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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

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Answer 8

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Answer 9

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Answer 10

I'm interested in describing the procedure for solving equations with one variable (finding symbolic solutions) by rearranging the equation.

Applying mathematical operations when solving equations means applying functions.

In the definition of a function $f$ with $f\colon x\mapsto f(x)$, the function term is $f(x)$.

Rearranging an equation means applying a mathematical operation to both the left-hand and the right-hand side of the equation so that the operation is the outer function of a function composition.

Here the operations applied to the left-hand and the right-hand side of the equation are functions with one and the same function term.

1.)

How can my above description be improved?

2.)

Is my statement correct that a polynomial equation $p_0+p_1x+p_2x^2=0$ ($p_0,p_1,p_2\in\mathbb{C}\setminus\{0\}$) cannot be solved for $x$ by rearranging by applying algebraic operations alone?

And how can we show this?

Applying the solution formula for quadratic equations is not rearranging.

3.)

Let $f_1,f_2$ be algebraically independent functions of a complex variable, and let $P(X_1,X_2)\in\mathbb{C}[X_1,X_2]$ have no univariate factor.

How can we show that the equation $$P(f_1(x),f_2(x))=0\tag{1}$$ cannot be solved for $x$ by rearranging by applying only the inverse relations of $f_1$ and $f_2$ and algebraic functions?

Answer 11

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Answer 12

(Editing just now please do not interfere -------------)
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 $S$ : $$ 2^S \ge \lceil N \cdot \beta \rceil $$

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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 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: $$ a \cdot b \cdot c = {a \over d}\left({ 3a+1 \over 2^A }\right) \left( { 3b+1 \over 2^B } \right) \left( { 3c+1 \over 2^C }\right) $$ and from this $$ 2^{A+B+C}=2^S= {a \over d}\left(3+{ 1 \over a }\right) \left( 3+{ 1 \over b } \right) \left( 3+{ 1 \over c }\right) $$

Answer 13

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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

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

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

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. Therefore:

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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

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Answer 17

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