6 ms·

For me it would be suprising only if the equation would be about real numbers only. Exponentiation of complex numbers is defined in a way that this is trivially

by nadam 7y ago

For me it would be suprising only if the equation would be about real numbers only. Exponentiation of complex numbers is defined in a way that this is trivially true. Or at least this is my perception as a non math-expert. My approach to Euler's equation (as a non-expert) was the following: 1. Try to understand the meaning of the operations in the equation. Search for the definition of exponential on complex numbers, because it was not trivial for me how it is defined on complex numbers. 2. I have read that it is defined by angles and the unit circle: e(alpha * i) is the point on the unit circle at angle alpha. 3: Looking at the equation: this is trivial, it basically says that cos(Pi) is -1. What am I missing? Edit: And why is exponentiation defined this way? Why is the base e? Why isn't it 2, like this: "2^(alpha * i) is the point on the unit circle at angle alpha." Would this definition lead to contradiction or some difficulties?

ben_w 7y ago
The surprise is that the definition you gave in 2. fully agrees with the definition of multiplication that comes naturally from the cartesian version of complex numbers: z = x + bi -> (x1 + b1i)(x2 + b2i) = x1x2 + x1b2i + x2b1i - b1b2 Vector multiplication has two very different forms (dot and cross), so multiplication (and by extension exponentiation) working out smoothly is neat.
mikorym 7y ago
I suspect it would spiral off the unit circle.
zests 7y ago
The derivative of e^ix is ie^ix. Multiplication by i is equivalent to rotation by 90 degrees. The derivative of velocity is acceleration. Acceleration perpendicular to velocity is circular motion. With base 2 the motion is not circular. All these pieces (and an initial value problem differential equation) come together to explain the identity.
obastani 7y ago
That is not actually the way to define e^x for complex numbers. There are a few ways you can do so. The most natural way is as the unique solution to the differential equation df/dx = f The solutions are e^x + const; requiring f(0) = 1 gives you the choice const = 0. Another way to define it is as the infinite series e^x = \sum_{i=0}^inf x^i/i! It is pretty easy to show this definition is equivalent to the previous one. A third way is as lim_{n -> inf} (1 + x/n)^n This definition comes from the intuition that e^x represents the limit of continually compounding interest. As before, it is pretty easy to show that it is equivalent to the previous two. In any case, all these definitions extend directly to complex values of x. The fact that e^{ix} = cos(x) + i sin(x) holds (and hence Euler's identity holds) is a consequence of a more natural definition, not the typical definition.
- Certhas 7y ago
  i x is x rotated left by pi/2 df/dx = f df(ix)/dx = i f (ix) so the "velocity" is always orthogonal to the "position". Thus the solution to this equation in the complex numbers has to be a rotation.
- cka 7y ago
  Further evidence that e^{ix} = cos(x) + i sin(x) is natural is that it fits in nicely with power series representations. If you define cos(x), sin(x), and e^x by their power series centered at 0 then it's straightforward to see that substituting ix into the power series for e^x yields the sum of the power series for cos(x) and i*sin(x) (as long as you accept that theorems about absolute convergence and rearranging terms extend to the complex numbers).
  - tonyarkles 7y ago
    The lecture in my Calculus IV course where we walked through that derivation is literally my only memorable math lecture in university. It just so beautifully ties so many different concepts together!
yiyus 7y ago
Complex exponentiation is not defined arbitrarily to make Euler identity true. The Euler identity is a consequence. I love the Euler identity in its typical form, but I think it is more surprising when you replace e and pi by their (approximate) numerical values and i by sqrt(-1)
dreamcompiler 7y ago
> Why is the base e? Why isn't it 2, like this One of the many things that makes e special is that e^x is the only function that is its own derivative. One of the things that make sin and cos special is that they are each other's derivatives (except for a change of sign in the case of dcos/dx). This doesn't prove anything on its own, but it suggests that there might be some relationship between e and the trig functions, and Euler's equation proves that indeed there is.

by nadam 7y ago

ben_w 7y ago

The surprise is that the definition you gave in 2. fully agrees with the definition of multiplication that comes naturally from the cartesian version of complex numbers: z = x + bi -> (x1 + b1i)(x2 + b2i) = x1x2 + x1b2i + x2b1i - b1b2 Vector multiplication has two very different forms (dot and cross), so multiplication (and by extension exponentiation) working out smoothly is neat.

mikorym 7y ago

I suspect it would spiral off the unit circle.

zests 7y ago

The derivative of e^ix is ie^ix. Multiplication by i is equivalent to rotation by 90 degrees. The derivative of velocity is acceleration. Acceleration perpendicular to velocity is circular motion. With base 2 the motion is not circular. All these pieces (and an initial value problem differential equation) come together to explain the identity.

obastani 7y ago

That is not actually the way to define e^x for complex numbers. There are a few ways you can do so. The most natural way is as the unique solution to the differential equation df/dx = f The solutions are e^x + const; requiring f(0) = 1 gives you the choice const = 0. Another way to define it is as the infinite series e^x = \sum_{i=0}^inf x^i/i! It is pretty easy to show this definition is equivalent to the previous one. A third way is as lim_{n -> inf} (1 + x/n)^n This definition comes from the intuition that e^x represents the limit of continually compounding interest. As before, it is pretty easy to show that it is equivalent to the previous two. In any case, all these definitions extend directly to complex values of x. The fact that e^{ix} = cos(x) + i sin(x) holds (and hence Euler's identity holds) is a consequence of a more natural definition, not the typical definition.

Certhas 7y ago
i x is x rotated left by pi/2 df/dx = f df(ix)/dx = i f (ix) so the "velocity" is always orthogonal to the "position". Thus the solution to this equation in the complex numbers has to be a rotation.
cka 7y ago
Further evidence that e^{ix} = cos(x) + i sin(x) is natural is that it fits in nicely with power series representations. If you define cos(x), sin(x), and e^x by their power series centered at 0 then it's straightforward to see that substituting ix into the power series for e^x yields the sum of the power series for cos(x) and i*sin(x) (as long as you accept that theorems about absolute convergence and rearranging terms extend to the complex numbers).
- tonyarkles 7y ago
  The lecture in my Calculus IV course where we walked through that derivation is literally my only memorable math lecture in university. It just so beautifully ties so many different concepts together!

yiyus 7y ago

Complex exponentiation is not defined arbitrarily to make Euler identity true. The Euler identity is a consequence. I love the Euler identity in its typical form, but I think it is more surprising when you replace e and pi by their (approximate) numerical values and i by sqrt(-1)

dreamcompiler 7y ago

> Why is the base e? Why isn't it 2, like this One of the many things that makes e special is that e^x is the only function that is its own derivative. One of the things that make sin and cos special is that they are each other's derivatives (except for a change of sign in the case of dcos/dx). This doesn't prove anything on its own, but it suggests that there might be some relationship between e and the trig functions, and Euler's equation proves that indeed there is.