User talk:GFauxPas/Archive1

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sin x over x
Your edits in Sandbox look promising! Graphic is superb. But we might want to rename the graphic once it has been finished to match the name of the proof it goes with. --prime mover 14:42, 23 September 2011 (CDT)


 * No problems with replying to my comment here by putting a response on my own talk page - but it is better to keep the conversation in one place by replying on the same page it starts. I make sure that all pages I edit are on my watchlist, so I am notified whenever a response is made on the same page I made the comment being replied to.


 * There's a mention of this issue on the main talk page, and I can see why it's been raised - if the conversation goes on a long time and spreads over several talk pages it gets difficult to follow it. --prime mover 16:51, 23 September 2011 (CDT)

How do I rename an uploaded file, do I have to reupload it with a new name?--GFauxPas 22:54, 24 September 2011 (CDT)


 * I believe you don't have the authorisation to rename pages. I've done the renaming of the file in question - see what's in the sandbox. --prime mover 03:19, 25 September 2011 (CDT)

Primemover, I can't find the accepted template for how to cite sources. How do I do it? Is there a standard way?--GFauxPas 17:19, 25 September 2011 (CDT)


 * There isn't a standard way of doing it. There's BookReference which references an entry in the Books section (but for that you need to have set up a page for the book in question), or there's just the technique of describing the entity and someone will go through and tidy it. For a link on the web just link to it. There are citation links to Planetmath and a couple of others. --prime mover 18:07, 25 September 2011 (CDT)

Alternate proof for FToC
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.

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

This guy $\uparrow$ needs proofreading --GFauxPas 10:51, 27 September 2011 (CDT)