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How to prove e^ix=cos x + i sin x

  1. Aug 4, 2006 #1
    How to prove e^ix=cos x + i sin x
     
  2. jcsd
  3. Aug 4, 2006 #2
    One way is to start with the taylor series for ex and then change x to ix and remembering that
    i2 = -1, i3 = -i, and i4 = 1 you can rearrange the series and show that this is equal to the other side of that equation.
     
  4. Aug 4, 2006 #3
    The proof depends on your definitions (for example, if you define [itex]\cos{x}[/itex] as [itex](e^{ix}+e^{-ix})/2[/itex] and [itex]\sin{x}[/itex] as [itex](e^{ix} - e^{-ix})/(2i)[/itex] then it's pretty easy!).

    It's also clear if you start from the Taylor series of the three functions about the origin as d_leet suggested.

    You can also start from the differential equation for e^x,

    [tex]y^\prime = y, y(0)=1.[/tex]

    Write

    [itex]e^{ix} = f(x) + ig(x),[/itex]

    (where f, g are real-valued)

    so that [itex]f(0) = 1[/itex] and [itex]g(0) = 0[/itex]

    then you must have (by the chain rule)

    [tex]ie^{ix} = f^\prime(x) + ig^\prime(x),[/tex]

    or, rearranging,

    [tex]e^{ix} = \frac{1}{i}f^\prime(x) + g^\prime(x) = g^\prime(x) - if^\prime(x).[/tex]

    Since [itex]e^0 = 1[/itex] that means [itex]g^\prime(0) = 1[/itex] and [itex]f^\prime(0) = 0[/itex]. Now differentiate again and you get

    [tex]ie^{ix} = g^{\prime \prime}(x) - if^{\prime \prime}(x)[/tex]

    or

    [tex]e^{ix} = -f^{\prime \prime}(x) - ig^{\prime \prime}(x).[/tex]

    Comparing the real and imaginary parts with our original [itex]e^{ix} = f(x) + ig(x)[/itex] you get two differential equations with initial conditions:

    [tex]f^{\prime \prime}(x) = -f(x), f^{\prime}(0) = 0, f(0) = 1[/tex]

    and

    [tex]g^{\prime \prime} (x) = -g(x), g^{\prime}(0) = 1, g(0) = 0.[/tex]

    It's easy to check that f(x) = cos(x) and g(x)= sin(x) satisfy these DEs, and by a theorem in the theory of DEs the solutions are unique.
     
    Last edited: Aug 4, 2006
  5. Aug 5, 2006 #4
    Let [tex] z = \rm{cos}\theta + i \cdot \rm{sin}\theta [/tex]

    Then derive z:

    [tex]\frac{\rm{d}z}{\rm{d}\theta} = - \rm{sin}\theta + i \cdot \rm{cos}\theta = -1 (\rm{sin}\theta - i \cdot \rm{cos}\theta) = i^2 (\rm{sin}\theta - i \cdot\rm{cos}\theta ) = i (i \cdot \rm{sin}\theta - i^2 \cdot \rm{cos}\theta) = i(\rm{cos}\theta + i \cdot \rm{sin}\theta) = i \cdot z[/tex]

    This is a differential equation:
    [tex]\frac{\rm{d}z}{\rm{d}\theta} = i z [/tex]
    whose solution is

    [tex] z = e^{i \theta} [/tex]

    Thus, [tex] \rm{cos}\theta + i \rm{sin}\theta = e^{i \theta} [/tex]
     
    Last edited: Aug 5, 2006
  6. Aug 8, 2006 #5
    Umm.. in fact i think that the "historical" development fo the formula came with euler when studying the "Harmonic (classical) oscilator"..

    [tex] y'' + \omega ^2 y =0 [/tex] trying y=exp(ax) then you get..

    [tex] a^2 +\omega ^2 =0 [/tex] so [tex] + \sqrt (-1) \omega =a [/tex] (and the same

    with a minus sign)

    -On the other hand the long-known solution for the Harmonic movement was [tex] y= Asin (\omega t) +B Cos(\omega t) [/tex] then if you set the initial condition [tex] y(0)=0 [/tex] you get the identity for sin. and using the well-known formula [tex] cos^2 (x) + sen^2 (x) =1 [/tex] oh..if today math were so easier....i would have became famous years ago.... :eek: :eek:
     
    Last edited: Aug 8, 2006
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