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Is this considered a closed expression for pi?

  1. Sep 10, 2010 #1

    this equation does not use limits or integrals, as you can see, but it does involve imaginary numbers. Does this make it an open expression, or does the fact that it uses i not matter?
  2. jcsd
  3. Sep 11, 2010 #2
    This is almost the definition of logarithm for complex numbers. You define something to be pi, then call it a closed form expression? Is pi itself a closed-form expression of pi?

    By the way, ln(-1)/i=pi is completely equivalent to the statement that "the angle between two parts of a straight line is pi", which is again almost the original definition of pi.
  4. Sep 11, 2010 #3
    It is an identity. The use of 'i' really not matter. It's just a constant. And it follows right from the Euler's Formula: e^ti = cos t + i sin t which follows from Taylor expansion for e, sin and cos. With t=pi you have e^(pi i) = -1, so ln(-1)/i=pi
  5. Sep 11, 2010 #4
    I think I see. thank you.
  6. Sep 11, 2010 #5


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    Doesn't the complex logarithm has an infinite number of branches?

    So [tex]e^{i\pi}=-1[/tex] can be extended to [tex]e^{i(\pi+2k\pi)}=-1[/tex] for k any integer.

    This means [tex]\frac{ln(-1)}{i}=\pi+2k\pi[/tex]. I'm not sure if we can strictly say that pi is equal to that expression, when an infinite number of values are equal to it as well.
  7. Sep 11, 2010 #6
    But.. We can say that -1 = e^i(pi+2kpi) , can't we..? So, i think that "problem" doesn't invalidate the identity. It's just that we are talking about complex numbers and extending functions can show us some "strange" things if we look at it like we look to the reals. Remember the famous sum: 1+2+3+... = -1/12. It is an identity, but if we look at that as a real sum it makes no sense.

    (Sorry if i can't make myself clear enough sometimes. My english is really not good. ;p)
  8. Sep 13, 2010 #7
    The word "closed expression" doesn't have a precise definition.

    Usually, a closed form helps simplify a difficult expression. I would argue that reducing an irrational number to an expression of complex logarithms isn't simplifying anything.
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