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Proof for e^ix

  1. Sep 23, 2004 #1
    How can I proof the identity

    e^ix = cos x + i sin x?
     
  2. jcsd
  3. Sep 23, 2004 #2

    CTS

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    Consider the infinite sums:
    e^x = 1 + x/1! + x^2/2! + x^3/3! + ...
    sin(x) = x/1! - x^3/3! + x^5/5! -x^7/7! + ...
    cos(x) = 1 - x^2/2! + x^4/4! -x^6/6! + ...

    e^ix = 1 + ix/1! + (ix)^2/2! + (ix)^3/3!...
    Notice the patterns with the powers of i and rearrange the terms to see how they relate to sin and cos.
     
  4. Sep 23, 2004 #3

    mathwonk

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    or check that both functions satisfy the same de and same initial conditions,

    i.e. f'' + f = 0 and f(0) = 1, f'(0) = i.
     
  5. Sep 24, 2004 #4
    You could use the function z = cos(x) + i*sin(x) with z(0)=1 and dz/dx = -sin(x) + i*cos(x) = i*cos(x) + i^2*sin(x) = i*(cos(x)+i*sin(x)) = iz. Which gives dz/dx = iz <=> dz/z = i dx integration gives [tex] \int \frac{dz}{z} = \int i \ dx [/tex] which gives ln(z) + C = ix + D => ln(z) = ix+E => z = e^(ix+E). Now we have that cos(x)+i*sin(x) = e^(ix+E), and with z(0)=1 it gives that e^(i*0+E)=e^E=1 => E=0 which yields e^(ix)=cos(x)+i*sin(x).

    Edit: E = D - C
     
  6. Sep 24, 2004 #5

    HallsofIvy

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    In other words, there are a number of different proofs, depending on what you already know and how you are defining the different functions.
     
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