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Deeper connection between exp() and cos()?

  1. Nov 11, 2008 #1
    Is there some mathematician who knows the "core reason" why exp() is connected with cos()?
    I mean I'm confident with "advanced maths for physicists", but maybe some abstract theory gives an answer to this.

    I was think of something like:
    therefore it's a unit circle.
    But also
    therefore the [itex]\phi[/itex] are have the mathematical structure required for angles. So lets define
  2. jcsd
  3. Nov 11, 2008 #2
    The "core reason" why exp and cos are related is that that's the only way a world could possibly exist.

    The cosine of a number is the real part of exp(xi). You can prove it through the Taylor series of exp. It's true by force of mathematical logic alone, and its truth is independent of physics or our world.

    I'm not sure what you mean by "mathematical structure required for angles". Angles in mathematical terms simply means distance. In particular, distance around a unit circle. You add the exponents because when considering the angles as arcs along a circle in the complex plane, the distance of those arcs is simply their sum.

    The connection is hauntingly deep, but no deeper than that.
  4. Nov 11, 2008 #3
    Thanks for your suggestions.
    I'm looking for a more abstract reasoning though. Provided we didn't know about cos() at all, it would have to come from somewhere with a reason. So I'm expecting some algebra, group theory or whatever reasoning here, because the standard stuff I probably know :)
  5. Nov 11, 2008 #4


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    It doesn't matter if you know why it's true, it's still true. A lack of proof of Fermat's Last Theorem didn't present an opportunity for a counterexample to come first. So if we didn't know about the cosine function, we'd probably be pretty boggled. Alternatively, we'd have alternative notation (since math would have evolved incredibly differently) and have a different description that suffices. I'm not sure how someone can describe how exp and cos are connected without assuming the existence of cos as a function
  6. Nov 11, 2008 #5
    Oh, but we have to know at least a little about cosine, or we couldn't make the connection!

    If you wanted a fuzzy explanation that appealed to the senses, what about this:

    Each of the complex numbers have a cute little mapping to a linear operator in R^2. Pure real numbers are scaling operations, changing the length of the number they multiply, but not the direction. If we want to change the direction, we must use the imaginary unit. Thus, all complex numbers z such that |z| = 1 corresponds to a rotation in the plane.

    Exponentiation, in both real and complex domains, is a dynamo for multiplication. So it's not strange that complex numbers, which do such fascinating things under multiplication, would have a special connection with it.

    I'm not sure there is any good explanation in abstract algebra or group theory, though there might be related concepts. The complex numbers are more specialized than either of those two subjects get. Though, what we're really doing here is using the complex numbers as a group with operation of multiplication. And the identity exp(x + y) = exp(x) exp(y) for imaginary x and y is a homomorphism over it. But I'll let someone else toy with that idea.
  7. Nov 11, 2008 #6
    So I thought about this a bit a couple years ago.

    I think the reason is because rotations have a special connection to exponentials. If you rotate a vector twice by angle theta, that is the same as rotating once by angle 2 theta. This recalls a similarity to exponentials, where squaring an exponential is the same as taking the exponential of twice the original argument.

    [tex]R_\theta^2 \vec{x} = R_{2 \theta} \vec{x}[/tex]

    [tex](e^\theta)^2 x_i= e^{2 \theta} x_i[/tex]

    To make the connection precise, you could derive the arbitrary 2D rotation matrix [itex]R_\theta[/itex] by diagonalizing the rotation by a known angle (eg: [itex]R_{\pi / 2}[/itex]) and taking powers. This will get you a matrix with sums of complex exponentials in each entry.

    On the other hand, you already know the rotation matrix as [cos(theta) sin(theta); -sin(theta) cos(theta)] (in some sense, the entries of this rotation matrix define sine and cosine), so you can equate these 2 to get the relationship between complex exp and cos.
    Last edited: Nov 11, 2008
  8. Nov 11, 2008 #7
    Yes, that's sort of what I meant. I thought maybe there is a freaky mathematical way to express that, so that no-one can understand or in fact read it :)

    The only other way to define cos() is by a functional equation?!

    (of course not Taylor, as cos() has a geometrical meaning and is not just a wild guess of a series)
  9. Nov 11, 2008 #8
    How many ways do you want to define it?
  10. Nov 11, 2008 #9
    Err, hmm maybe I didn't explain enough.

    You can define: "cosine(theta) is the x component of the vector (1,0) after rotation by angle theta", and "sine(theta) is y component of that vector". Then that is equivalent to saying "cosine theta is defined to be the first [1,1] component of the 2D rotation matrix by angle theta".

    On the other hand, you can derive the rotation matrix from first principles. Since rotation by multiples of an angle is the same as raising rotations to powers, you can construct the generic 2D rotation matrix by diagonalizing a known rotation matrix and taking powers of it. The eigenvalues on the diagonal will be complex numbers of length 1, and you are raising them to powers. The result is the rotation matrix (as derived from first principles) is filled with sums of complex exponentials.
  11. Nov 11, 2008 #10
    Any way that makes the geometrical meaning obvious.

    Functional equation sort of do.
    Differential equations - not sure.

    I really mean that cos() should have that one geometrical meaning. I know that in principle there are many ways to define it, but then it's hard to prove that these ways make sense.
  12. Nov 11, 2008 #11

    Ben Niehoff

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    There seem to be a lot of questions about this lately.

    You can prove [itex]e^{i\phi} = \cos \phi + i \sin \phi[/itex] without appealing to Taylor series. Just use the Cauchy-Riemann equations. Then define exp(z) as the unique analytic continuation of exp(x) to the complex plane.
  13. Nov 11, 2008 #12


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    If you get to pick definitions, there's an even easier to prove Euler's formula: just use a definition in which things are defined by it. :smile:
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