Second degree of smallness

>>8270436
(dx)^2 is the smallest number possible that's higher than 0

that small

>>8270450

OP here

I thought that was dx?

So is there an axiom or something that says [math](dx)^2 = 0[/math] always?

>>8270436
The square of a positive number less than 1 must be less than that number

>>8270436
>My question is, how can one rigorously disregard (dx)2(dx)2 in this case?
This isn't a rigorous treatment of calculus and it shouldn't be regarded as such. This is merely an argument of intuition.

Say we have a small number 1/a, where a > 1. If a is large, 1/a is small. Then (1/a)^2 is even smaller.

So if dx is "small," (dx)^2 is even smaller. The argument the author is making is that (dx)^2 is so much smaller than 2x*dx that you can disregard it. There's nothing rigorous about it. It's not based on any axiom. It's just a convenience.

Most treatments of calculus are like this. Real analysis is the rigorous treatment, but it's more difficult to study and should be held off until you understand the intuition behind the concepts.

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>>8270468

Ah, I see. I already have an intuitive understanding of derivatives (through the limit definition) but I guess this type of understanding could be useful in some areas.

>>8270464
Yup, so if dx is the smallest positive number higher than zero, intuitively/gut feeling [math](dx)^2[/math] ought to be zero. I get it now.

Thanks broskis

>>8270436
[math] y+\Delta y=(x+\Delta x)^2\\
\Delta y=2x\ \Delta x+(\Delta x)^2\\
\frac{\Delta y}{\Delta x}=2x+\Delta x\\
\lim_{\Delta x \to 0}\frac{\Delta y}{\Delta x}=\frac{\mathrm{d}y}{\mathrm{d}x}=2x\\ [/math]

>>8270436
It is not really a rigorous treatment, unless the author states you're working in [math]\frac{{\mathbb{R}\left[ {\operatorname{dx} } \right]}}{{\left\langle {{{\operatorname{dx} }^2}} \right\rangle }}[/math]. But I doubt that.

>>8270436
What the fuck is this?

>>8271448
underage and/or American spotted

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>>8270436
>https://www.gutenberg.org/ebooks/33283
OP that book is absolute diarrhea. Ditch it and start an easy topology book. Learn the basic set operations and what a topology is. Proceed then with a rigorous analysis book.

>>8270436
First, you have to rigorously define limits, continuous functions, Cauchy sequences and real numbers. Then, you can do this:
d/dx of x^2 = lim h->0 [(x+h)^2 - x^2]/[x+h-x]
= lim h->0 [x^2 + 2xh + h^2 - x^2]/[h] = lim h->0 [2xh+h^2]/h
= lim h->0 2xh/h + h^2/h = lim h->0 2x + h = 2x

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>>8271458
>underage and/or American spotted
Retard who thinks that book is anything but toilet paper spotted. That book is going to fuck up OP's mind for good.

>>8271462
>start learning topology before knowing fuck all about math

kek

>>8270436
>https://www.gutenberg.org/ebooks/33283

Ultimately in calculus, we use the symbol dx to represent as infinitely small, inverse of infinity.

So, they are trying to use a loose simile with numbers to get you comfortable with that concept, and more importantly that dx is not merely a place holder which has mathematical meaning.

>>8271497
And this whole (dx)^2 stuff would be a double integration and never described as such, but rather dxdx.

Because the author is using a simile to analogize the concept he applies (dx)^2 which is only practical in the example and not actual calculus. Multiply two numbers is easy to understand (hence only in the specific example), but multiplying infinity twice is beyond calc 1.

>>8270436
Different order of infinities
2∞^2 > ∞^2 + ∞ ≋ ∞^2 > 2∞ > ∞+1 ≋ ∞ > 2 > 1 +1/∞ ≋ 1 > 1/∞ > 1/∞^2 ≋ 1/∞^2 + 1/∞^3 .... >0

All you care about is the leading magnitude term when you finish the calculation. When you divide by dx you get
dy/dx = 2x + dx
o(dy/dx)=o( 1/∞ / 1/∞ ) = o( ∞/∞ ) = o(1)
o(2x)=o(1)
o(dx)=o(1/∞)
Compared to something of finite magnitude, the 1st order infinitesimal is negligible.

You can do it in the previous step with the other terms being order 1/∞ and dx^2 being 1/∞^2