# Thread: infinitesimal sketch to "show" d/dx (e^ix)

1. ## infinitesimal sketch to "show" d/dx (e^ix)

I'm not sure if this should be in Calculus or pre-Calculus, since the problem is a lead-in to a presentation of a calculus concept, and hence implicitly contain calculus concepts, but the lead-in itself will not explicitly use calculus.

This is something of a pedagogical problem. I am trying to draw a diagram with my "infinitesimal microscope" to "show", without using the standard chain rule derivation (or the series expansion of e^ix, or Euler's identity), the instantaneous change of a rotation is that rotation plus a quarter turn
(i.e., d/dx (e^ix) = i*(e^ix), without using these symbols).

To show that the instantaneous change is perpendicular to the rotation is easy, with infinitesimal secants as the difference between vectors. However, to show that the size of the change will be the same as the size of the original rotation is not as clear, since the secants are clearly smaller than my radius in my diagram.

While I am doing it, it would also be useful if it was shown to be most convenient to represent the rotation in radians.

This is supposed to be a heuristic diagram, not a formal proof.
Any suggestions? Thanks for anything.

2. Try to solve, using standard approach, the linear first order DE...

$\displaystyle x^{'}(t) = i\ x(t)$ (1)

Kind regards

$\chi$ $\sigma$

3. Grazie, chisigma, but not quite what I was looking for, unless I am missing something. That is, it is not just the result that counts here, but also the method which I outlined : essentially using a rough-and-ready non-standard analysis for students without much background, including them having never seen a DE before.

I am trying to draw a diagram with my "infinitesimal microscope" to "show", without using the standard chain rule derivation (or the series expansion of e^ix, or Euler's identity)
Without Euler's identity, how can you even explain that e^ix is a rotation?

5. Originally Posted by snowtea
Without Euler's identity, how can you even explain that e^ix is a rotation?...
It is weel known from the basic theory of complex numbers that if $a$ is any complex number, then $b= i\ a$ is $a$ counterclockwise rotated of $\frac{\pi}{2}$, and the DE...

$y^{'} = i\ y$ (1)

... the solution of which is $y= c\ e^{i x}$, expresses just this concept! It is remarkable the fact that the (1) can be solved without the Euler's identity...

Kind regards

$\chi$ $\sigma$

6. Originally Posted by chisigma
It is weel known from the basic theory of complex numbers that if $a$ is any complex number, then $b= i\ a$ is $a$ counterclockwise rotated of $\frac{\pi}{2}$, and the DE...

$y^{'} = i\ y$ (1)

... the solution of which is $y= c\ e^{i x}$, expresses just this concept! It is remarkable the fact that the (1) can be solved without the Euler's identity...

Kind regards

$\chi$ $\sigma$
Right, agreed. But the original post is trying to prove $\frac{d e^{ix}}{dx} = i e^{ix}$ assuming that $e^{ix}$ is a rotation (and no chain rule).

7. Actually the question is really about a pedagogical device invented by Jerome Keisler and his students at Wisconsin in the 1970’s. It is the infinitesimal microscope used with hyperreal numbers in a non-standard analysis approach to teaching calculus. Here is a link to a free download of his textbook.
It is on page 24 of the edition I have.

8. Thanks, Plato, snowtea, and chisigma.

First, Plato, thanks for the book, looks very helpful. I shall be checking it out to find the answer to my question.

Secondly, thanks snowtea, for both posts: the second post making that point to chisigma; the first post was a good question; the answer is: graphically. True, the idea of a rotation would implicitly contain Euler's identity, but I meant to do this without explicitly using the identity.

Thirdly, thanks chisigma for trying again, but as snowtea pointed out, this wasn't the question.