ProofWiki:Sandbox

Graphical Proof
This proof utilizes the Squeeze Theorem.

Let $\theta$ be an angle in the unit circle, measured in radians. Define the following points:

$A = (0,0)$

$B = (1,0)$

$C = (cos\theta,sin\theta)$

$D = (1, tan\theta)$

and consider all $\theta$ in the open interval $(0..\frac{\pi}{2})$



$\triangle$$ABC$ has an area of $\frac{1}{2}⋅b⋅h = \frac{1}{2}⋅1⋅sin\theta = \frac{1}{2}sin\theta$

The region formed by $arcBC$ subtending $A$ is $\frac{\theta}{2\pi}$ths of the entire circle and so its area is given by

$\frac{\theta}{2\pi}⋅πr^2 = \frac{\theta}{2\pi}⋅π⋅1^2 = \frac{1}{2}\theta$

Clearly this wedge cannot be smaller in area than $\triangle$$ABC$ and so we have the inequality

$\frac{1}{2}sin\theta \le \frac{1}{2}\theta,\theta\in(0..\frac{\pi}{2})$

I don't like how the $\theta\in(0..\frac{\pi}{2})$ is all up in the inequality's grill, how do I move it more to the right and put a bit of a space between the inequality and the domain? Or any other idea of how to make it look better?

$\triangle$$ABD$ has an area of $\frac{1}{2}⋅b⋅h = \frac{1}{2}⋅1⋅tan\theta = \frac{1}{2}tan\theta$

$\triangle$$ABD$ is clearly not smaller than the wedge. We now have the triple inequality,

$\frac{1}{2}sin\theta \le \frac{1}{2}\theta \le \frac{1}{2}tan\theta, \theta\in(0..\frac{\pi}{2})$

Uch that looks terrible, I'll figure out how to play with stretching the inequality out a bit horizontally.

Were I to reflect any of the plane regions about the $x$-axis, the magnitudes of the signed areas would be the same. Our inequality, then, will hold in quadrant IV if I take the absolute value of all terms.

$|\frac{1}{2}sin\theta| \le |\frac{1}{2}\theta| \le |\frac{1}{2}tan\theta|, \theta\in(-\frac{\pi}{2}..0)\cup(0..\frac{\pi}{2})$

$\implies$

$\frac{1}{2}|sin\theta| \le \frac{1}{2}|\theta| \le \frac{1}{2}\frac{|sin\theta|}{|cos\theta|},  \theta\in(-\frac{\pi}{2}..0)\cup(0..\frac{\pi}{2})$



$ \newcommand{\Re}{\mathrm{Re}\,}  \newcommand{\pFq}[5]{{}_{#1}\mathrm{F}_{#2} \left( \genfrac{}{}{0pt}{}{#3}{#4} \bigg| {#5} \right)}$We consider, for various values of $s$, the $n$-dimensional integral\begin{align}  \label{def:Wns}  W_n (s)  &:=   \int_{[0, 1]^n}     \left| \sum_{k = 1}^n \mathrm{e}^{2 \pi \mathrm{i} \, x_k} \right|^s \mathrm{d}\boldsymbol{x}\end{align}which occurs in the theory of uniform random walk integrals in the plane, where at each step a unit-step is taken in a random direction. As such, the integral \eqref{def:Wns} expresses the $s$-th moment of the distance to the origin after $n$ steps.By experimentation and some sketchy arguments we quickly conjectured and strongly believed that, for $k$ a nonnegative integer\begin{align} \label{eq:W3k}  W_3(k) &= \Re \, \pFq32{\frac12, -\frac k2, -\frac k2}{1, 1}{4}.\end{align}Appropriately defined, \eqref{eq:W3k} also holds for negative odd integers. The reason for \eqref{eq:W3k} was long a mystery, but it will be explained at the end of the paper.

How long do the "newcommands" survive?

Definition
Let $C\subseteq\mathbb C$ be some domain in the complex plane such that $z\!+\!1\in C \forall z \in C$

Let $D\subseteq\mathbb C$ be some domain in the complex plane.

Let $F: D\mapsto D$ be holomorphic function.

Let $H: C\mapsto D$ be holomorphic function.

Let $H(F(z))=F(z+1)$ for all $z\in C$.

Then $F$ is called superfunction of function $H$, and function $H$ is called transfer function of $F$.

Project of Definition:Tetration by Kouznetsov 07:33, 20 May 2011 (CDT)
Let $b\in \mathbb R$, $b \ge \exp(1/e)$

Let $L\in \mathbb C$ be fixed point of $\log_b$ such that $\Im(L) \ge 0$

Let $C = \mathbb C \backslash \{x \in \mathbb R : x\le -2 \}$

Let $\mathrm{tet}_b: C\mapsto \mathbb C$ be superfunction of $z\mapsto b^z$ such that
 * $\mathrm{tet}_b(0)=1$
 * $\mathrm{tet}_b(z^*)= \mathrm{tet}_b(z)^* \forall z \in C$
 * $\displaystyle \lim_{y\rightarrow +\infty} \mathrm{tet}_b(x+\mathrm i y)=L \forall x\in \mathbb R$

Then function $\mathrm {tet}_b$ is called tetration to base $b$.

NOWIKI works?
$$\lim_{z\rightarrow z_0} f(z)=f(z_0)$$ $$\lim_{z\rightarrow z_0} f(z)=f(z_0)$$

Allowed??
P.S. I look at the update of Definition:Superfunction; as I understand, the only one source is allowed... Ok, let it be so.