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Answer 1

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

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

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

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

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

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

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

Answer 2

Reach the solution by applying Discriminant-like method .

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Observe that :

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

Then, you have :

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

which completes the answer .

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$\rm {Explanation:}$

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

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

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

which is a method of completing the square .

Answer 3

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

Answer 4

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

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

Answer 5

$$ \begin{align} { 2^S \over 3^N } &=\exp( S \cdot l2 - N \cdot l3) \\ &= \exp( l2 \cdot (1+N \cdot ld3-\{N \cdot ld3\}) - N \cdot l3) \\ &= \exp( l2 \cdot (1-\{N \cdot ld3\}) ) \\ \end{align} $$ This answer is free for anyone to use

Answer 6

Let

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

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

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

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

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

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

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

Answer 7

This answer is free for anyone to use

Answer 8

Here's an alternate solution.

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

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

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

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

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

Answer 9

This answer is free for anyone to use.

Answer 10

This answer is free for anyone to use.

Answer 11

This post free for anyone to use

Answer 12

(Editing just now please do not interfere -------------)
part4

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start
In following the remark of L.Garner in my first long-comment: "...if the number of odd terms encountered in the Collatz sequence is not too great..."
I give some examples which first may rise doubt that "$\sigma(n) = \kappa(n)$ in all cases", but we seem to arrive at some curious limit with larger $N \gt \gt 1e6$ which might come out significant in further analysis.
First the general derivation in my notation.

part1
A formula for the orbit of a Collatz-sequence of, for instance, 3 steps can be written using the Syracuse-form: $$\mathcal U: \begin{array} {} b={ 3a+1 \over 2^A } & c={ 3b+1 \over 2^B } & d={ 3c+1 \over 2^C } \end{array} $$ where I use the capital letters in the exponents of the denominators related to the sequence-elements in the numerator $ A=val_2(3a+1)$, $B =val_2(3b+1)$ and so on.

Moreover, I use $N$ for the number of all odd steps (= length of the orbit) and $S$ for the sum-of-exponents/number-of-divisions-by-$2$.

^{$\qquad$ Note: my $S$ is also the length of the orbit in the notation-style $$\mathcal T: b = \left\{ \begin{matrix} {3a+1\over 2} & \text{if $a$ is odd} \\ {a \over 2} & \text{if $a$ is even} \end{matrix} \right.$$ $\qquad$ because it simply counts the divisions by $2$ which is mostly taken as length in terms of that notation.}

part2
Now, if we have -for easy example- a 3-step-orbit, we might write a product-expression as equation: $$ b \cdot c \cdot d = \left({ 3a+1 \over 2^A }\right) \left( { 3b+1 \over 2^B } \right) \left( { 3c+1 \over 2^C }\right) $$ and to form it a bit nicer for the following massage: $$ a \cdot b \cdot c = {a \over d}\left({ 3a+1 \over 2^A }\right) \left( { 3b+1 \over 2^B } \right) \left( { 3c+1 \over 2^C }\right) $$ From this we get $$ 2^{A+B+C}=2^S= {a \over d}\left(3+{ 1 \over a }\right) \left( 3+{ 1 \over b } \right) \left( 3+{ 1 \over c }\right) $$ part3
and then $$ {2^S \over 3^N} = {a \over d}\left(1+{1 \over 3a }\right) \left(1+ { 1 \over 3b } \right) \left( 1+{1 \over 3c }\right) $$ If we discuss a cycle, we assume $d=a$ and the term $a/d=1$ can be omitted. If we discuss the "dropping time" then we assume $d \le a$ which means $1 \le a/d \lt 2$ because we need only the first division by $2$ after $2a>(3c+1)/2^{C-1}>a$.

Our first argument is now, that $2^S \gt 3^N$ is required, because the rhs is larger than $1$. That means, for the cycle (where $a/d=1$!) the smallest possible value for $S$ is $ S \ge N \beta $ ^{(where we always use $\beta = \log_2(3)$) in the following)} and because we use only integral values for $N$ and for the exponents and thus for their sum $S$ : $$ S \ge \lceil N \cdot \beta \rceil $$ This is, what everybody already knows. But the questions is, can it happen, for some sequence of $a_k$ in the rhs, that it becomes equal or even larger than the lhs $2^{S-N \cdot log_23}$?

Answer 13

This answer is free for anyone to use.

Answer 14

The lower bounds of distances of perfect powers $\Delta=\mid a^A-b^B \mid$ is one of those attractive, simple-to-state, but difficult-to-solve problem. For me it occured in the question of cycles in the Collatz-problem, where we ask for the lower bound of distances of perfect powers of $2$ and $3$: $$\Delta(N,S)=|2^S - 3^N|$$ or $$\Lambda(N,S)=|S \cdot \ln2-N \cdot \ln 3| \qquad .$$
What I've found so far is

• the estimate by J.W.Ellison formulated in the 1970'ies (adressing $\Delta(N,S)$), and
• the estimate by G.Rhin formulated in 1987 (adressing $\Lambda(N,S)$), as reported in J.Simons & deWeger[2007] on the existence of "m-cycles" in the Collatz-problem.

The lower bounds from those estimates are -in the view of an amateur- astonishingly far away from the actual lowest distances -as checked for $N$ of about $1$ million digits. (Of course, the only relevant $N$ are selected according to the convergents of the continued fraction of $\log_2(3)$.)

Often discussions of this topic state the general form of estimation-formulae, but don't give explicite values for the constants involved - for instance in the well-visited blog of T. Tao on the "separation of powers of 2 and 3". Naive functional estimates for such lower bounds based on heuristics would suggest small values for such constants, for instance while G.Rhin gives something like $1/457/N^{13.3}$ a naiive fitting suggests something like $.../N^1$ or $.../N^{1.01}$ instead, valid for $N$ in my mentioned range of search.
A recent guess, using $.../N/\ln(N)$ instead, leaving aside any $1+\varepsilon$-exponent at all seems to be a really good estimate for the lower bounds. However, the famous estimate for the number of primes below some number $n$ as value of the integrallogarithm $Li()$ as proposed by the young K.F.Gauss has later been proven to be not really a lower bound, but where the first counterexample could only be upper-limited by something like $10^{10^{10^{34}}}$(WP:Skewes-Number) (modern computation reduced that upper bound to $1.39 \cdot 10^{316}$, see there).

Actual computations with large exponents diverge extremely far from that given estimates so I once tried to find an own algebraical pathway to such an lower-bound estimate. My current attempt provides a lower bound which is much nearer to the empirical values than the two other estimates - unfortunately this is not based on an analytical argument but only on heuristic finding.
Here I intend to create a "biglist" of immediately usable such lower bounds and invite other readers to add information and / or alternatives to the two established estimates and to try to improve my proposed estimate with analytical arguments.

Preliminaries:
While Ellison immediately looks at $\Delta(N,S) = |2^S - 3^N|$ and gives $$\Delta(N,S) \gt \exp(S \cdot (\ln 2 - 0.1)) \qquad S\gt 27 \tag {\text{EL}}$$

G. Rhin looks at the difference of logarithms $\Lambda(N,S)=| S \ln2 - N \ln 3 |$ and gives $$\Lambda(N,S) \gt {1\over 457 N ^{13.3} }| \tag {\text{RH}}$$ I've looked as well at $\Lambda(N)$ and propose $$\Lambda(N,S) \gt {1\over 10 N \ln N } \tag {\text{HE}}$$ To compare that bounds, we must norm for the data of $\Delta()$ or for that of $\Lambda()$. The best adaption is surely dependent on the question that one has - either using a formula involving the exponential expression or one involving the logarithmic expression. So the discussion of this should be done in the single answers of the big-list where we reference some question in MSE.
To trigger the interest, here some pictures, how the three estimates behave in contrast to the empirical values. Here I adapted the Ellison-estimate to a $\Lambda()$-version.

We see, that the empirical $\Lambda()$ jitter between $0$ and $1/2 \cdot \ln 2$ (blue dots), the values at $N=\{2,5,12,53,...\}$ show very small distances, and even decreasing towards zero. The idea is to have a continuous function $f(N)$ which can supply a lower bound, such that we can say $\Lambda(N,S) \gt f(N)$ .
The red curve for the Ellison-estimate is below of the empirical values only for $N \gt 17$ while the green curve for the Rhin-estimate is so near the zero-line, that we barely see it. My estimate shown by the brown curve is always below the $\Lambda()$ .
To see a bit more detail, and to get aware of the basic characteristics of the estimates, a picture in log()-scalings is more appropriate. Picture 2: x,y logarithmic scaling Here we see the characteristics better. All three methods of estimates work for $N$ towards $=200$. Ellison took a much different shape compared with Rhin's, and while Rhin's lower bound is unfortunate small over the whole range, the Ellison's is initally too large, then better than Rhin's but from about $N \approx 700$ it decreases much faster than Rhin's. So, for the larger $N$, Rhin's lower bound should be preferred, but for the moderate values in $N$ Ellison's estimate gives a simpler formulation and a better lower bound.
But the "LBLam"-function has the best characteristic; it is nearly parallel to a lower hullcurve, and also is valid from the smallest $N=2$ on. That this property holds forever, that means the formula is analytically useful for all $N$, has not been proved, but empirical heuristic show that it holds for all $N$ up to $1$ million decimal digits and is always tight to the most extreme small values of $\Lambda()$...
A last picture, for the aesthetical impression is the following. I rescaled the values of $\Lambda(N,S)$ to the open interval $-1[...]1$ by $w(N,S)=4 \cdot \Lambda(N,S) / \ln 2 -1$ and then show the values $Y(N,S)=\tanh^{-1}(w(N,S))$ and the accordingly rescaled values for the three lower-bound-estimates. The slightly jittering effect in the Ellison-curve is because its function uses the somehow jittering values $S$ instead that of $N$ for its argument.

Answer 15

$\def\sinc{\operatorname{sinc}} \def\J{\operatorname J}\def\Re{\operatorname{Re}}$

Rewriting:

$$\sinc(x)=a\mathop\implies^{x=\frac{\sin(w)}a}\begin{matrix}\sin(w)+aw=(2k+1)\pi a\\\sin(w)-aw=2k\pi a\end{matrix},k\in\Bbb Z$$

lets us invert $\sinc(x)$ via complex Fourier series; Fourier sine series entails a longer derivation. Unfortunately, the inverse of $\sin(x)-ax,a>0,\left[-\frac\pi2,\frac\pi2\right]$ is not a function and Fourier series on smaller intervals than this one gives us the rest of the branches of the inverse function, but the result would be more complicated. Let $f(w)=\sin(w)+aw,\left[-\frac\pi2,\frac\pi2\right]$ and $f^{-1}(w)=g(w),[-L,L]=\left[-f\left(\frac\pi2\right), f\left(\frac\pi2\right)\right]=\left[-\frac{\pi a}2-1,\frac{\pi a}2+1\right]$:

$$\frac{\sin(g(x))}a=\sum_{n\in\Bbb Z}e^\frac{\pi i n x}L \frac1{2L}\int_{-L}^L \frac{\sin(g(t))}a e^{-\frac{\pi i n t}L}dt\mathop=^{t\to g(t)}\Re\sum_{n=1}^\infty e^\frac{\pi i n x}L\int_{-\frac\pi2}^\frac\pi2\frac{\sin(t)}{aL}e^{-\frac{\pi i n}Lf(t)}df(t)$$

Now integrate series coefficients by parts and sum over its constant term:

$$\frac{\sin(g(x))}a=\frac x{aL}-\Re\sum_{n=1}^\infty e^\frac{\pi inx}L\frac i{\pi a n }\int_{-\frac\pi2}^\frac\pi2 \cos(t)e^{-\frac{\pi i n}L f(t)}dt$$

Unfortunately, this integral likely has no closed form, so a Jacobi-Anger expansion suffices:

$$\int_{-\frac\pi2}^\frac\pi2 \cos(t)e^{-\frac{\pi i n}L(\sin(t)+at)}dt=\sum_{m\in\Bbb Z}(-1)^m J_m\left(\frac{\pi n}L\right)\int_{-\frac\pi2}^\frac\pi2\cos(t)e^{\left(m-\frac{\pi a n}L\right)it}dt=-2\sum_{m\in\Bbb Z} J_m\left(\frac{\pi n}L\right)\frac{(-1)^m\cos\left(\frac{\pi v}2\right)}{v^2-1},v=m-\frac{\pi a n}L$$

Expanding $\sum\limits_{m\in\Bbb Z}$ into $\sum_\limits{m=1}^\infty$ complicates the result and $\Re\left(i e^\frac{\pi i n x}L\right)=-\operatorname{Im }\left(e^\frac{\pi i n x}L\right)=-\sin\left(\frac{\pi n x}L\right)$. $\frac{sin(g(w))}a$ inverts $w+\sin^{-1}(a w)$/[]

[]

note

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

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

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

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

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

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

Saks and Sygmund [8, pp. 201–202])

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

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

Repeated Maclaurin Expansion:

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

Stirling Expansion:

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

Coefficients for All Branches:

[Evaluate?]

Answer 16

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

Answer 17

multi-valued algebra.

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

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

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

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

How do you axiomatize ZFC + Worldly Cardinals?

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

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

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

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

Here's an axiomatization of ZFCW.

Let $P$ be a list of parameters.

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

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

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

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

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

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

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transfinite induction:

Let $A$ be a set of ordinals.

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

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

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

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Is the "ordinalification" of ZFC equiconsistent with ZFC?

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

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

and the following function:

• $ (,) : E \times E \to E $

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

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

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

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

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

Consistency of ZFC implies consistency of $T$.

Assume ZFC is consistent.

ZFC has a model $A$.

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

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

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