ProofWiki:Sandbox

Alternate proof for FToC
--GFauxPas 10:13, 27 September 2011 (CDT)

Let $f$ have a primitive $F$ continuous on the closed interval $\left[{a..b}\right]$.

$[a..b]$ can be divided into any number of subintervals of the form $[x_{k-1}..x_k]$ where

$ a = x_0 < x_1 ... < x_{k-1} < x_k = b $

By repeatedly adding and subtracting like quantities,

$F(x_k) \underbrace{- F(x_{k-1}) + F(x_{k-1})}_{0}... \underbrace{- F(x_1)+ F(x_1)}_{0} - F(x_0)$

$\implies$

$(A) \quad F(b) - F(a) = \sum_{i=1}^{k} F(x_i) - F(x_{i-1}) $

Because $F = f'$, $F$ is differentiable. Because $F$ is differentiable, $F$ is continuous. By the mean value theorem, in every subinterval $[x_{k-1}..x_k]$ there is some $c_i$ where $F'(c_i) = \frac{F(x_i)-F(x_{i-1})}{\Delta x_i}$

where $\Delta x_i \equiv x_{i} - x_{i-1}$

I know it's messy, I'm just getting the framework down.

If I multiply both sides by $\Delta x_i$ I get

$F'(c_i)\Delta x_i = F(x_i) - F(x_{i-1})$

Substituting $F'(c_i)\Delta x_i$ into $(A)$ we get

$F(b) - F(a) = \sum_{i=1}^{k} F'(c_i)\Delta x_i$

Because $F' = f$

$F(b) - F(a) = \sum_{i=1}^{k} f(c_i)\Delta x_i$

Because $f$ and $F$ are both continuous, we can take the limit $||\Delta||→0$ of both sides, where $||\Delta||$ is the magnitude of the largest subinterval $[x_{k-1}..x_k]$

$\lim_{||\Delta|| \to 0} F(b) - F(a) = \lim_{||\Delta|| \to 0} \sum_{i=1}^{k} f(c_i)\Delta x_i$

The $LHS$ is a constant and is unchanged by taking the limit of it. The $RHS$ is the definition of the integral.

$F(b) - F(a) = \int_{a}^{b}f(x)\mathrm{d}{x}$



$ \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.