Why can't people think for themselves?

[mrCage]

This is going way off topic. But it reminds me of Euclidean geometry. It says that through a point outside a straight line we have exactly ONE straight parallell to that line. There are 2 other geometries possible. We have no such line or we have infinitely many such lines ... Not in ANY way cube/ing related!!

Per

 

[riffz]

LMAO at Chris's post on S-slope. Seriously, you put so much detail and effort into everything you do.

 

[Stefan]

LMAO at Chris's post on S-slope. Seriously, you put so much detail and effort into everything you do.

I just wish he put a little more effort into making it short enough for me to read...

 

[cmhardw]

I just wish he put a little more effort into making it short enough for me to read...

Stefan I knew it would probably go past your threshold That's, in part, why I added the tl;dr warning.

I wanted to prove every claim I had made to a fairly high degree of formality, or at least give a proof sketch. I gave a proof sketch for my claim of the bijection between S-slope and m-slope, as making the claim without the proof is meaningless since it's a core part of my argument that m-slope and S-slope are basically no different from each other. I also wanted to show a walkthrough proof of how S-slope can still be used to calculate S-slope derivatives in a meaningful way, since that was a mighty big claim for me to make without providing a proof of it. On the next page Aron's question leads to a general proof of how to calculate S-slope derivatives for any function, but at the time I had only done this for the S-slope power rule.

How else could I have shortened it, other than removing the commentary which I added to let people follow along who maybe are not as familiar or comfortable with these sorts of math topics? My audience is made up of people who both understand all the math I am showing, and those who understand only some of the math I am showing. I was trying to accommodate as many people as possible.

The flow beyond the tl;dr warning was basically the following. Define S-slope. Show an example for two points on a vertical line, then any vertical line. Define the mapping between S-slope and m-slope. Provide a proof sketch of the bijection. Walk through a proof of the S-slope power rule for derivatives. The extra wording, I hope, is only the commentary to explain that to the lay person who is only casually following the argument.

 

[Rpotts]

In fact, the two concepts form a bijection: for all intensive purposes, they are the exact same concept, just appearing differently

HAHAHAHAA I had to, your posts were too perfect, and no one pointed it out yet. lolz

Spoiler

I thought that it was intensive purposes for the majority of my life, it wasn't til I saw it written out that I had my epiphany

 

[Stefan]

How else could I have shortened it

I guess I'd have to read it to answer that

Well, also has to do with the topic, not just the length. I'm just not interested enough in an alternative slope definition.

 

[cmhardw]

For those who do not want to read the longer version of my post, here are the Cliff notes.

Definitions
Let m-slope be defined by:
\( m = \frac{y_2-y_1}{x_2-x_1} \)

Let S-Slope be defined by:
\( S = \frac{(x_2-x_1)+(y_2-y_1)}{(x_2-x_1)-(y_2-y_1)} \)

Any m-slope \( m = \frac{b}{a} \) maps into S-slope via the mapping \( S = \frac{a+b}{a-b} \)
This mapping can be shown to be a bijection between m-slope and S-slope.

Calculus with S-slope
Let f(x) be a continuous function in a neighborhood around point x. The instantaneous S-slope at the point \( \left(x,f(x)\right) \) is defined by:

\( \lim_{c \to 0} \frac{[(x+c)-x]+[f(x+c)-f(x)]}{[(x+c)-x]-[f(x+c)-f(x)]} \)
\( \lim_{c \to 0} \frac{c+f(x+c)-f(x)}{c-f(x+c)+f(x)} \)

Derivatives in S-slope
Let \( \left(\frac{d}{dx}\right)_S f(x) = \lim_{c \to 0} \frac{c+f(x+c)-f(x)}{c-f(x+c)+f(x)} \)

\( \left(\frac{d}{dx}\right)_S f(x) = \lim_{c \to 0} \left[\left(c+f(x+c)-f(x)\right) * \left(c-f(x+c)+f(x)\right)^{-1}\right] \)
\( \left(\frac{d}{dx}\right)_S f(x) = \lim_{c \to 0} \left[\left(\frac{c+f(x+c)-f(x)}{c}\right) * \left(\frac{c-f(x+c)+f(x)}{c}\right)^{-1}\right] \)
\( \left(\frac{d}{dx}\right)_S f(x) = \lim_{c \to 0} \left[\left(1+\frac{f(x+c)-f(x)}{c}\right) * \left(1-\frac{f(x+c)-f(x)}{c}\right)^{-1}\right] \)
\( \left(\frac{d}{dx}\right)_S f(x) = \left[1+f'_m(x)\right] * \left[1-f'_m(x)\right]^{-1} \)
\( \left(\frac{d}{dx}\right)_S f(x) = \frac{1+f'_m(x)}{1-f'_m(x)} \)

Where \( f'_m(x) \) is defined as the m-slope (standard) derivative of f with respect to x.

 

[mrCage]

Just for those who do not what a bijection is!!

Per