Question

Thing I'm showing @ikram002p to extend even/odd functions (2n, 2n+1) to 3n, 3n+1, and 3n+2 functions.

Answer 1

im here , go ahead :)

Answer 2

Every number has a radius portion from 0 to infinity and a direction portion from rotating, and the direction part is: \[\LARGE e^{i \theta}\] so when theta = 0 or 2pi we are pointing to positive numbers while pi points us 180 degrees backwards.

So I'm calling the real numbers a special case, they're just two-directional-numbers (positive and negative) gives us even and odd, so notice that we divide the full rotation (2 pi) by 2.\[\LARGE e^{i \frac{2 \pi}{2}*0}=1 \\ \LARGE e^{i \frac{2 \pi}{2}*1}=-1\]

Similarly let's look at three-directional numbers:
\[\LARGE e^{i \frac{2 \pi}{3}*0}=1 \\ \LARGE e^{i \frac{2 \pi}{3}*1}=\frac{-1}{2}+i \frac{\sqrt{3}}{2} \\ \LARGE e^{i \frac{2 \pi}{3}*2}=\frac{-1}{2}-i \frac{\sqrt{3}}{2}\] So as you might expect, each one of these is just a "unit" vector and they're all pointing three apart: and here I'm saying \[\LARGE \omega = \frac{-1}{2}+i \frac{\sqrt{3}}{2}=e^{i \frac{2\pi}{3}}\] and I'm now trying to remove the "idea" of complex numbers so now instead of marking numbers as positive or negative we are marking them with 1, w, and w^2.

Basically I'm saying that when we discovered negative numbers we just assumed they were 180 degrees apart and didn't talk about complex numbers. So I'm pretending we don't know about complex numbers here either and that we just know how to go along these three paths...

Ok... Now I'll show you the functions!

Answer 3

sounds good , keep going :)

Answer 4

So in the past we could say every function is the sum of an even and an odd function like this: \[\LARGE f(x)=\frac{f(x)+f(-x)}{2}+\frac{f(x)-f(-x)}{2}\]

we can say the same for 3n, 3n+1, and 3n+2 functions.

\[\small f(x)=\frac{f(x)+f( \omega x)+f(\omega^2 x)}{3}+\frac{f(x)+\omega f( \omega x)+\omega ^2f(\omega^2 x)}{3}+\frac{f(x)+\omega ^2f( \omega x)+\omega f(\omega^2 x)}{3}\]

Just like the derivative of an even function is an odd function and the derivative of an odd is an even, we have the same stuff going on here with these three guys.

So in simple terms I'm saying that because this is the second one, it's the form of a 3n+1 function: \[\LARGE g(x)= \frac{f(x)+\omega f(\omega x)+\omega ^2f(\omega ^2x)}{3}\] and it's fairly easy to check that it's the derivative of a 3n function and its derivative will be a 3n+2 function.

So when we had an even and odd, that means they satisfied f(x)=f(-x) and f(x)=-f(-x).

Now I'm saying for 3n, 3n+1, and 3n+2 are: \[\LARGE f(x)=f(\omega x)=f(\omega ^2 x) \\ \LARGE f(x)=\omega f(\omega x)=\omega^2 f(\omega ^2 x) \\ \LARGE f(x)=\omega^2 f(\omega x)=\omega f(\omega ^2 x)\]

Ok great lol. So all this build up for pretty much nothing I guess? Well I am talking to my professor right now since she walked in sorry for the delay!

Answer 5

hehe i liked ur idea pretty neat :O

Answer 6

Ok, but remember, we're pretending there is no space inbetween these are our new "real" numbers. Just like negative numbers sticks you on the track from left to right, these numbers will also restrict you to the three-way track here.

So just like the square (even function) of a negative is a positive, we can think similarly here as well that the (3n) of a function will always map to the +x direction no matter if we go down the wx or w^2x direction.