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Math Help - Euler's formula

  1. #1
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    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
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  2. #2
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    I assume your problem is between these two steps?

    Quote Originally Posted by Colin View Post

    z=cosx + i sinx

    dz = i(cosx + i sinx) dx
    You can look at it like this,

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

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

    Now integrate both sides with respect to <br />
\displaystyle x

    \displaystyle\int  \frac{dz}{dx} ~dx=\int  i(cosx + i sinx)~dx
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  3. #3
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    Quote Originally Posted by Colin View Post
    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 dz= i(-\frac{1}{i}sin(x)+ cos(x)) dx and, since \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.

    = 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.

    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.
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  4. #4
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    Thank you very much for the quick replies!

    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!
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