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E to i(pi) = -1

  1. Nov 5, 2008 #1
    Why should it be obvious that:


    e raised to i (pi) = -1
     
    Last edited: Nov 5, 2008
  2. jcsd
  3. Nov 5, 2008 #2

    Hootenanny

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    It follows directly from Euler's formula

    [tex]\exp\left(i\theta\right) = \cos\theta + i\sin\theta[/tex]
     
  4. Nov 5, 2008 #3

    mgb_phys

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    It's easier to see in polar form http://en.wikipedia.org/wiki/Image:Euler's_formula.svg
    Remember that pi (rad) is 180deg so the arrow moves 180deg anti clockwise and points to -1 on the real axis.
     
  5. Nov 6, 2008 #4

    arildno

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    It is not obvious at all.
     
    Last edited by a moderator: Nov 6, 2008
  6. Nov 6, 2008 #5

    HallsofIvy

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    It is after you have Euler's formula!

    I once actually did have a professor once say "now it is obvious that" as he was writing something on the board, stop and say "now why is that obvious?", think for a minute and then continue, "Yes, it is obvious!"
     
    Last edited: Nov 6, 2008
  7. Nov 6, 2008 #6

    arildno

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    Hmm..why did you edit into my comment something I didn't say in this context, HallsofIvy?

    Sure,if you define the complex exponential by Euler's formula, thenitis "obvious" that the result follows.

    However, it still remains unobvious, prior to proving, say, De Moivre's formula, that exponentials should have anything to do with trig functions in their complex forms.
     
  8. Nov 6, 2008 #7

    HallsofIvy

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    I am so sorry. I meant to edit MINE and accidently clicked on the wrong button!

     
  9. Nov 6, 2008 #8
    I wonder if you can explain it in english. Gauss said that it would be obvious to person with a future in math (I don't even have a past in math). But it seem like 2.71 to 3.14 = about 23. How does using the imaginary number make it equal -1?
     
  10. Nov 6, 2008 #9

    rbj

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    personally, i think that De Moivre's follows from Euler's. the more fundamental formula is Euler's.

    anyway, outside of calculus, it is unobvious that exponential functions have any relationship to trig functions. but once you start thinking about derivatives, that the derivative of an exponential is another exponential (with the same "[itex]\alpha[/itex]" inside) and the derivative of a sinusoidal function is another sinusoid (with the same "[itex]\omega[/itex]" inside), that you might start to wonder that there is a connection. then, once you get to Taylor or Maclaurin Series, and you compare the series for sin() and cos() and ex, then it becomes less and less unobivious.

    but someone had to have the insight for seeing it first, and Euler, whom some folks think is the "Einstein" of mathematics, was the first to see it. now, it's sorta obvious.
     
  11. Nov 6, 2008 #10

    rbj

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    even though sometimes it is inaccurate (but not this time), Wikipedia can be your friend. please check out the proofs in http://en.wikipedia.org/wiki/Euler's_formula . you need to remember that i is constant, that i2=-1, and what the derivatives of the exponential and sine and cosine functions are. then pick either of the three proofs shown.
     
  12. Nov 6, 2008 #11
    So we can rearrange the formula to look like:
    [tex] \[e^{i\pi} = i^2 \][/tex] and therefore
    [tex] \[i = e^{\frac{i\pi}{2}}\] [/tex]

    [tex] \[\frac{\ln i}{i} = \frac{\pi}{2}\][/tex]
     
  13. Nov 7, 2008 #12

    arildno

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    A) It remains unobvious for every new generation.
    B) That it becomes obvious as you progress in your mathematical understanding, does not lessen its initial unobviosity
    C) The standard induction proof for De Moivre's formula is simple to perform, but its result is surprising. Hence, it is unobvious.
     
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