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Evil, evil proof

  1. Sep 6, 2005 #1
    okay, so i'm really confused, and i'm not sure where to even start. any direction or advice, would be most appreciated.

    for an exercise, we were asked to proove 'euler's formula' (not the polyhedron one)

    [tex]e^{-i \phi} = cos \phi + i sin \phi[/tex]


    for the 'hint' we were given

    [tex]e^x = \sum_{n=0}^\infty \frac{x^n}{n!}[/tex]

    so yah, like i said any direction would be most appreciated.
     
  2. jcsd
  3. Sep 6, 2005 #2
    You could expand it as a taylor series. You could also do the following:

    [tex]z=\cos{\theta}+i\sin{\theta}[/tex]

    [tex]\frac{dz}{d\theta}=i\cos{\theta}-\sin{\theta}=i\left(\cos{\theta}+i\sin{\theta}\right)=iz[/tex]

    [tex]\frac{dz}{z}=i d\theta\implies\ln{z}=i\theta\implies z=e^{i\theta}[/tex]
     
  4. Sep 6, 2005 #3

    lurflurf

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    Homework Helper

    I think you mean
    [tex]e^{i \phi} = cos \phi + i sin \phi[/tex]
    Expand both sides in series and show they are equal.
    For such a proof to be valid you must
    -define exp(z) for complex z (or at least imaginary)
    -show that the power series for exp(z) holds when z is complex (or at least imaginary)
    -show that the series converges absolutely so that rearangment is allowed
     
  5. Sep 7, 2005 #4
    okay yah i have no clue. taylor series i might be able to manage, i don't know. i'll check the calc 2 book tommorow when i'm not drunk and have electrodynamics in six hours. i have no clue about imaginary numbers, it's been almost ten years since i was in highschool working with them. where does z come from, and where can i get some clue as to this i business. i can follow the differentiation (almost). there may very well be no negative, i have had calculus 1 & 2 and that's it. working on 3 now, and this book on modern mathematical physics is mildly terrifying.

    thanks so much, i really appreciate the effort afforded.

    cheers
     
  6. Sep 7, 2005 #5

    lurflurf

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    Homework Helper

    Worry not those into math books for science usually only assume one year of calculus. z is just the variable commonly used for complex numbers. Complex numbers are just a feild extension of reals. That means your familar field axioms hold and you have an element i such that i^2=-1. The formal manipulation such books like is something like
    [tex]e^{i\phi}=\sum_{n=0}^\infty \frac{(i\phi)^n}{n!}=\sum_{n=0}^\infty \left((-1)^n\frac{\phi^{2n}}{(2n)!}+(-1)^n\frac{\phi^{2n+1}}{(2n+1)!}i\right)=\cos(\phi)+i\sin(\phi)[/tex]
    Another way to do the same thing is to note that is two analytic functions (those that can be represented by taylor series) have derivatives of all orders (0,1,2,...) equal at a point, then they are equal.
    [tex]f(x):=e^{i\phi}[/tex]
    [tex]f^{(n)}(x)=i^n e^{i\phi}[/tex]
    [tex]f^{(n)}(0)=i^n[/tex]
    [tex]g(x):=\cos(\phi)+i\sin(\phi)[/tex]
    [tex]g^{(n)}(x)=\cos(\phi+n\pi/2)+i\sin(\phi+n\pi/2)[/tex]
    [tex]g^{(n)}(0)=\cos(n\pi/2)+i\sin(n\pi/2)[/tex]
    [tex]g^{(n)}(0)=i^n[/tex]
    so they are equal
    For any of that to have meaning as I mentioned before we need
    -define exp(z) for complex z (or at least imaginary)
    -show that the power series for exp(z) holds when z is complex (or at least imaginary)
    -show that the series converges absolutely so that rearangment is allowed
     
  7. Sep 9, 2005 #6
    okay, thanks again for the help. i was quite fond of the differentiating/integrating approach outlined above. i didn't understand it at first, but then i looked at what was going on and derived the steps myself.

    as far as the series approach goes, i'm stuck at step two, which is pretty sad. well, it's not exactly step two, but i did all the steps at once. anyway:

    (sorry the tex effort would be horrendous and i can't seem to post an image)
    here it is

    anyone adept at the series math want to help me out from here? just a hint on what to do next would be appreciated over the solution. maybe i made a mistake, i really do need to brush up on the chapter on series from calc 2. thanks again for any help already given and any to be afforded.

    cheers

    edit i did forget the [tex](-1^n)[/tex]
    wow i also forgot a [tex](2^n + 1)[/tex] in the denominator
     
    Last edited: Sep 9, 2005
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