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I read on Wiki:

Intuitively, smooth infinitesimal analysis can be interpreted as describing a world in which lines are made out of infinitesimally small segments, not out of points.

I tried to find infinitesimally small segments for graph $y = x^2$.

At first I obtained that point $A (dx,dx^2)$ is upon on $X$ axis - $(dx,0)$.

$(dr)^2 = (dx)^2 + ((dx)^2)^2$ (Pythagorean equation)

$(dr)^2 = (dx)^2$

$dr=dx$

In this case $dx$ is nilsquare infinitesimal ($(dx)^2=0$).

Then I tried to find infinitesimally small segment for $[dx,2dx]$ but I can't:

$dy=2xdx$

for $x=dx$: $dy = 2(dx)^2$

And

$(dr)^2 = (dx)^2 + (2(dx)^2)^2$ (Pythagorean equation)

$dr = \sqrt{(dx)^2 + (2(dx)^2)^2}$

$dr = dx\sqrt{(1+4(dx)^2}$

$dr = dx$

Where did I go wrong? How to calculate $dr$?

Thanks.

Mikhail Katz
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Mike_bb
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1 Answers1

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The "infinitesimally small" segment of the curve $y=x^2$ that starts at the point $(x, x^2)$ ends at the point $$ \begin{align} (x+dx, (x+dx)^2) &= (x+dx, x^2 +2x dx + dx^2) \\ & \approx (x+dx, x^2 +2x dx). \end{align} $$ The length of that segment is $$ \sqrt{1 + 4x^2} dx . $$

Ethan Bolker
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  • Thx. $dx$ is nilsquare infinitesimal when $dr=dx$, but in another cases $dx$ isn't nilsquare. Is it right? – Mike_bb Feb 27 '23 at 05:52
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    I have never encountered the term "nilsquare". In an expression where $dx$ and i$dx^2$ occur in separate terms you can (usually) ignore the term with the square since the square of a small number is a very small number. This answer and the comments there about infinitesimals might interest you: https://math.stackexchange.com/questions/1991575/why-cant-the-second-fundamental-theorem-of-calculus-be-proved-in-just-two-lines/1991585#1991585 – Ethan Bolker Feb 27 '23 at 14:48