# Euler's formula

• Jan 25th 2011, 12:04 PM
Colin
Euler's formula
I was hoping someone might be able to clear up a fundamental issue that I am having trying to follow a proof for the Euler Formula, as follows.

(Sorry I'm not sure how to make the equations look pretty)

z=cosx + i sinx

dz = i(cosx + i sinx) dx
= i z dx
integrating both sides.
[dz/z] = [i dx]
lnz = ix

From this point on I fully follow the proof however I don't think I understand the notation of dy/dx.

My current understanding is that when you differentiate a function y with respect to x you are calculating the gradient function and that is simply called (dy/dx). It could be called f'(x) or something else. However in this proof it's being treated like a fraction.

If it is being treated like a fraction then what is dy, and what is dx? Isn't dx an infinitely small distance, when delta x has tended to 0?

Any help is much appreciated
• Jan 25th 2011, 12:12 PM
pickslides
I assume your problem is between these two steps?

Quote:

Originally Posted by Colin

z=cosx + i sinx

dz = i(cosx + i sinx) dx

You can look at it like this,

$\displaystyle \displaystyle z=\cos x + i \sin x$

$\displaystyle \displaystyle \frac{dz}{dx} = i(cosx + i sinx)$

Now integrate both sides with respect to $\displaystyle \displaystyle x$

$\displaystyle \displaystyle\int \frac{dz}{dx} ~dx=\int i(cosx + i sinx)~dx$
• Jan 25th 2011, 12:17 PM
HallsofIvy
Quote:

Originally Posted by Colin
I was hoping someone might be able to clear up a fundamental issue that I am having trying to follow a proof for the Euler Formula, as follows.

(Sorry I'm not sure how to make the equations look pretty)

z=cosx + i sinx

dz = i(cosx + i sinx) dx

Well, dz= (-sin(x)+ i cos(x))dx. Of course, if we factor out i, we get $\displaystyle dz= i(-\frac{1}{i}sin(x)+ cos(x)) dx$ and, since $\displaystyle \frac{1}{i}= -i$, that is, in fact, the same as dz= i(-(-i)sin(x)(+ cos(x))dx= (cos(x)+ i sin(x))dx again.

Quote:

= i z dx
integrating both sides.
[dz/z] = [i dx]
lnz = ix

From this point on I fully follow the proof however I don't think I understand the notation of dy/dx.

My current understanding is that when you differentiate a function y with respect to x you are calculating the gradient function and that is simply called (dy/dx). It could be called f'(x) or something else. However in this proof it's being treated like a fraction.
dy/dx is NOT fraction but it is a limit of fractions. You can show that it can always be treated like a fraction by going back "before" the limit, using the fraction properties and then taking the limit. Since it can be "treated like" a fraction, it is convenient to "formalize" that by defining the "differentials", dy and dx, by "dy= (dy/dx)dx= (f'(x))dx. They cannot be defined separately and should never be used separately.

Quote:

If it is being treated like a fraction then what is dy, and what is dx? Isn't dx an infinitely small distance, when delta x has tended to 0?

Any help is much appreciated
dy/dx can be defined (and originally was by Newton and Leibniz) as a ratio of "infinitesmals". However, it turns out to be very difficult to give a rigorous definition of "infinitesmal" (Bishop Berkeley famously refered to infinitesmals as "ghosts of vanished quantities"!) and it was not until recently that that was done ("non-standard Analysis").
Better to think of dy and dx as purely symbols connected by dy= (f'(x))dx.
• Jan 25th 2011, 01:12 PM
Colin
Thank you very much for the quick replies!

http://www.mathhelpforum.com/math-he...f6a7420d09.png This is a clearer way for me to picture what is happening as opposed to missing out steps.

I think I now need to get my head around these 2 sentences,

"it is convenient to "formalize" that by defining the "differentials", dy and dx, by "dy= (dy/dx)dx= (f'(x))dx. They cannot be defined separately and should never be used separately."

I haven't thought about it this way before and I am not sure what the implications are from a first principles point of view (need time to think about it) but you have helped a lot thank you!