Im reading https://www.math.wisc.edu/~keisler/calc.html.
If you open up the chapter $2$ pdf...
The $2$ diagrams
(1st on page $14$ of the pdf (not the text book),
2nd on page $15$) have me confused. Keisler says the accurate picture is the 2nd... but to me this doesn't make sense... (i.e. I think the 1st one makes sense).
Infinitesimals (though non commensurable with reals) have the same behavior and properties as reals... So why wouldn't the infinitesimal "magnified" also have curvature?
Infinitesimals are just really tiny fractions... and (e.g.) for $y=x^{1\over 2}$ ($0 < x < 1$; where $x$ is real) the fractions have curvature. So then for $y=x^{1\over 2}$ ($0 < x <$ smallest finite real; i.e. where $x$ is hyper-real) why wouldn't this have curvature ?
When I start putting in values of ${1 \over H}$ (where $H$ is infinite), ${2 \over H}$, ${3 \over H}$ into the function above why would this not have curvature?
Answer
Look at what he says immediately after giving the Increment Theorem in the form $\Delta y=dy+\epsilon\,dx$:
The Increment Theorem can be explained graphically using an infinitesimal microscope. Under an infinitesimal microscope, a line of length $\Delta x$ is magnified to a line of unit length, but a line of length $\epsilon\,\Delta x$ is only magnified to an infinitesimal length $\epsilon$.
He goes on to point out that when $f'(x)$ exists, the following two things are true:
- The differential $dy$ and the increment $\Delta y=dy+\epsilon\,dx$ are so close to each other that they cannot be distinguished under an infinitesimal microscope.
- The curve $y=f(x$ and the tangent line at $(x,y)$ are so close to each other that they cannot be distinguished under an infinitesimal microscope; both look like a straight line of slope $f'(x)$.
Figure $2.2.3$ shows the operation of a single infinitesimal microscope, one that expands the infinitesimal $dx=\Delta x$ to unit length. In that picture you can clearly see $\Delta y-dy$: it’s roughly $40$% of $dx$. But in fact $\Delta y-dy=\epsilon\,dx$, where $\epsilon$ is an infinitesimal, if the expanded $\Delta x(=dx)$ in the diagram has unit length, we shouldn’t be able to see $\Delta y-dy$ at all: it’s infinitesimal. Moreover, that means that we should see no separation between the tangent line and the curve; and since the tangent line is unquestionably straight, that means that what little of the curve appears in the infinitesimal microscope should also appear to be straight. In short, Figure $2.2.3$ grossly exaggerates that difference between $dy$ and $\Delta y$ and the visual curvature of the part of $y=f(x)$ that appears in the microscope.
In Figure $2.2.4$, on the other hand, the first infinitesimal microscope expands $dx=\Delta x$ to unit length and labels only $\Delta y$, not $dy$. If $dy$ were labelled in that picture, it would be visually identical to $\Delta y$, because at the scale of that microscope the difference $\Delta y-dy$ is infinitesimal (and hence invisible): $\Delta x$ appears in that microscope to be of unit length, so $\epsilon\,\Delta x$ appears to be of length $\epsilon$, too small to be seen. And the tangent line and the curve are shown as being visually indistinguishable at that scale, as we just saw that they should be.
The second infinitesimal microscope then expands $\epsilon\,\Delta x$ to unit length, so that we can see it. That means that we can also see the curve and the tangent line as distinct lines. Moreover, at that magnification the separation between them doesn’t change visibly over the field of view of the microscope, so both look straight.
We’re actually dealing with three size levels here: ordinary finite real numbers (e.g., $1$), infinitesimals comparable in size to $dx$, and infinitesimals comparable in size to $\epsilon\,dx$. Where $dx$ is infinitesimal compared with $1$, say, $\epsilon\,dx$ is infinitesimal not just in comparison with $1$, but in comparison with the infinitesimal $dx$.
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