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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 Maybe someday we can get rid of this outdated crutch.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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I am still working on this. Please do not delete it until the end of August or at the least by June 5th! Thanks! – Tyma Gaidash (remark added by G. Helms due to comment of TymaGaidash)


I just realized you can have all four x’s. The other non inverse with all four x’s is 0. Must redo!!


Here is another integral over an function of four arguments, this is not the inverse with respect to all four arguments of the Generalized Regularized Incomplete Beta Function, but rather one argument, or integration bound based on its definition, explained below. Wolfram Alpha cannot approximate this value too well. Let: $$\mathscr {I_2}\mathrm{=\int_0^\frac 12I^{-1}_{(x,x)}(x,x)dx=\int_0^\frac12I^{-1}_{I_x(x,x)+x}(x,x)= \int_0^\frac12I^{-1}_{\left(\frac12-x,\frac12-x\right)}\left(\frac12-x,\frac12-x\right)dx= \int_0^\frac12I^{-1}_{I_{\frac12-x}\left(\frac12-x,\frac12-x\right)+\frac12-x}\left(\frac12-x,\frac12-x\right)=.37234...}$$

One identity with the argument change included is that $$\mathrm{I^{-1}_{(x,x)}(x,x)=I^{-1}_{I_x(x,x)+x}(x,x)\implies x=I_{\left(x, I^{-1}_{I_x(x,x)+x}(x,x)\right)}(x,x) \implies \frac{xΓ^2(x)}{Γ(2x)}=\int_x^{ I^{-1}_{I_x(x,x)+x}(x,x)}(t-t^2)^{x-1}dt}$$

which uses the (Inverse of the)(Generalized)Regularized Incomplete Beta Function. The link includes each of these combinations.

Here is a graph of the inverse and the visual representation of this constant:

enter image description here

enter image description here

Here is the reflected version with x$\to \frac12-\mathrm x$ and the function rewritten in terms of the three argument non-generalized form: enter image description here

This shows what would have to be done using the $DI$ method with the integrand being $D$ and 1 being $I$. The integrand derivative is here which seems like a bad idea on solving this problem as it is long and complicated because of the multivariable chain rule. Simply click “step-by-step solution” to see this principle in action. From this source, this is implied: $$\mathrm{x=I_{\left(x,I^{-1}_{(x,x)}(x,x)\right)}(x,x)=\frac{\int_x^ {I^{-1}_{(x,x)}(x,x)}(t-t^2)^{x-1}dt}{B(x,x)}\implies xB(x)= \int_x^{I^{-1}_{(x,x)}(x,x)}{I^{-1}_{(x,x)}(x,x)}(t(1-t))^{x-1}dt}$$

Beta(x,x)=B(x,x)=B(x) is many times defined as such which is the inspiration of the shortening of the function name here and for the subscript of 2 in the constant definition:

$$\pmb{Link!}$$

It can be applied that the Lagrange Inverse Formula cannot be used nor the formula for the definite integral of an inverse function because the generalized $I^{-1}$ function is only the inverse of one of the arguments and not the fully written out function in the source link as this one is multidimensional and this would make a complicated inverse function in 4 “argumented” space.

An evaluation in exact form not using the definition of the Riemann Sum nor any $n^{th}$ derivatives for the Taylor Series of this function nor approximation as these are trivial solutions. A more well defined evaluation is required. Please give me feedback and correct me!

The decimal precision here is ok, but here are some possible closed forms for this constant.

The integral of this function is known, but only with respect to the variable it is the inverse of. In this case it is the upper bound of the regularized incomplete beta function. This is also the right variable in the “subscripted” variables. However, this one is in terms of this inverse beta function and a hypergeometric function. Also, Even Wolfram Alpha cannot integrate this, but the same goes for the Sophomore’s Dream which it cannot integrate despite a simple series expansion making anyone be able to integrate using a gamma function. -Ty.Ga

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  • $\begingroup$ I am still working on this. Please do not delete it until the end of August or at the least by June 5th! Thanks! $\endgroup$ – Tyma Gaidash May 26 at 1:41
  • $\begingroup$ 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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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

*** This box is free for everyone to use ***

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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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I know that for a 0Ω wire of uniform cross sectional area, the potential difference across its ends is zero

Yes but only trivially so. Why should there be any potential difference across you resistor at all? Is it in a circuit? Is there a potential across it? Is there a current flowing through it? You haven't mentioned any of this. A $0$ ohm resistor lying diconnected from any sources can have whatever the potential we want at its ends depending on ambient electrical environment.

as no electric field is required to move charge with constant velocity (necessary to keep current constant throughout the circuit) as no resistive force is offered by the wire.

Of course there is no potential difference needed, you set the resistance to $0$! Your resistor isn't a resistor at all - its just a plain old wire.

Also on a more deeper note, the electric field isn't zero in the sense that you are reasoning. Think about it. If there is no field,anywhere in a wire, how does current flow through it at all? If you say charges move with uniform velocity, that is wrong. If you consider the simple Drude model, current is conducted in wires because of drift. Between collisions, on average, charges are accelerated by the field. In the absence of any field, they won't move with constant velocity but instead thermalize quite quickly. More on this below.

But what if the cross sectional area of that 0Ω wire increases on moving? Do we still say potential difference across its ends need to be zero?

Yes, when in a circuit, definitely. Its dimensions don't matter. Its material doesn't matter. Its temperature doesn't matter. Once something has $0$ resistance, its cannot develop a potential difference across it with only a constant current passing through it.

Yes from Ohm's law it needs to be zero,

No. Ohm's law isn't valid here as $R=0$. The current going through a $0$ ohm resistor is independent of the potential across it, which as you yourself said, is always $0$. The current in the resistor is determined by other elements of the circuit.

But if charge continues to move with constant velocity and area increases on moving ahead then current through wire will increase. $I=neAV_d$

No. The formula you use for your reasoning uses $V_d$. This is the average drift velocity that I alluded to earlier. Since you have claimed, that there is no electric field, $V_d=0$ which sets $I=0$. This should give you pause. There's no flow of current, so what charges moving with constant velocity are you talking about?. As remarked prior, in the absence of field, charges don't move with uniform velocity, they thermalize - the average velocity becomes zero and so does the current.

So why have you landed in this quagmire? That is because you are thinking about the degenerate case of a resistor - when its $0^1$. When such an element is put in a circuit with other non-zero resistors, voltage sources and electrical elements, a current is establsihed through it. However, since no potential can develope across it, how does the current flow through it?

The fallacy is to think there's no potential across it. True the finite extent of the resistor, the individual component that existed before being put in the circuit wasn't really capable of sustaining a potential difference. But when attached to wires on either end in a live circuit, it $becomes$ the wire. Again, a $0$ ohm resistor is a wire. So its no different from the wires it connects to. Lets follow what exists at the end of these wires.

The two ends of a wire in any live circuit ultimately end up at the two terminals of an effective potential source. Its the potential difference across these terminals that sets up a net electric field that drives the current through the circuit.

In the real world wires do carry current because they have a miniscule but non-zero resistance. But in theory they can't since each individual piece doesn't have any potential gradient across it. In theory, we use wires as convenience for connecting components in diagrams.


$^1$ there's another degenerate case: when its $\infty$.

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Could you review my edited question?$^1$

In your eqn. $2$, $P_C$ is known to be $P_a$ by Pascal's law. $P_C$ is therefore always constant i.e. $\forall$ values of $h$ or $v_a$.

Also, in your eqn. $3$, when you try to determine what $P_C$ should be, you did the following:

  1. set $v_a$ and therefore $h$ to zero.
  2. then $P_A$=$P_C$.

So when $v_a$ and $h$ are zero, the equality holds, but not otherwise. So when you ultimately solve $h$ for a non-zero $v_a$, you can't use the equality you just derived - $P_A$ must be treated as an unknown. Eqn. $4$ therefore never follows.

So let's say you set $P_C=P_a$. How do you solve your eqn. $3$? What is $P_A$? One must look to your eqn. $1$.Due to initial conditions, the RHS of your eqn. $1$, can be determined to be $P_a$.

But this is always $P_B$ right? So is $P_B$ always at atmospheric pressure? This doesn't make any sense.

It took me a while to understand, but this a deeper mistake. A conceptual one. The Bernoulli eqn., in my opinion can not be applied to points $A$ and $B$ i.e. $$P_A+\rho_a v_{a,A}^2+\rho_a g h_A \ne P_B+\rho_a v_{a,B}^2+\rho_a g h_B \tag {0}$$

such that $v_{a,A}$ is the air velocity at $A$ with which one blows.

Mathematically, this is reflected in the fact that while the LHS of eqn. $0$ is always $P_a$ because of the initial condition, the RHS approximated to just $P_B$ obviously isn't - there is definitely a pressure drop.

Physically its because the LHS and RHS of eqn. $0$ represent different flows - the horizontal cloud of blown air that moves over the straw and the air within the straw, resp. One is clearly blowing, while the other is stagnant (at equilibrium). These flows have $A$ as the common point. The Bernoulli equation is derived by assuming a flow element conserving its energy as it moves within the flow stream. This assumption doesn't hold when two different flows merge.


So here's a correct derivation

using your notation.(all heights measured wrt. $C$)

At $A$,

$$P_A+\rho_a g h_{A}+\rho_a v_{a,A}^2/2=const.=c \tag{1}$$

This equation is taking $A$ as part of the 'blow flow'. As the blown air moves over the straw's cross-section, there is hardly any change in its height. Hence eqn. $1$ reduces to

$$P_A+\rho_a v_{a,A}^2/2=const.=c' \tag{2}$$

and, as stated earlier, with the initial condition $v_{a,A}= 0,P_A=P_a$, becomes

$$P_A+\rho_a v_{a,A}^2/2=P_a \tag{3}$$

At $A$, the following is also satisfied

$$P_A+\rho_a g h_A+\rho_a v_{a,A}^2/2 =P_B+\rho_a g h_{B}+\rho_a v_{a,B}^2/2 \tag {4}$$

This eqn. treats $A$ and $B$ as part of the 'stagnant flow', the air column inside the straw. In this way, its consistent with the refutal in eqn. ($-1$).

Since, the air column is stagnant $v_{a,A},v_{a,B}=0$. Also since $\rho_a$ is pretty small to cause gauge pressure at $B$ we can ignore it, but lets keep it anyways. Eqn. $4$ then reduces to

$$P_A =P_B+\rho_a g (h-H) \tag {5}$$

where I have taken $h_A=H, h_B=h$.

From Pascal's law, $P_C=P_a$ therefore

$$P_C=P_B+\rho_w g h=P_a\tag{6}$$

Substituting $P_A$ from eqn. $3$ and $P_B$ from eqn. $6$ into eqn. $5$ and solving for $h$ gives $$ \begin{align} h&=\frac{\rho_a}{\rho_w-\rho_a}\left(\frac{v_{a,A}^2}{2g} -H\right)\\ &\approx\frac{\rho_a}{\rho_w}\frac{ v_{a,A}^2}{2g}\\ &\approx \frac{v_a^2}{20000} m \tag{in SI units} \end{align} $$

This expression seems too small to enable suction by blowing over a straw. I am not sure what the problem is. e.g. blowing at say $10\,ms^{-1}$ only generates $5\,mm$ of lift. Maybe using another straw to blow instead of blowing directly would increase the speed.


$^1$ I re-wrote the answer to stress more on your query - the contradiction - and also to avoid notational confusion, and some corrections.

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(Deutsch: MathJax: LaTeX Basic Tutorial und Referenz)

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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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Test newly available syntax using data from this question.

$n$ approximation solution
$1$ $2.84071 +1.69496 \,i$ $2.84550 +1.68429 \,i$
$2$ $3.34357 +1.44741 \,i$ $3.35044 +1.43963 \,i$
$3$ $3.64315 +1.33808 \,i$ $3.65025 +1.33195 \,i$
$4$ $3.85317 +1.27397 \,i$ $3.86019 +1.26882 \,i$
$5$ $4.01366 +1.23065 \,i$ $4.02051 +1.22614 \,i$
$6$ $4.14297 +1.19882 \,i$ $4.14966 +1.19475 \,i$
$7$ $4.25095 +1.17409 \,i$ $4.25749 +1.17037 \,i$
$8$ $4.34346 +1.15414 \,i$ $4.34987 +1.15068 \,i$
$9$ $4.42428 +1.13756 \,i$ $4.43056 +1.13432 \,i$
$10$ $4.49595 +1.12348 \,i$ $4.50212 +1.12041 \,i$
$20$ $4.95332 +1.04533 \,i$ $4.95880 +1.04314 \,i$
$30$ $5.21027 +1.00868 \,i$ $5.21539 +1.00685 \,i$
$40$ $5.38829 +0.98573 \,i$ $5.39318 +0.98411 \,i$
$50$ $5.52410 +0.96940 \,i$ $5.52881 +0.96791 \,i$
$60$ $5.63366 +0.95690 \,i$ $5.63824 +0.95551 \,i$
$70$ $5.72535 +0.94687 \,i$ $5.72982 +0.94556 \,i$
$80$ $5.80411 +0.93855 \,i$ $5.80849 +0.93730 \,i$
$90$ $5.87307 +0.93147 \,i$ $5.87738 +0.93027 \,i$
$100$ $5.93438 +0.92534 \,i$ $5.93862 +0.92418 \,i$
$200$ $6.32917 +0.88910 \,i$ $6.33302 +0.88818 \,i$
$300$ $6.55383 +0.87067 \,i$ $6.55748 +0.86985 \,i$
$400$ $6.71067 +0.85862 \,i$ $6.71420 +0.85787 \,i$
$500$ $6.83097 +0.84980 \,i$ $6.83439 +0.84908 \,i$
$600$ $6.92840 +0.84289 \,i$ $6.93176 +0.84222 \,i$
$700$ $7.01021 +0.83726 \,i$ $7.01351 +0.83661 \,i$
$800$ $7.08066 +0.83252 \,i$ $7.08391 +0.83189 \,i$
$900$ $7.14249 +0.82845 \,i$ $7.14570 +0.827838 \,i$
$1000$ $7.19756 +0.82489 \,i$ $7.20073 +0.824289 \,i$
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$$\left{\align{1 & 2 \\ 3 & 4}\right.$$

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