1. Jul 10, 2015

Mappe

I want to have a simple and intuitive explanation of why the sin and cos waves have such a simple repetitive values for their derivatives at a specific point. Their derivative values are also periodic in respect to the derivative order. For example, e^-x is also periodic, but its derivatives are never zero. Is there a good explanation of this without involving complex numbers? Or if not, is there an easy answer to why complex numbers adds their angles when multiplied? Because that kind of answers the same question. The simpler the proof the better!

2. Jul 10, 2015

Dr. Courtney

sine and cosine have repeated values and repeated derivatives because you are going around the unit circle as the arguments of each increase.

A circle is symmetric and repeats.

3. Jul 10, 2015

EM_Guy

$e^{-x}$ is not periodic (unless x is imaginary, in which case you have a real periodic function and an imaginary periodic function).

As Dr. Courtney said, check out angles of inscribed triangles within the unit circle. You keep going around and around. $1$ revolution for every $2\pi$ radians.

4. Jul 10, 2015

Mappe

Dr Courtney, I understand but your statement implies that we know that e-i2π = 1, and from the definition of eix by its taylor expansion, how can we see on its derivatives that its gonna be periodic and perhaps also how do we see from this definition that it describes a circle? The same question phrased differently can be "is there a simple proof that doesn't use knowledge about sin and cos, that shows us that complex numbers add their angles when multiplied, and multiplies their norm"?

5. Jul 10, 2015

WWGD

If you accept the result $e^{i\theta}= cos(\theta)+(isin\theta)$ , then this is automatic. And if you take the polar representations ( in a region where valid) of two complex numbers $z_1=e^{i\theta_1}, z_2 = e^{i \theta_2 }$ , then $z_1 z_2=e^{i(\theta_1+ \theta_2)}$ , which takes $z_1=e^{i \theta_1}=cos\theta_1+i sin \theta_2$ to $z_1z_2= cos(\theta_1+ \theta_2)+ isin(\theta_1+ \theta_2)$ , which is a rotation by an angle of $\theta_2$

But it ultimately depends on your starting point/assumptions.

6. Jul 18, 2015

homeomorphic

You should read Visual Complex Analysis. This stuff might be in the free preview here:

http://usf.usfca.edu/vca//

He just illustrates it by example. Multiplying by i is rotating by 90 degrees, so you can draw a picture of it with that it mind for some example like (4+3i)(1+i), thinking of the complex numbers as vectors. Here, you would get 4(1+i), which which is 1+i scaled by a factor of 4 and then you get 3i(1+i), which is the 1+i rotated by 90 degrees and scaled by a factor of 3, and then you add those vectors together. So you you end up with a triangle similar to the triangle formed by 4+3i, with its real imaginary components being the other sides of the triangle that gets put on top of the one representing 1+i. When you stack the two triangles, you can see that the angles are added, and by similar triangles, the modulus gets multiplied.

It's a little cumbersome to describe verbally without a picture, so you really have to draw it and work it out for yourself.

Also, for the sine and cosine derivatives, you ought to think of taking the velocity vector of a particle that is moving around a unit circle. Some ideas along these lines are presented in Visual Complex Analysis, as well.

7. Jul 18, 2015

homeomorphic

Oh, and incidentally, if you want to understand exactly why the sine and cosine functions look the way that they do, just get take a look at some Cavatappi noodles.

http://www.fabios.co.za/15-short-cut-extruded-pasta/trafilata-861-cavatappi/

A Cavatappi noodle can be interpreted as a 3-d plot (x,y,t) of particle moving around in a circle in the x-y plane, with time being the other coordinate. Viewed from the top, they'd look circular, but viewed from the side, they are sine waves (ignoring their thickness).