Euler can prove summation but not derivative

[Glom]

The Euler formula can be used to cut out a huge amount of crap that is normally necessary to prove the summation of angle trigonometric identities (cos(A+B) etc) as well as the trigonometric derivatives.

Let's first state Euler.

cos A + i sin A = e^iA

Now prove the summation formulae.

cos (A+B) + i sin (A+B) = e^i(A+B)

= e^iAe^iB

= (cos A + i sin A)(cos B + i sin B)

= cos A cos B - sin A sin B + i (sin A cos B + cos A sin B)

Compare coefficients. Taking the real part of both sides we get:
cos (A+B) = cos A cos B - sin A sin B
Taking the imaginary part, we get:
sin (A+B) = sin A cos B + cos A sin B

QED

Now prove the derivatives. This takes even fewer lines.

d/dA (cos A + i sin A) = d/dA e^iA

= ie^iA

= i cos A - sin A

Taking the real part
d/dA cos A = -sin A
d/dA sin A = cos A

QED

Now why can't you use Euler to prove the derivative? Well because Euler comes from the Taylor expansions of sin, cos and exp, which uses the derivatives so we need to know the trigonometric derivatives before we can prove them in this way. Hence we'd be going round in circles.

See! I remember all this stuff! Who says I didn't?

[AndreH]

.....See! I remember all this stuff! Who says I didn't?

Never forget, there is always someone arround who has by now forgotten more than you will ever know! ;-)

[Disinfo Agent]

Now why can't you use Euler to prove the derivative?

Which derivative?

[Ivan Viehoff]

Now why can't you use Euler to prove the derivative? Well because Euler comes from the Taylor expansions of sin, cos and exp, which uses the derivatives so we need to know the trigonometric derivatives before we can prove them in this way. Hence we'd be going round in circles.

I think this is right, but only if I make it more precise.

My first thought, misled by the imprecision, was the following. So, one happens to use the Taylor expansion to prove that exp(iA) = cosA + i sinA. So what. Once one has proven it, you can just use it, you don't have to go back and look at how you proved it in order to proceed. But that isn't right. One doesn't "prove" that exp(iA) = cosA + i sinA in the conventional sense, as the following shows.

I think the true question is "what does exp z actually mean, for complex z?" We have the real function exp (A) for real A. How do we extend this to be meaningful for complex numbers, and generally behave like we expect exponentiation to behave? As you say, one way to do that is to define it would be through its Taylor expansion. We can't just take for granted any known properties of the real exp would extend to the complex exp, rather they are subject to verification.

For example, we had a recent thread in which we were reminded (by me) that exponentiation in general has some rather different meanings in the complex numbers. In particular, z ^ (1/3) is a multifunction with three possible answers, as we understand general exponentiation for complex numbers. But we do not give that meaning to exp (1/3) for the complex exp function. So it is clear that at some point someone assigned a meaning, it came to be generally accepted, and the consequences of that meaning need to be checked.

So we would need to check that (d/dx) exp (zx) = z exp (zx) for complex z. One would verify this by appealing to the original definition. If the Taylor expansion was the definition one was using, one would have to be careful in relation to convergence; as you say, that would be equivalent to proving the derivatives of the trig functions on the way. Another way of doing it would be to define exp (iA) as cos A + i sin A, and check that has the desired properties. But in fact then you need to know how to differentiate sin and cos in order to check the derivative.

[Glom]

Wow!

Any real number raised to the ith power becomes a root of unity!

And any root of unity raised to the ith power becomes a real number!

Profound!

The proof:

Take a real number, x. It can be written as e^{ln x}. Therefore,
xⁱ = e^{i ln x},
which is a root of unity.

Take a root of unity, e^iA. Raised to the power of i, it becomes e^-A, which is a real number.

[Ivan Viehoff]

Basically what you are saying is that mod (x^i) =1 for arbitrary real x, and arg (x^i) = ln x. That is correct.

As it happens, your argument is fine, but in general you have to be careful when interpreting one number raised to another in the complex numbers, you can easily get to "paradoxes". You can't just raise one number to a power, and then say it is the same as the exp function if the number being raised happens to be e. The exp function is a well-defined single valued function, whereas general exponentiation can be multi-valued. To compare it to a common deceit in the real numbers, you can't just take square roots of an equation, and then if the square root of e pops up say it is the same as exp(0.5), because it might be the negative square root.