Sandbox for drafts of long, complex posts

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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.

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Answer 1

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Answer 2

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Answer 3

Existence of a one-step-cycle: $$ 2^{S_2} 3^{S_3} = \left(5+ \frac1{a_1} \right) \text{ in } [5,6] \\ \implies (S_2,S_3)=(1,1) \implies a_1=1 $$ Existence of a two-step-cycle: $$ 2^{S_2} 3^{S_3} = \left(5+ \frac1{a_1} \right) \left(5+ \frac1{a_2} \right)\\ \implies \text{ lhs in } [25,36] \implies \text { solutions for lhs } \in \{27,32,36 \} \\ \to (a) \qquad 2 = 5(\frac1{a_1}+\frac1{a_2})+ \frac1 {a_1a_2} \\ 4a_m^2 - 20a_m +25 = + 27\\ a_m = (\sqrt{27}+5)/2 $$ Existence of a three-step-cycle: $$ 2^{S_2} 3^{S_3} = \left(5+ \frac1{a_1} \right) \left(5+ \frac1{a_2} \right) \left(5+ \frac1{a_3} \right) \\ \text{ rhs in } (125,216) \\ \text{ lhs } \in \{128,144,162,192,216 \} \\ $$

$$ 2^{S_2} 3^{S_3} = \left(5+ \frac1{a_1} \right) \left(5+ \frac1{a_1} \right) \cdots \left(5+ \frac1{a_N} \right) $$

Answer 4

Euler's identity, $e^{i\pi}+1=0$, is a special case of Euler's formula: $$ e^{i\theta} = \cos(\theta) + i\sin(\theta) \, . $$ If you plug in $\theta=\pi$ into the above equation, then it gives us $e^{i\pi}$. So to understand Euler's identity, it is first necessary to understand Euler's formula. Here is an intuitive explanation. Consider a particle moving around the complex plane. After $t$ seconds, it occupies the position $(\Re(e^{it}),\Im(e^{it}))$. Since velocity is the derivative of position, we find that the velocity of the particle is $$ \frac{d(e^{it})}{dt}=ie^{it} \, . $$ Written another way, $$ d(e^{it}) = ie^{it} dt \, . $$ This statement can be unpacked even further. Say after $t$ seconds, the particle occupies the position $(x,y)$. Then, after a further $\Delta t$ seconds have elapsed, the changes to the real and imaginary components of the particle are $\Delta x$ and $\Delta y$ respectively. What the formula $d(e^{it}) = ie^{it} dt$ means is that $$ (\Delta x,\Delta y) \approx i(x,y) \cdot \Delta t \, . $$ In other words, the changes to the real and imaginary components of the particle should be roughly proportional to the time elapsed, with $i(x,y)$ being the proportionality constant. Simplifying the formula further, it becomes $$ (\Delta x,\Delta y) \approx \Delta t(-y,x) \, . $$ And since $$ \text{Change in position} = \text{Velocity} \cdot \text{Time} \, , $$ the velocity of the particle is $(-y,x)$ units per second. This can also be understood graphically:

In this example, $t=0.1, x=0.9950$, and $y=0.0998$. After $0.05$ seconds, $\Delta x = -0.0062$ and $\Delta y = 0.0496$. If we substitute these values into the equation $(\Delta x,\Delta y) \approx \Delta t(-y,x)$, we get $$ (-0.0062,0.0496) \approx 0.05(-0.0998,0.9950) $$ which is not a bad approximation in the least.

In the above diagram, I could not help but give away the path that the particle traces. It looks like the motion of the particle is circular, and this can be verified if we solve the differential equation $$ \frac{dy}{dx} = -\frac{x}{y} \, . $$ Again, it is important to recognise what this differential equation represents. In this case, we are comparing the change in $y$ with the change in $x$ of the particle (rather than comparing the change in both $x$ and $y$ with the change in $t$). Recall that at the point $(x,y)$, the instantaneous velocity of the particle is $(-y,x)$: $$ \frac{d(e^{it})}{dt}\biggr|_{(x,y)} = (-y,x) \, . $$ This means that the derivative of $x$, with respect to $t$, is $$ \frac{dx}{dt} = -y \, . $$ Furthermore, the derivative of $y$ with respect to $t$ is $$ \frac{dy}{dt} = x \, . $$ Putting these equations together, we get that $$ \frac{dy}{dx} = \frac{dy}{dt} \div \frac{dx}{dt} = -\frac{x}{y} \, . $$ Another, perhaps more intuitive way of making sense of this is by considering that multiplication by $i$ produces a $90$ degree rotation about the origin. If the particle occupies the position $(x,y)$, then its velocity is $i(x,y)$. Again, a graph might help:

Answer 5

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Answer 6

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.

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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.

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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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Question title: Has it become too hard to write self answer questions?

Recently I posted a self answer question which I had been working on in the Sandbox for drafts of long, complex posts for $2$ months (August $10^{th}$ to October $10^{th}$), which underwent $44$ revisions before it was posted. I also vetted the answer in the logic room to make sure it was correct before posting.

Despite the effort involved the question still got closed, and received $4$ downvotes within a short timeframe.

I was able to get the question reopened by addressing the critical feedback in the comments. The process of learning new content to edit the question, asking for it to be reopened, and now this meta post has taken a lot of time. I also went to constructive feedback to ask for possible explanations regarding the downvotes, since no comments were left.

Should self answer questions really require this much involvement? I learnt a lot from writing this question and from the feedback I got, but the closure and downvotes were rather unwelcome and time consuming.

Answer 7

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Answer 8

(4) Now let us look for which (odd!) $n$ the expression $f(n)$ contains small primefactors.

• Assume $p=3, \lambda=2$, find $n$ :
  Remember: $$\begin{align} \{f(n),p\} &= (\{2n : \lambda\} &-\{n : \lambda\})&\cdot (1 + \{n,p\})&-\{n,p\} \\ \end{align}$$ For every odd $n$ the leading parenthese must evaluate to $1$, so $2n$ must be divisible by $\lambda=2$ but not $n$, so simply $n$ must be odd: $$\begin{align} n \text{ odd:}\\ \{f(n),3\}&= (\{2 \cdot n : 2\} &-\{n : 2\})&(1 + \{n,3\})&-\{n,3\} \\ &= (1&-0)&(1 + \{n,3\})&-\{n,3\} \\ &= &&1 + \{n,3\}&-\{n,3\} \\ &= 1 \end{align}$$ and the smallest $n$ is simply $n=1$. The next $n$ are from $(1,3,5,7,9,...)$ and always $\{f(n),3\}=1$ which is the same as $f(n) = 3 \cdot x$ where $x$ does not contain the primefactor $3$.

• Assume $p=5$, $\lambda=4$ The first parenthese $(\{2 \cdot n : 4\} -\{n : 4\}) $evaluates to $1$ when $\{n,2\}=1$, so for $n \in \{2,6,10,... \}$ we have $ \{ f(n),5\} = 1 $ and for odd $n$ we have $\{f(n),5\}=-\{n,5\}$. We already know, that $n$ must be odd, so we will never have $5$ as primefactor in the result.

• Assume $p=7$, $\lambda=3$ The first parenthese $(\{2 \cdot n : 3\} -\{n : 3\}) $ evaluates always to $0$ so we have always $\{f(n),7\}=-\{n,7\}$ and we will never have $7$ as primefactor in the result.

• Assume $p=11, \lambda=10=2\cdot 5$
  For the leading parenthese to equal $1$, $2n$ must be divisible by $\lambda=10$ but not $n$, so simply $n$ must be divisible by $5$ and be odd. $$ \begin{align} \text{for } n = 5+ k \cdot 10 : \\ \{f(n),3\}&= (\{2 \cdot n : 2\} &-\{n : 2\})&(1 + \{n,3\})&-\{n,3\} \\ &= (1&-0)&(1 + \{n,3\})&-\{n,3\} \\ &= &&1 + \{n,3\}&-\{n,3\} \\ &= 1 \end{align}$$ and the smallest $n$ is simply $n=5$.

Answer 9

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Answer 10

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Answer 11

I've had myself difficulties to decode the WP-entry on Ramanujan-summation, but with some additional explanations and intermediate steps made explicite I think it becomes clearer. Note, that G.H.Hardy already mentioned that the set of definitions by Ramanujan does not make his solution unambiguous and that at least after that remark slightly different versions are underway - but I don't want to take this into the scope of this short edit.

WP starts with a version of the Euler-MacLaurin formula: $$\begin{array} {} \frac12 f(0)&+f(1)+f(2)+...+f(n-1)+\frac12f(n)\\ &=\frac12[f(0)+f(n) ] + \displaystyle \sum_{k=1}^{n-1} f(k) \\ & \displaystyle =\int_{t=0}^{n} f(t) d t + \sum_{k=1}^p { B_{k+1} \over (k+1)!}[f^{(k)}(n)-f^{(k)}(0)] + R_p \end{array}\tag {1a}$$ I do not know from why the term $\frac12[f(0)+f(n) ]$ has been introduced here, but noticing that we have $n$ as well as $n-1$ in the upper bound of the sum-indices I find it easier to adapt this first to write $$\begin{array} {} \frac12[f(0)-f(n) ] + \displaystyle \sum_{k=1}^{n} f(k) & \displaystyle =\int_{t=0}^{n} f(t) d t + \sum_{k=1}^p { B_{k+1} \over (k+1)!}[f^{(k)}(n)-f^{(k)}(0)] + R_p \end{array}$$ and take that expression to the right side $$\begin{array} {} \displaystyle \sum_{k=1}^{n} f(k) & \displaystyle =\int_{t=0}^{n} f(t) d t + \frac12[f(n)-f(0) ] + \sum_{k=1}^p { B_{k+1} \over (k+1)!}[f^{(k)}(n)-f^{(k)}(0)] + R_p \end{array}$$ and subsume it to the sum-of-Bernoullinumbers, assuming $B_1=-\frac12$: $$\begin{array} {} \displaystyle \sum_{k=1}^{n} f(k) & \displaystyle =\int_{t=0}^{n} f(t) d t + \sum_{k=0}^p { B_{k+1} \over (k+1)!}[f^{(k)}(n)-f^{(k)}(0)] + R_p \end{array} \tag {1b} $$ In the following derivations it is always assumed, that in the limit $p \to \infty$ we can ignore $R_p$ (or: that we apply this all only at functions, where this is meaningful!), and a small reorganisation of the factorials in the denominators in the rhs-sum expression makes the "basic notation"
$$ \sum_{k=1}^{n} f(k) =\int_{t=0}^{n} f(t) d t + \sum_{k=0}^\infty { B_{k+1} \over k+1}{f^{(k)}(n)-f^{(k)}(0)\over k! } \tag 2$$

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We see in (2) that the sum has a constant expression, namely the derivatives of the $f(0)$ . We might separate this $$ \sum_{k=1}^{n} f(k) =\int_{t=0}^{n} f(t) d t + \sum_{k=0}^\infty { B_{k+1} \over k+1}{f^{(k)}(n)\over k! } - \sum_{k=0}^\infty { B_{k+1} \over k+1}{f^{(k)}(0)\over k! } \tag {3a}$$
and Ramanujan took this out and called that constant part $C$ with definition $C=- \sum_{k=0}^\infty { B_{k+1} \over k+1}{f^{(k)}(0)\over k!} $ so that we find often the notation $$ \sum_{k=1}^{n} f(k) = C + \int_{t=0}^{n} f(t) d t + \sum_{k=0}^\infty { B_{k+1} \over k+1}{f^{(k)}(n)\over k! } \tag {3b}$$

and also called this a "center of gravitation of the series". The key is here, that he takes this value as finite value representation for the divergent series, computable by the rearranged equation: $$ C = \sum_{k=1}^{n} f(k) - \int_{t=0}^{n} f(t) d t - \sum_{k=0}^\infty { B_{k+1} \over k+1}{f^{(k)}(n)\over k! } \tag {4}$$

Answer 12

question:see MSE $$\begin{array} {} 2^{a}-3^{b}&=C & \text{assume we have a current solution}\\ 2^{a+a_1}-3^{b+b_1}&=C & \text{assume we have a second solution}\\ 2^{a+a_1}-2^a &= 3^{b+b_1}-3^b & \text{subtract }\\ {2^{a_1}-1 \over 3^b} &= {3^{b_1} -1\over 2^a} &\text{must be existent with } a_1,b_1 \gt 0 \end{array}$$ Then by divisibility conditions, $a_1=2.3^{b-1}.a_2$, $b_1=2^{a-1}.b_2$ $$\begin{array} {} {2^{2.3^{b-1}.a_2}-1 \over 3^b} &= {3^{2^{a-1}.b_2} -1\over 2^a} \end{array}$$ For the following analysis using primefactorization of the lhs and rhs we write now $$\begin{array} {} lhs:& {2^{2.3^{b-1}.a_2}-1 \over 3^b} &= p_1^{e_1}\cdot x_1 \\ rhs:& {3^{2^{a-1}.b_2} -1\over 2^a} &=p_1^{e_1}\cdot y_1 \end{array}$$ Because lhs and rhs shall be equal, of course $x_1$ and $y_1$ must as well be equal. But we proceed by assuming one $b$ and $a_2$ then look at the smallest primefactor $p_1$ in the lhs, and configure $a$ in the rhs, such that $p_1^{e_1}$ occurs in the rhs as well.
Example. Let's choose $C=5$. A known solution is $2^5-3^3=5$ such that $a=5$ and $b=3$. Then, assuming for $a_2=1$ the smallest nonzero value we have $$\begin{array} {} lhs:& {2^{2.3^2.a_2}-1 \over 3^3} &= p_1^{e_1}\cdot x_1 &= 7.19.73\\ rhs:& {3^{2^4.b_2} -1\over 2^5} &=p_1^{e_1}\cdot y_1 \end{array}$$

Answer 13

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Answer 14

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Answer 15

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(3) The more involved problem of $f(n) = {2^n+1 \over n}$ being integer gives even more restrictions. We have

• $n$ must be odd, because otherwise the denominator would not divide the numerator
• since $n$ is odd, $\lambda(p)$ must be even with exactly one primefactor $2$ : $\lambda(p) = 2 x$ where $x$ is odd. This defines a restriction on allowable primes: $\small (p,\lambda(p))=(3,2),(11,10),(13,6),(19,18),\cdots$ are allowed but not $\small (5,4),(17,4),\cdots$

We have moreover that -if the expression $f(n)$ is integer- it is also squarefree because: $$ \left\{{2^n+1 \over n},p\right\}= (1+ \{n,p\})-\{n,p\}=1$$

$ \qquad \qquad$ as far as is does not contain Wieferich primes.

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(2) Now for the question on $2^n+1$ we refer to that $2^n+1=(2^{2n}-1)/(2^n-1)$ and since we look only at odd and non-wieferich primes $p$ we can write $\alpha=1$: $$\small \begin{align} \{2^n+1,p\} &=\{2^{2n}-1,p\} - \{2^{n}-1,p\} \\ &= \{2n : \lambda\}\left(1+ \{2n,p\}\right) - \{n : \lambda\}\left(1+ \{n,p\}\right) \qquad \text{ can be simplified } \\ &= \left(\{2n : \lambda\}-\{n : \lambda\}\right) \left(1+ \{n,p\}\right)\\ \end{align}$$ This gives three little observations:

• primes p with odd $\lambda$ cannot occur in $2^n+1$ : we have then $\left(\{2n : \lambda\}-\{n : \lambda\}\right)=0$ because if odd $\lambda$ divides $2n$ then it also divides $n$ and the parenthese evaluates to zero.The same result of course if odd $\lambda$ does not divide $2n$ .

• if $n$ is odd, then $\lambda=2 c$ for some odd $c$ and having $p=1+k \lambda$

• if $n$ is even, say $n=2^a c$ then $\lambda=2^{a+1} d$ for some odd $c,d$

For a prime $p$ which actually occurs in $2^n+1$ its exponent in the primefactorization is exactly $1+ \{n,p\}$.

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(1) First notice that $$ 2^n+1 = {2^{2n}-1\over 2^n-1 }$$ Then for divisibility of $2^n-1 $ by a primefactor $p$ there are nice little rules due to Fermat's little theorem, Euler's totient rules, and the so-called LTE ("Lifting the exponent")-rules. In an essay where I looked at this I preferred the following notations for better writing: $$\begin{array} {}\text{divisibility: } \qquad &\{a : p \} = \begin{cases} 1 & \text{ if } p \mid a \\ 0 & \text{ if } p \not \mid a \end{cases} \\ \text{valuation: } & \{a , p \} = \nu_p(a) \\ \text{order: } & \lambda_2(p) \underset{def}= \text{minimal } m>0 \text{ such that } \{2^m-1 : p \}=1 \\ \text{offset (of expon.):} & \alpha_2(p) \underset{def}= \{2^{\lambda_2(p)}-1,p\} \ge 1 \end{array} $$ For better readability, if the index and the argument of $\lambda_2(p)$ and $\alpha_2(p)$ are clear by the formula where they occur, I'll omit their reference for instance in $ \alpha = \{2^\lambda-1,p\}$

Then, for the exponent of a prime $p$ in $ 2^n-1 $ we have the general rule $$ \{2^n-1, p\} = \{n : \lambda\}\left(\alpha+ \{n,p\}\right) \qquad \text {with } \alpha \ge 1 \tag 1 $$ Note, that for base $2$ in $2^n-1$ all $\alpha_2(p)$ are $1$ with up to today two known exceptions $p=1093$ and $p=3511$ ("Wieferich primes") where $\alpha_2(p)=2$. It is an open problem whether there exist more Wieferich-primes and what $\alpha$ they have.
Examples:

$$ \begin{align} \{2^n-1,3\} &= \{n:2\}\left( 1+ \{n,3\} \right)&(\lambda,\alpha)=(2,1)\\ \{2^n-1,5\} &= \{n:4\}\left( 1+ \{n,5\} \right) &(\lambda,\alpha)=(4,1)\\ \{2^n-1,7\} &= \{n:3\}\left( 1+ \{n,7\} \right) & (\lambda,\alpha)=(3,1)\\ \{2^n-1,19\} &= \{n:18\}\left( 1+ \{n,19\} \right) & (\lambda,\alpha)=(18,1)\\ \vdots & \text{ the two known exceptional} & \text{ Wiefer.primes} \\ \{2^n-1,1093\} &= \{n:364\}\left( 2+ \{n,1093\} \right) & (\lambda,\alpha)=(364,2)\\ \{2^n-1,3511\} &= \{n:1755\}\left( 2+ \{n,3511\} \right) & (\lambda,\alpha)=(1755,2)\\ \vdots \end{align}$$

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$$ \{2^n+1,p\} = \{2^{2n}-1,p\}-\{2^n-1,p\} $$ $$\begin{array}{} \{2^n-1,3\} &= \{ n :2 \}(1+\{n,3\}) \\ \{2^n+1,3\} &= \{ 2n :2 \}(1+\{2n,3\})-\{ n :2 \}(1+\{n,3\})\\ &= (1-\{n :2 \}) (1+\{n,3\}) \end{array}$$ $$\begin{array}{} \{2^n-1,19\} &= \{ n :18 \}(1+\{n,19\}) \\ \{2^n+1,19\} &= \{ n :9 \}(1+\{n,19\})-\{ n :18 \}(1+\{n,19\})\\ &= \{n:9\}(1-\{n :2 \}) (1+\{n,19\}) \end{array}$$

Answer 16

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Answer 17

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