the 'observer' and the other for the 'system'), nor to the idea that measurement changes what is measured in a indescribable way (or even brings into existence), nor to the 'quantum potential', 'pilot waves', ect. of the hidden variable theorists. These no more than Reichenbach's 'universal forces'.

Precisely what did he mean by this? In particular, what did he mean by the "idea that measurement changes what is measured in a indescribable way (or even brings into existence)"? How does one clearly define the the issue raised in the first point, and has it been resolved today?

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From this you have bounds for your $r(a_1)$-parameter: $$ r(a_1) = { E \over N+E } = 1-{ 1 \over E/N+1 } \implies $$

$$E_{min}=1-\{N \gamma_1 \} + N \gamma_1 \\ E_{max}=N $$

$ \displaystyle \lim_{N \to \infty} r(a_1)_{E_{min}} = 1-{ 1 \over (1-\{N \gamma_1 \} + N \gamma_1 )/N+1 } = 1-\frac 1{\gamma_1+1} = 1- \log_3(2) = \log_3(1.5) $

$ \displaystyle \lim_{N \to \infty} r(a_1)_{E_{max}} = { E_{max} \over N + E_{max} } = { N \over 2N } = \frac 12 $

$$

{\lceil N \cdot \gamma_1 \rceil \over N+E} \le \frac E{N+E} = r(a_1) \le \frac N{N+E} $$


Let us assume the "Syracuse"-style notation of the Collatz-iteration $$ a_{k+1}= {3a_k+1\over 2^{A_1}} \qquad a_k \small \text{ from the odd integers} $$ and for a $N$-fold iterated transformation the short, vectorial, notation: $$ a_{N+1}=T(a_1;[A_1,A_2,...,A_N]) $$ So let $N$ denote the (N)umber of steps $3x+1$ and $S$ denote the (S)um of the exponents $A_k$, which is also the number of $x/2$-steps.


Then, to convert this into the version of $(3x+1)/2$ and $x/2$ -stepping, we introduce $E$ the number of even steps without the $(3x+1)/2$ steps, so $E=S-N$.

With that, I understand your $r(a_1)$ as $r(a_1)=E/(N+E) = (S-N)/S = 1- N/S$.


We can observe,

  • that the trival cycle $1 = T(1;[2,2,2,...2])$ to any length $N$, has the values $S=2N$ and $E=N$ and $r(a_1)= 1-N/S = 1-N/(2N)=1/2 $
  • that the first cycle in the negative numbers $-1= T(-1;[1,1,1,1,...,1])$ to any length $N$, has $S=N$, $E=0$, and $r(a_1)= 0$
  • that the second cycle in the negative numbers $-5=T(-5;[1,2,1,2,1,2,...,1,2])$ to any even length $N=2n$, has $S=3n$, $E=n$ and $r(a_1) = E/(N+E)= n/(3n) = 1/3 $

Now to have a cycle of any length, and other than $T(a_1;[2,2,2,...,2])$ we can use the well known multiplication-formula for the $N$ members of an expected cycle $a_k$ ($k=1..N$) $$ a_2 \cdot a_3 \cdot ... \cdot a_N \cdot a_1 = \left({3a_1+1\over 2^{A_1}}\right) \left({3a_2+1\over 2^{A_2}}\right) \cdots \left({3a_N+1\over 2^{A_N}}\right)$$ This can be rearranged to $$ 2^S = 2^{A_1+A_2+...A_N} =\left(3+{1\over a_1}\right) \left(3+{1\over a_2}\right) \cdots \left(3+{1\over a_N}\right)$$ We see, that the rhs must be at least as large as the smallest perfect power of $2$ larger than $3^N$, but at most as $4^N = 2^{2N}$ so we get for the lhs (writing $\gamma=\log_2(3)$, and further below $\gamma_1=\log_2(3)-1$): $$ 2^{\lceil N \cdot \gamma \rceil} \le 2^S \le 2^{2N} $$ which in terms of $S$ means $$ \lceil N \cdot \gamma \rceil \le S \le 2N \qquad \text{where } S \in \mathbb N^+ $$ and in terms of $E$ instead $$ \lceil N \cdot \gamma \rceil -N =\lceil N \cdot \gamma_1 \rceil \le E \le N $$ From this you have bounds for your $r(a_1)$-parameter: $$ r(a_1) = { E \over N+E } \implies \\ {\lceil N \cdot \gamma_1 \rceil \over N+E} \le \frac E{N+E} = r(a_1) \le \frac N{N+E} $$ Well, this formula, in which $E$ must be evaluated after $N$ is given and might be checked for calculation of $r(a_1)$ between $\lceil N \gamma_1 \rceil$ and $N$, looks not very nice to me, so I'd reconsider the choice for the ratio-parameter $r()$.

FREE REAL ESTATE $$ {}{}{}{}{} $$

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$$=\underbrace{4n^2(n-1)(n+1)}_{\equiv ~0~(\text{mod}~~ 48)}+\underbrace{(n-1)n(n+1)(n+2)}_{\equiv ~0~(\text{mod}~ 24)}$$

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Feel free to improve my post:

(This post is an attempted answer to the question: Has research involving the generalization of the arithmetic mean been done for the following?)

Defining First Measure

If $S\subseteq A$:

  • $\ell$ is the length of an interval
  • $\left(I_{n}\right)_{j=1}^{n}$ and $\left(J_{k} \right)_{k=1}^{m}$, for $m\in\mathbb{N}$, are a sequence of open intervals where $\left(I_1\right)=\left(I_2\right)=...=\left(I_{n}\right)=g\in\mathbb{R}^{+}$, $\;\ell(J_1)=...=\ell(J_m)=g\in\mathbb{R}^{+}$ and the infimum in the equations below are taken over all possible $I_j$ and $J_k$

\begin{align} \mathcal{M}(g,S)=\begin{cases} g\cdot\inf\left\{m\in\mathbb{N}: {S^{\prime}\text{ is countable}, \; \left(S\setminus S^{\prime}\right)\subseteq\bigcup\limits_{k=1}^{n}I_{n}}\right\} & A \text{ is uncountable}\\ g\cdot\inf\left\{m\in\mathbb{N}: {S\subseteq\bigcup\limits_{k=1}^{n}I_{n}}\right\} & A \text{ is countable} \end{cases} \end{align}

and

$$\mathcal{N}(S,A)=\lim\limits_{g\to 0}\frac{\mathcal{M}\left(g,S\right)}{\mathcal{M}\left(g,A\right)}$$

then if

\begin{align} \mathcal{O}(g,S)= \small{g\cdot\inf\left\{m\in\mathbb{N}: {S_{j}\subseteq S, \; \mathcal{N}(S_{j},A)=0,\;\bigcup\limits_{j=1}^{\infty}{S}_{j}=S^{\prime},\;\left(S\setminus S^{\prime}\right)\subseteq\bigcup\limits_{k=1}^{m}J_{k}}\right\}} \end{align}

(I added $\mathcal{O}(g,S)$ so when $A$ is an intervals of finite lengths, $\mathcal{O}(g,S)$ gives the Lebesgue Measure of $S$ divided by the Lebesgue Measure of $A$.)

therefore the outer measure is

$$\mathcal{P}^{*}(S,A)=\lim\limits_{g\to 0}\frac{\mathcal{O}(g,S)}{\mathcal{O}(g,A)}$$

And the inner measure is $\mathcal{P}_{*}(S,A)=1-\mathcal{P}^{*}(A\setminus S,A)$, which means measure $\mathcal{P}(S,A)$ exists when $\mathcal{P}^{*}(S,A)=\mathcal{P}_{*}(S,A)$.

Now note $S$ and $A$, when $\mathcal{P}(S,A)$ is undefined, can also have a measure when $S$ and $A$ have a algebraic notation written in terms of subsets of $\mathbb{R}$.

Defining Algebraic Form

For arbitrary set $A$, if

  • $p_{1,1},...,p_{w,q_{w}}\subseteq\mathbb{R}$ and $z_{1,1},...,z_{w,v_w}\subseteq\mathbb{R}$ (For example the subsets can be $\mathbb{Z}$, the Cantor Set, or finite length intervals and:)
  • $w=[1,...,l\in\mathbb{N}]$
  • $T_{1,1},...,T_{l,w}:p_{1,1,1} \times...\times p_{l,w,q_{l,w}}\times z_{1,1},...,z_{w,v_w}\to\mathbb{R}$

then the algebraic form of $A$, or $\alpha$ is:

\begin{align} & \alpha=\bigcup_{w=1}^{l}\bigcap\limits_{u_{w,1}\in z_{w,1}}...\bigcap\limits_{u_{w,v_{w}}\in z_{w,v_{w}}}\bigcup\limits_{s_{w,1}\in p_{w,1}}...\bigcup\limits_{s_{w,q_{w}}\in p_{w,q_{w}}}\left\{T_{l,w}\left(s_{w,1},...,s_{w,q_{w}},u_{w,1},...,u_{w,v_{w}}\right)\right\} \end{align}

To illustrate, when set $A$ has an algebraic form $\alpha_1=\bigcup\limits_{m\in\mathbb{Z}}\bigcup\limits_{n\in\mathbb{Z}}\left\{\frac{m}{8n+4}\right\}$ where

  • $l\in \left(k_1=\left\{1\right\}\right)$
  • $w\in \left(k_2=\left\{1\right\}\right)$
  • $s_{1,1,1}=m$ and $p_{1,1,1}=\mathbb{Z}$
  • $s_{1,1,2}=n$ and $p_{1,1,2}=\mathbb{Z}$

other algebraic forms must be equivelant to $\alpha_{1}$. For example, another algebraic form of $A$ is $\alpha_2=\bigcup\limits_{m_1\in\mathbb{Z}}\bigcup\limits_{m_2\in\mathbb{Z}}\left\{\frac{2m_1+1}{8m_2+4}\right\}\cup\bigcup\limits_{m_3\in\mathbb{Z}}\bigcup\limits_{m_4\in\mathbb{Z}}\left\{\frac{m_{3}}{12m_4+6}\right\}$ where

  • $w\in k_1=\left\{1\right\}$
  • $l\in k_2=\left\{1,2\right\}$
  • $s_{1,1,1}=m_1$ and $p_{1,1,1}=\mathbb{Z}$
  • $s_{1,1,2}=m_2$ and $p_{1,1,2}=\mathbb{Z}$
  • $s_{1,2,1}=m_3$ and $p_{1,2,1}=\mathbb{Z}$
  • $s_{1,2,2}=m_4$ and $p_{1,2,2}=\mathbb{Z}$

Since $\alpha_1=\alpha_2$, $\alpha_2$ is an algebraic form of $A$.

Moreover, note we can define the set of all algebraic forms of $A$ as $\mathcal{A}(A)$ where $\alpha\in\mathcal{A}(A)$. For example:

$$\mathcal{A}(\mathbb{Q})=\left\{\left\{\frac{m}{n}:m,n\in\mathbb{Z}\right\},\left\{\frac{m}{2n}:m,n\in\mathbb{Z}\right\},...,\left\{\frac{m}{n^2}:m,n\in\mathbb{Z}\right\},...\right\}$$

And an $\alpha\in\mathcal{A}\left(\mathbb{Q}\right)$ is $\alpha=\left\{\frac{m}{n}:m,n\in\mathbb{Z}\right\}$

Therefore, we can now define a "Folner-like" Sequence inorder to define a measure

Defining New Measure

Going back to $\alpha$. If

  • $\Phi_{1,w}=\left\{1,...,q_{w}\in\mathbb{N}\right\}$
  • $\Phi_{2,w}=\left\{1,...,v_{w}\in\mathbb{N}\right\}$
  • $i_{1,w},r_{w}\in\Phi_{{1,w}}$
  • $i_{2,w}\in\Phi_{2,w}$
  • $t_{l,w}\in\Phi_{l,w}\setminus\{r_{l,w}\}$

then $\alpha$ becomes

\begin{alignat}{2} &\mathcal{F}(w,\alpha,\psi,r_{l,w},\epsilon,\omega)= \\ &\bigcup\limits_{i_{l,w}\in\Phi_{l,w}}\bigcap\limits_{\large{u_{i_{w}}\in z_{i_{w}}}}\bigcup\limits_{\large{s_{w,i_{w}}\in p_{i_{w}}}}&&\left\{T_{w}\left(s_{w,1},...,s_{w,q_{w}},u_{w,1},...,u_{w,v_{w}}\right):\left|s_{w,t_{w}}\right|\le \omega, |u_{w,i_{w}}|\le\eta\right. \\ & &&\left. \left|T_{w}\left(s_{w,1},...,s_{w,q_{w}},u_{w,1},...,u_{w,v_{w}}\right)-\psi \right|\le\epsilon \right\} \end{alignat}

Note if all $p_{w}$ are subsets of $\mathbb{Z}$ and $\omega\to\infty$ for every $\eta$ when $\eta\to\infty$, then each $s_{r_{w}}$ would give different Folner Sequences. However, this forces us to choose a particular $r_{w}$.

Also note we add $\left|T_{w}\left(s_{w,1},...,s_{w,q_{w}},u_{w,1},...,u_{w,v_{w}}\right)-\psi \right|\le\epsilon$ so when $A$ is dense in an interval and $S\subseteq A$ is dense almost nowhere, the density of $S$ using Folner-esque sequences is always zero.

To avoid choosing between different $r_{w}$, take the intersection of all of them and take the Folner Sequences of the $l$ and $w$'s which helps obtain a measure that matches $\mathcal{P}$ when $\mathcal{P}(S,A)$ is defined.

\begin{align} \alpha_{\mathcal{F}}\left(\alpha,\psi,\epsilon,\omega,\eta\right)=\bigcup\limits_{w=1}^{l}\bigcap\limits_{r_{w}\in\Phi_{w}}\mathcal{F}\left(w,\alpha,\psi,r_{w},\epsilon,\omega,\eta\right) \end{align}

(Note we take $k_1$ less than $\eta$ so when $A=\mathbb{Q}$ one can derive a unique Folner Sequence which gives a unique measure.)

However, there are multiple algebraic forms of $\alpha$ in $A$ which prevent us from obtaining such a measure. Therefore we add additional criteria:

If $\alpha_1,\alpha_2\in\mathcal{A}(A)$ we solve

$$\beta=\inf\left\{\lim\limits_{\epsilon\to 0}\lim\limits_{\eta\to\infty}\lim\limits_{\omega\to\infty}\mathcal{P}\left(\alpha_{\mathcal{F}}(\alpha_1,\psi,\epsilon,\omega,\eta),\alpha_{\mathcal{F}}(\alpha_1\cup\alpha_2,\psi,\epsilon,\omega,\eta)\right)\right\}$$

and take the set of all $\alpha_1$ that gives $\beta$.

$$\mathcal{A}_{\beta}(A)= \left\{\alpha_1:\left(\forall\alpha_{2}\right)\left(\lim\limits_{\epsilon\to 0}\lim\limits_{\eta\to\infty}\lim\limits_{\omega\to\infty}\mathcal{P}\left(\alpha_{\mathcal{F}}(\alpha_1,\psi,\epsilon,\omega,\eta),\alpha_{\mathcal{F}}(\alpha_1\cup\alpha_2,\psi,\epsilon,\omega,\eta)\right)=\beta\right)\right\}$$

For example, when $A=\mathbb{Q}$, the Folner Sequence

$$\lim\limits_{\epsilon\to 0}\lim\limits_{\eta\to\infty}\lim\limits_{\omega\to\infty}\bigcap_{k\in\mathbb{N},|k|\le\eta}\left(\bigcup\limits_{n\in\mathbb{Z},|n|\le\omega}\bigcup_{m\in\mathbb{Z}}\left\{\frac{m}{n^{k}}:\left|\frac{m}{n^k}-\psi\right|\le\epsilon\right\}\cup\bigcup\limits_{n\in\mathbb{Z}}\bigcup_{m\in\mathbb{Z},|m|\le\omega}\left\{\frac{m}{n^{k}}:\left|\frac{m}{n^k}-\psi\right|\le\epsilon\right\}\right)$$

Should be in $\mathcal{A}_{\beta}(A)$

(We can picture the triple-limit as follows: For every $\epsilon>0$ and $\eta>0$, take $\omega$ as $\omega\to\infty$ then for every $\epsilon>0$ take $\eta$ as $\eta\to\infty$. As $\epsilon$ is evaluated for smaller values, the limit should converge.

Then take the union of all $\alpha$ in $\mathcal{A}_{\beta}(A)$ to get:

$A_{\mathcal{F}}\left(\psi,\epsilon,\omega\right)=\bigcup\limits_{\alpha\in\mathcal{A}_{\beta}(A)}\alpha_{\mathcal{F}}(\alpha,\psi,\epsilon,\omega)$

(Presumably), we are not only dealing with countable sets. Hence instead of taking the density of $S\subseteq A$ using Folner Sequences:

\begin{align} \lim\limits_{\epsilon\to 0}\lim\limits_{\omega\to\infty}\frac{\left|S\cap A_{\mathcal{F}}\left(\psi,\epsilon,\omega\right)\right|}{\left|A_{\mathcal{F}}\left(\psi,\epsilon,\omega\right)\right|} \end{align}

We take

\begin{align} d(S,\psi)=\lim\limits_{\epsilon\to 0}\lim\limits_{\omega\to\infty}\mathcal{P}\left(S\cap A_{\mathcal{F}}\left(\psi,\epsilon,\omega\right),A_{\mathcal{F}}\left(\psi,\epsilon,\omega\right)\right) \end{align}

Now, even if $d(S,\psi)$ exists, it only takes the density over $[\psi-\epsilon,\psi+\epsilon]$. Therefore, we broaden the density to all $A$ by taking the weighted average of $d(S,\psi)$ using $\mathcal{P}$. In formal terms:

Divide $[0,1]$ ($0\le d(S,\psi)\le 1$) into partitions $0=c_1\le \cdot\cdot\cdot\le c_i \le \cdot\cdot\cdot\le c_u=1$ where if $x_i\in[c_{i-1},c_i]$ and $C_i=[c_{i-1},c_i]$, then our final outer measure is

\begin{align} & \mathcal{U}^{*}(S,A)= \lim\limits_{u\to\infty}\sum\limits_{i=1}^{u}x_i\cdot\mathcal{P}^{*}\left(\left\{\psi:d\left(S,\psi\right)=C_i\right\},A\right) \end{align}

And our final inner measure is

\begin{align} & \mathcal{U}_{*}(S,A)= 1-\lim\limits_{u\to\infty}\sum\limits_{c=1}^{u}x_i\cdot\mathcal{P}^{*}\left(\left\{\psi:d\left(A\setminus S,\psi\right)=C_i\right\},A\right) \end{align}

Therefore, $\mathcal{U}(S,A)$ exists when:

$$\mathcal{U}^{*}(S,A)=\mathcal{U}_{*}(S,A) $$

Generalizing the Measure

We can generalize our measure to $\mu$:

\begin{equation} \mu(S,A)=\begin{cases} \mathcal{P}(S,A) & \mathcal{P}^{*}(S,A)=\mathcal{P}_{*}(S,A) \\ \mathcal{U}(S,A) & \mathcal{P}^{*}(S,A)\neq\mathcal{P}_{*}(S,A), \; \mathcal{U}^{*}(S,A)=\mathcal{U}_{*}(S,A) \end{cases} \end{equation}

Defining The Average

To define the average, first split $[a,b]=[\min(A),\max(A)]$ into sub-intervals using partitions $x_i$

$$a= x_0 \le \dots \le x_i \le \dots \le x_r =b$$

Here, if $1\le i \le r$, then $[x_{i-1},x_i]$ are sub-intervals of $[a,b]$.

For every $i$, choose a $v_i\in A\cap[x_{i-1}, x_{i}]$ and define $A_i=A\cap[x_{i-1},x_i]$. Next we define set $P$, such that $i \in P\subseteq\left\{1,...,r\in\mathbb{N}\right\}$ when $A\cap[x_{i-1},x_i]\neq\emptyset$. This gives

\begin{align} & \lim_{r\to\infty}\sum_{i\in P} f(v_i) \times \mu(A_i,A) \end{align}

(This may look tedious but we can use shortcuts to simplify the sum. If I write it out the post will be too long.)

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$$\begin{align}\lim_{(x,y) \to (0,0)} \dfrac{1 - \cos x}{x+y} &=\lim_{(x,y) \to (0,0)} \dfrac{1 - \cos (x+y-y)}{x+y} \\ &\,=\lim_{(x,y) \to (0,0)} \dfrac{\,1 - \cos (x+y)\cos y-\sin (x+y)\sin x\,}{x+y} \\ &=\lim_{(x,y) \to (0,0)} \dfrac{1 - \cos (x+y)}{x+y}\\ &= 0.\end{align}$$

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(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.

  1. 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).

  2. There are in 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.

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

  4. 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} $x_{i^2}$.

  5. 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 $\lfloor x \rfloor$.

  6. 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 $\idotsint$.

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

  8. Different Fonts

  • $\mathbb{C}$, $\mathbb{R}$, $\mathbb{Q}$, $\mathbb{Z}$,
  • $\mathcal{CHNQRZ}$
  • $\mathscr{CHNQRZ}$
  • $\mathfrak{CHNQRZ}$.
  1. 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.

  2. Some special functions such as $$\sin,\cos, \tan, \cot, \arcsin,\arccos, \arctan$$ $$\sinh,\cosh, \tanh, \coth, \arsinh,\arccos, \sec$$

$$\ln, \log,\lg,\log_2,\log_{16}$$ 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) $\operatorname{foo}(x)$.

  1. 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.

  1. 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.

  1. 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.

  2. 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

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This box is free for everyone to use

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This answer is free for anyone to use

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  1. A Pythagorean Triple (PT) is a triple $(a,b,c)$ of positive integers for which $a^2+b^2 = c^2$. Pythagorean triples can be parameterized by positive integers $s$ and $t$ by $$(a,b,c) = (2st, s^2-t^2, s^2+t^2)$$

  2. An arithmetic progression of squares (APS) is a triple $(a^2, b^2, c^2)$ for which $b^2 - a^2 = c^2 - b^2$. This is equivalent to $a^2+c^2 = 2b^2$. The value of $b^2 - a^2 = c^2 - b^2$ is called the common difference.

  3. If $(a,b,c)$ is a PT, then $$((b-a)^2, c^2, (b+a)^2)$$ is an APS. The common difference is $2ab$.

  4. Since $(2st, s^2-t^2, s^2+t^2)$ is a PT, then $$((s^2-t^2 - 2st)^2, (s^2+t^2)^2, (s^2-t^2+2st)^2)$$ is an APS. The common difference is $4st(s^2-t^2)$


We are given the APS $(205^2, 425^2, 565^2)$ with a common difference of $138600$. The corresponding PT is $(385, 180, 425)$

The common difference is $425^2 -

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Notice that removing any edge $e\in T\setminus T'$, from $T$ itself, will result in two connected components: $V,U$. Since $T'$ is a tree, it is connected. Therefore, there must be some edge $e'\in T'$ between the components $V$ and $U$.

I claim, the three following things:

  1. $e'\notin T$
  2. By removing $e$ from $T$ and adding instead $e'$, we get a tree $\hat T$
  3. The tree $\hat T$ we get is a minimum spanning tree

Lets prove those statements one by one:

  1. Assume towards contradiction that $e'\in T$. Then, since $e\notin T'$ but $e'\in T'$ we can conclude that $e\neq e'$. Therefore, removing $e$ from $T$ will still keep $e'$ in $T$, and thus $V$ and $U$ will still be connected - contradiction.

  2. both $V$ and $U$ must be connected, since $T$ was. Adding the edge $e'$ will connect $V$ and $U$, hence the resulting graph must be connected. Additionally, the number of edges in the resulting graph is $|T|-1+1=|T|=n-1$ where $n$ is the number of vertices. Thus, the resulting graph must be a tree.

  3. This is the trickiest one to prove. Denote by $V',U'$ the subgraph of $T'$ on the vertices of $V,U$. It can be easily shown that $w(V')=w(V)$ ($w(\cdot)$ is the weight of the subgraph), since otherwise either $T$ or $T'$ would not be an MST. The same can be applied for $w(U')=w(U)$. Now, notice that $T'$ is a tree, and thus there is only one single edge between $V'$ and $U'$ that is also in $T'$, and this edge is $e'$ (if there were two edges, the graph would contain a cycle). Clearly, \begin{equation}{w(V)+w(U)+w(e)=w(T)=w(T')=w(V')+w(U')+w(e')}\end{equation} Substitute into this equation the fact that $w(V)=w(V')$ and $w(U)=w(U')$, and subtruct from both sides of the equation them. We finally get that $w(e)=w(e')$. Now, by the definition of $\hat T$, we have $w(\hat T)=w(V)+w(U)+w(e')=w(V)+w(U)+w(e)=w(T)$. And since $w(T)$ is minimal, then also $w(\hat T)$ is. Thus $\hat T$ is a minimum spanning tree.

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$j=5^{(5^2)}$ $j-Z(j!)=j-($ $(j-1)-Z((j-1)!)=(j-1)-Z((j-1)!)$

$(5^1-5^0) - Z((5-5^0)!) = 4- 0 = 4$ $5^1 - Z(5!) = 5- 1 = 4$

$5^5 - Z(5^5!) = 5^5- (5^4 + 5^3 + 5^2 + 5 + 1)$ $(5^5-5)- Z((5^5-5)!) = (5^5-5)- (5^4-1 + 5^3-1 + 5^2-1 + 5 -1+ 1-1)=5^5- (5^4 + 5^3 + 5^2 + 5 + 1)$ $(5^5-5^1-5^0)- Z((5^5-5^1-5^0)!) = (5^5-5-1)- (5^4-2 + 5^3-1 + 5^2-1 + 5 -1+ 1-1)=5^5- (5^4 + 5^3 + 5^2 + 5 + 1)$

$(5^{25}-5) - Z((5^{25}-5)!) = (5^{25}-5)- (5^{24}-1 + 5 ^{23}-1 + 5^{22}-1 + ... +(5^4-1) + (5^3-1) + (5^2-1) + (5-1) + (1-1))$ $(5^{25}-5^1-5^0) - Z((5^{25}-6)!) = (5^{25}-5 -1 )- (5^{24}-2 + 5 ^{23}-1 + 5^{22}-1 + ... +(5^4-1) + (5^3-1) + (5^2-1) + (5-1) + (1-1))$ $(5^{25}+5^2-5^1) - Z((5^{25}+24)!) = (5^{25}+24 )- (5^{24}+4 + 5 ^{23} + 5^{22} + ... +5^4 + 5^3 + 5^2 + 5 + 1)$ $(5^{25}+5^2+5^1) - Z((5^{25}+26)!) = (5^{25}+26 )- (5^{24}+5 + 5 ^{23} + 1 + 5^{22} + ... +5^4 + 5^3 + 5^2 + 5 + 1)$

2: $5^1-5^0, 5^1\mid\mid 5^1-5^0$ First time a number appearing two times: $5^1-5^0$ $(5^1-5^0+4)$

3: $5^5-5^1-5^0, 5^5-5^1, 5^5 \mid\mid 5^5- (5^4 + 5^3 + 5^2 + 5 + 1)$ First time a number appearing three times: Two numbers: $5^5- (5^4 + 5^3 + 5^2 + 5 + 1) + 0$ $5^5- (5^4 + 5^3 + 5^2 + 5 + 1) + 4$

4: $5^{25}-5^1-5^0, 5^{25}-5, 5^{25}+5^2+5^1,5^{25}+5^2+5^1\mid\mid(5^{25}+26 )- (5^{24}+5 + 5 ^{23} + 1 + 5^{22} + ... +5^4 + 5^3 + 5^2 + 5 + 1)$ First time ??? a number appearing four times: Five numbers: $5^{25}- (5^{24}+ 5 ^{23}+ 5^{22} + ... +5^4 + 5^3 + 5^2 + 5 + 1)+0??????$ a mild speculation Is this a missing one? $5^{25}- (5^{24}+ 5 ^{23}+ 5^{22} + ... +5^4 + 5^3 + 5^2 + 5 + 1)+4$ $5^{25}- (5^{24}+ 5 ^{23}+ 5^{22} + ... +5^4 + 5^3 + 5^2 + 5 + 1)+8$ $5^{25}- (5^{24}+ 5 ^{23}+ 5^{22} + ... +5^4 + 5^3 + 5^2 + 5 + 1)+12$ $5^{25}- (5^{24}+ 5 ^{23}+ 5^{22} + ... +5^4 + 5^3 + 5^2 + 5 + 1)+16$ $5^{25}- (5^{24} + 5 ^{23} + 5^{22} + ... +5^4 + 5^3 + 5^2 + 5 + 1)+20$

Some number near $5^{125}-(5^{125}+ 5^{124}+ 5 ^{123}+ 5^{122} + ... +5^4 + 5^3 + 5^2 + 5 + 1)$ appears five times as generated by a number near $5^{125}-(5^{3}+ 5^{2}+ 5 ^{1} + 1)$

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